Analysis and Design of a Continuous R. C. Raker Beam using Eurocode 2

8/17/2019 Analysis and Design of a Continuous R. C. Raker Beam using Eurocode 2

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Analysis and Design of Stadium Raker Beam Using EC2 Ubani Obinna U. (2016)

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Analysis and Design of a Continuous Reinforced Concrete Raker

Beam for Stadium Using Eurocode 2

Ubani Obinna Uzodimma

Works/Engineering Services Department

Ritman University, PMB 1321, Ikot Ekpene, Akwa Ibom [email protected]

Abstract

A continuous intermediate raker beam in the first tier of a football stadium was analysed using elastic

method and designed using Eurocode 2. The raker beam was analysed for permanent and variable

actions due to crowd load and permanent loads only. Due to its inclination, it was subjected to

significant bending, axial, and shear forces. However, design results show that the effect of axial

force was not very significant in the quantity of shear reinforcement required. Asv/Sv ratio of 1.175

(3Y10mm @ 200 c/c) was found to satisfy shear requirements. The greatest quantity of longitudinal

reinforcement was provided at the intermediate support with a reinforcement ratio of 1.3404%. The

provided reinforcement was found adequate to satisfy ultimate and serviceability requirements.

1.0 Introduction

The most common construction concept of sports stadiums today is a composite type where usually precast

concrete terrace units (seating decks) span between inclined (raker) steel or reinforced concrete beams and rest

on each other, thereby forming a grandstand (Karadelis, 2012). The raker beams are usually formed in-situ with

the columns of the structure, or sometimes may be preferably precast depending on site/construction constraints.

This arrangement usually forms the skeletal frame of a stadium structure.

In this paper, a raker beam isolated from a double tiered reinforced concrete grandstand that wraps around a

football pitch has been presented for the purpose of structural analysis and design. A repetitive pattern has been

adopted in the design which utilizes a construction joint of 25mm gap between different frame units. By

estimate, each frame unit is expected to carry a maximum of 3600 spectators, under full working conditions.

With ten different frames units, the stadium capacity is about 35000 after all other reservations have been taken

into account. Each grandstand frame has precast L-shaped seating terrace units that span in between reinforced

concrete raker beams inclined at angles between 20° - 22° with the horizontal. Crowd load and other loads are

transferred from the seating units to the raker beams, which then transfers them to the columns and then to the

foundations. Load from the service areas and concourse areas are also transferred using the same method.

Figure 1.1: 3D skeletal structure of each grandstand frame units (slabs and sitting areas removed)

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The three dimensional view of the skeleton of the grandstand is shown in Figure 1.1, while a sectionthrough the grandstand is shown if Figure 1.2. Section through the L-shaped seating unit is shown inFigure 1.3.

In simple horizontal beams, vertical forces will produce vertical reactions only. However, once a beam is

statically indeterminate and inclined, vertical forces will produce both vertical and horizontal reactions and as a

result, axial forces which may be compressive or tensile in nature will be induced in the beam. In the design of

horizontal floor beams in normal framed structures, the effect of axial force in the shear force capacity of the

section is usually neglected. This is largely due to the fact that these forces are usually compressive, and in

effect tends to increase the concrete resistance shear stress (Vc) and (VRd,c) of the section in accordance with BS

8110-1:1997 and EC2 respectively. It is a well known phenomenon that compressive axial force increases the

concrete resistance shear stress of a section, while tensile axial force will reduce the concrete resistance shearstress. So this neglect can be justified in terms of it being a conservative design which can only err in economy.

However for inclined beams members in a frame (as in the case of a grandstand), axial force behaviour can vary

greatly especially when the load is applied in the global direction (which is the prevalent scenario). In other

words, based on the structural configuration of the structure and the loading, it is even common to see the

nature of axial forces moving from positive (tensile) to negative (compressive) in the same span of an

inclined member. If the load is however resolved and applied in the local direction of the inclined member, the

axial forces will be absent. A good design will therefore require the use of less shear reinforcement in the axial

compression zone, and more shear reinforcement in the axial tension zone. While the effect of axial forces may

be neglected in horizontal floor beams under axial compression, it may be unsafe to neglect it in inclined beams

because more often than not, some sections are usually under axial tension.

Figure 1.2: Section through the grandstand

Figure 1.3: Section of the precast seating unit



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1.2 Materials, Model, and Loading

Structural design of stadiums is critical and this becomes more obvious when EN 1990 (Eurocode – Basis for

Structural Design) classified it under ‘Consequence Class 3’ which by description means high consequence for

loss in terms of human life, economy, environmental considerations, and otherwise if failure should occur.

Several codes of practice across different countries and bodies have provisions made in them for the design of

structures subjected to crowd loading (for example stadiums), but the level of expertise often associated with the

processes in terms of analysis, design and construction is often perceived to be something left to a limited few.

The application of static crowd imposed loads according to both BS 6399-1:1997( Loading on buildings - Code

of practice for dead and imposed loads) and EN 1991-1: ( Action on structures: General actions - Densities,

self-weight, imposed loads for buildings) are given in Table 1 below:

Table 1.1: Values of variable actions on grandstands from BS 6399 and EN 1991

CODE CATEGORY(DESCRIPTION) IMPOSED

LOAD/VARIABLE

ACTION (KN/m2)

CONCENTRATED

LOAD (KN)

BS 6399-1:1996 C5 (Areas susceptible to

overcrowding e.g. grandstands)

5 3.6

EN 1991-1-1

C5( Areas susceptible to large

crowds, e.g

sports halls including stands)5.0 – 7.5 * 3.5 – 4.5*

*Exact range of value to be set by various national annex

Raker beams in stadiums usually support precast seating terrace units which may be L-shaped, or extended into

a more complex shape (see Figure 1.5). These seating terrace units are designed as simply supported elements

spanning between the raker beams (Karadelis, 2012, Salyards et al 2005). The crowd loading is supported

directly by these terrace units, which then transfer the load to the raker beams through the bearings. This

construction concept has been adopted in the design of Cape Town Stadium (South Africa) for the 2010 FIFAworld Cup (Plate 1.5). The picture in (Figure 1.4) below shows the formwork and construction of in-situ raker

beams at San Diego State University Student Activity Center (Steele and Larson 1996).

In this design, each L-shaped seating unit is 7m long, which means that the raker beams are spaced at 7m centre

to centre. The crowd loading is supported by the terrace seating units, which is then transferred to the raker

beams through the end shears. The raker beams can be analysed as sub-frames or as full 3D structures in order

to get the most realistic behaviour of the structure.

Figure 1.4: Typical formwork and

reinforcement for in-situ raker beam (Steele

and Larson 1996).

Plate 1.5: Precast seating units being installed

on raker beams at Cape Town Stadium (2010)



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1.2.1 Partial Factor for load

The partial factor for all permanent actions (dead load) Gk is 1.35 while the partial factor for all variable actions

(imposed load) Qk is 1.5. No reduction factor was applied in the analysis, and the effect of wind was neglected.

1.2.2 Material Properties for the design

Design compressive of concrete f ck = 35 N/mm

2

Yield strength of steel f yk = 460 N/mm2

Table 1.2: Values used in the computation of loading

Load Value

Density of concrete 25 KN/m3

Imposed load/variable action 5 KN/m2

Weight of finishes, rails, seats, stair units 2 KN/m2

1.2.3 Concrete cover

Exposure class = XC1

A concrete cover of 40mm is adopted for the section

1.2.4 Design equations according EC2

From EC2 singly reinforced concrete stress block;

MRd = FCz ------------ (1)

FC =.

. 0.8 ; z = d – 0.4x -------------- (2)Clause 5.6.3 of EC2 limits the depth of the neutral axis to 0.45d for concrete class less than or equal to C50/60.

Therefore for an under reinforced section (ductile);

x = 0.45d ----------------- (3)

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

MRd = 0.167

---------------------- (4)

Also from the reinforced concrete stress block;

MEd = FSz ------------------ (5)

FS =. ------------------ (6)

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

= . -------------------- (7)The lever arm z in EC2 is given ;

z = d [0.5 (0.25 0.882) ] ---------------------- (8)where K =

---------------- (9)

1.2.4.1 For doubly reinforced sections;

Area of compression reinforcement AS2 =−

.(− ) ----------------------- (10)

Area of tension reinforcement = . + AS2 -------------------- (11)



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Where z = d [0.5 (0.250.882 ′) ] where K’ = 0.1671.2.4.2 Check for deflection (Clause 7.4.2)

The limiting basic span/ effective depth ratio is given by;

L/d = K 111.5 3.2 1 ⁄ if ≤ --------------------------- (12)

L/d = K 111.5 − ⁄ if > ---------------------- (13)

Where;

L/d is the limiting span/depth ratio

K = Factor to take into account different structural systems

= reference reinforcement ratio = 10−

= Tension reinforcement ratio to resist moment due to design load

′ = Compression reinforcement ratio

1.2.4.3 Shear design

In EC2, the concrete resistance shear stress without shear reinforcement is given by;

VRd,c = [CRd,c.k. (100 ) + k 1. ]bw.d ≥ (Vmin + k 1. ) ----------------------- (14)CRd,c = 0.18/; k = 1+ < 0.02 (d in mm); = < 0.02 (In which is the area of tensile reinforcementwhich extends ≥ ( ) beyond the section considered; Vmin = 0.035 K 1 = 0.15;

= NEd /Ac < 0.2fcd (Where NEd is the axial force at the section, Ac = cross sectional area of the

concrete), fcd = design compressive strength of the concrete.

1.2.5 Load Analysis

1.2.5.1 Loading on precast seating unit

Permanent Actions

Self weight of the 7m precast seating deck (see Figure 1.4)

(GK1) = (25 × 0.25 × 0.15 × 7) + (25 × 0.95 × 0.15 × 7) = 31.5 KNWeight of finishes, rails, seats (G

K2) = (2

× 0.95

× 7) = 13.3 KN

Variable Actions

Imposed load for structural class C5 (QK ) = (5 × 0.95 × 7) = 33.25 KNTotal action on L-shaped seating terrace unit at ultimate limit state by Eurocode 2

(FE) = 1.35∑(GKi) + 1.5QK = 1.35(44.8) + 1.5(33.25) = 110.355 KN

1.2.5.2 Loading on the raker beams

Height of beam = 1200mm

Width of beam = 400mm



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Self weight of raker beam

Concrete own weight (waist area) = 1.2m × 0.4m × 25 KN/m3 = 12.00 KN/m (normal to the inclination i.e. inthe local direction)

Height of riser in the raker beam = 0.4m; Width of tread in the raker beam = 0.8m; Angle of inclination (

)=

20.556°

Stepped area (risers) = 1 2⁄ × 0.4 × 25 = 5 KN/m (in the global direction)For purely vertical load in the global y-direction, we convert the load from the waist of the beam by;

UDL from waist of the beam = (12.00 × cos 20.556°) = 11.236 KN/m Total self weight (Gk) = 11.236 + 5 = 16.235 KN/m

Self weight of raker beam at ultimate limit state;

n = 1.35∑(GKi) = 1.35 × 16.235 = 21.917 KN/mLoad from precast seating units

End shear from precast seating unit = FE/2 = 110.355 = 55.1775 KN

Total number of the precast seating units on the beam = 24/0.8 = 30 unitsFor an intermediate beam supporting seating units on both sides;

Total number of precast seating units = 2 × 30 = 60 unitsTherefore, total shear force transferred from the seating units to the raker beam = 55.1775 × 60 = 3310.65 KNEquivalent uniformly distributed load in the global direction at ultimate limit state =

. = 137.94 KN/m

Total load on intermediate raker beams at ultimate limit state in the global direction = 137.94 + 21.917 =

159.857 KN/m

1.3 Structural Analysis

A full 3D elastic analysis of the whole stadium was performed using Staad Pro with all elements

loaded at ultimate limit state. Also, the raker beam is isolated as a subframe and also analysed. Theresults from the two models are very comparable.

Figure 1.6: 3D Modelling of the grandstand on Staad Pro



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The internal stresses on the intermediate raker beams from the analysis of the frame at ultimate limit

state are shown in Figures 1.7 to 1.9.

Figure 1.7: Bending Moment Diagram

Figure 1.8: Shear Force Diagram

Figure 1.9: Axial Force Diagram



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1.3.1 Summary of Analysis Results

The summary of the analysis result of the raker beams is shown in Table 1.3.

Section Moment

(KN.m)

Section Shear Force

(KN)

Section Axial Force

(KN)

MA 1967.54 QAB 934.62 NAB 380.061(C)

MABspan 948.078 QBA 983.88 NBA 339.376(T)

MB 2283.18 QBC 999.52 NBC 510.767(C)

MBCspan 1249.787 QCB 918.98 NCB 208.670(T)

MC 1565.63

1.4 Structural Design

The structural design of the of the raker beam using EN 1992-1-1has been carried out and all the

parameters used in the, and steps followed are shown below in the subsequent sections.

Design compressive of concrete f ck = 35 N/mm2

Yield strength of steel f yk = 460 N/mm2

bw = 400mm; h = 1200mm; Cc = 40mm

1.4.1 Flexural Design of span AB (MABspan)

MABspan = 948.078 KNm

d = h – Cc – ϕ/2 – ϕlink

d = 1200 – 40 – 16 – 10 = 1134mm

k =

=. ×

× × = 0.0527 Since k < 0.167 No compression reinforcement required

z = d [0.5 (0.25 0.882) ] = z = d [0.5 (0.25 0.882(0.0527) ] = 0.95d

= . =. ×

. × × . × = 2199 mm2 Provide 5Y25mm BOT (ASprov = 2450 mm2)

To calculate the minimum area of steel required; (TABLE 3.1 EC2)

fctm = 0.3 × ⁄ = 0.3 × 35 ⁄ = 3.2099 N/mm2ASmin = 0.26 × mFyk × bw × d = 0.26 × 3.2099 ×400 ×1134 = 822.962 mm2 Check if ASmin < 0.0013 × bw × d (589.68 mm2)Therefore, ASmin = 822.962 mm2

Check for deflection;

K = 1.5 for beam fixed at both ends

L/d = K 111.5 3.2 1

⁄

if ≤

Table 1.3: Analysis Results of the Raker Beam



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L/d = K 111.5 −

⁄ if >

= = × = 0.00540 < 10−√ 35

L/d = 1.5 111.5√ 35 × .. 3.2√ 35 .. 1 ⁄ = 1.5(20.695 + 0.5333) = 31.842 Modification factor = = = × × × = 278.241 N/mm2 = . =1.11 Since the span is greater than 7m, allowable span/depth ratio = × 31.842 × = 1.11 × 31.842× = 19.374Actual deflection L/d =

= 11.301

Since 11.301 < 19.374, deflection is ok.

1.4.2 Flexural Design of support A (MA);

MA = 1967.54 KNm

k = 0.1093; la = 0.8919; AS1 = 4861 mm2; ASmin = 822.9785 mm2

Provide 4Y32mm + 4Y25mm TOP (ASprov = 5180 mm2)

1.4.3 Flexural Design of support B (MB);

MA = 2283.18KNmk = 0.1268; la = 0.8717; AS1 = 5772 mm2; ASmin = 822.9785 mm2

Provide 6Y32mm + 4Y20mm TOP (ASprov = 6080 mm2)

1.4.3 Flexural Design of span BC (MBCspan)

MBCSpan = 1249.787 kNm

k = 0.0694; la = 0.9345; AS1 = 2947mm2; ASmin = 822.9785 mm2

Provide 5Y25mm + 2Y20mm BOT (ASprov = 3083 mm2)

Check for deflection

= = × = 0.00679 > 10−√ 35

L/d = K 111.5 −

⁄ if >

L/d = 1.5 111.5√ 35 × .. − 0 = 28.066Modification factor =

= = × × × = 299.039 N/mm2



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= . = 1.0366Since the span is greater than 7m, allowable span/depth ratio = × 28.066 × = 1.0366 × 28.066 × = 15.89Actual deflection L/d = = 11.301Since 11.301 < 15.89, deflection is ok.

1.4.4 Flexural Design of support C (MC);

MC = 1565.63 KNm

k = 0.0870; la = 0.9163; AS1 = 3765 mm2; ASmin = 822.9785 mm2

Provide 5Y32mm TOP (ASprov = 4020 mm2)

Provide Y16 @ 200mm c/c on both faces as longitudinal side bars

1.4.5 Shear Design

1.4.5.1 Support A

VEd = 934.62 KN; N = 380.061 KN (Compression)

Taking shear at the centreline of support; VEd = 934.62 KN

VRd,c = [CRd,c.k. (100 ) + k 1. ]bw.d ≥ (Vmin + k 1. ) bw.d CRd,c = 0.18/ = 0.18/1.5 = 0.12k = 1+ = 1+ = 1.4199 < 2.0Vmin = 0.035 = Vmin = 0.035 ×(1.4199) ×(35) = 0.3504 N/mm2

= = × = 0.011419 < 0.02; K 1 = 0.15 = NEd /Ac < 0.2fcd (Where NEd is the axial force at the section, Ac = cross sectional area of the concrete),fcd = design compressive strength of the concrete.)

= . × × = 0.7917 N/mm2

VRd,c = [0.12 × 1.4199 (100×0.011419 ×35 ) + 0.15 ×0.7917 ]400 ×1134 = 318111.948 N = 318.11 KNSince VRd,c < VEd , shear reinforcement is required.

Assume strut angle = 21.8°Let us now investigate the compression capacity of the strut;v1 = 0.61 = 0.61 = 0.516f cd =

Taking = 0.85; fcd = . × . = 19.833 N/mm2; z = 0.9d

VRd,max =( + ) = × . × × . × .(. + .) 10− = 1440.64 KN > VEd

Since VEd < VRd,max

VEd,s = cot = 934620 N



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=

(. × ×. × × .) = 0.9153

Trying 3Y10mm @ 200mm c/c (235/200 = 1.175)

1.175 > 0.9153 Hence shear reinforcement is ok.

Following the steps described above;

1.4.5.2 Support B; Shear at VBA

VEd = 983.88 KN; N = KN 339.376 (Tension)

Note that due to the tensile axial force in the section, the second term of VRd equation assumes a negative value.

= 0.0134; = 0.7070 N/mm2; vmin = 0.3504 N/mm2; VRd = 230.6532 KNSince VRd,c < VEd , shear reinforcement is required

Assuming that the strut angle = 21.8° v1 = 0.5160; f cd = 19.8450 N/mm

2

; z = 0.9d = 1020.6 mm; VRDmax =1440.64 KNSince VRDmax > VEd

= 0.9635

Trying 3Y10mm @ 200mm c/c (235/200 = 1.175)


1.4.5.3 Support B; Shear at VBC

VEd = 999.52 KN; N = KN 510.767 (Compression)

Note that due to the tensile axial force in the section, the second term of VRd equation assumes a negative value. = 0.0134; = 1.0641 N/mm2; vmin = 0.3504 N/mm2; VRd = 351.161 KNSince VRd,c < VEd , shear reinforcement is required

Assuming that the strut angle = 21.8° v1 = 0.5160; f cd = 19.8450 N/mm2; z = 0.9d = 1020.6 mm; VRDmax =1440.64 KN

Since VRDmax > VEd

= 0.9789

Trying 3Y10mm @ 200mm c/c (235/200 = 1.175)


1.4.5.4 Support C; Shear at VCB

VEd = 918.98 KN; N = KN 208.670 (Tension)

Note that due to the tensile axial force in the section, the second term of VRd equation assumes a negative value.

= 0.0089; = 0.4347 N/mm2; vmin = 0.3504 N/mm2; VRd = 213.2707 KNSince VRd,c < VEd , shear reinforcement is required

Assuming that the strut angle = 21.8° v1 = 0.5160; f cd = 19.8450 N/mm2; z = 0.9d = 1020.6 mm; VRDmax =1440.64 KN

Since VRDmax > VEd



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= 0.9000

Trying 3Y10mm @ 200mm c/c (235/200 = 1.175)


1.5 Discussion and Conclusion

It is very easy to see that the influence of axial force was not very pronounced in the results produced. It would

have been very significant using BS 8110. The maximum reinforcement was seen at support B due the high

magnitude of moment at that section. This phenomenon is consistent with horizontal continuous beams. See

detailing sketches in Figure 2.0.

Figure 2.0: Reinforcement detailing sketches (not to scale)



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References

[1] BS 6399 part 1: 1996: Loading for Building code of practice for dead and imposed loads.British Standards Institution.

[2] BS 8110 – 1:1997: Structural use of concrete Part1: Code of practice for design andconstruction. British Standard Institutions.

[3]

EN 1991-1-1 (2002): General Actions- Densities, self weight, imposed loads for buildings

[4] EN 1992-1-1 (2004): Design of concrete structures: General Rules and rules for building

[5] Jeff Steele, Mark Larsen (1996): Raker-Beam Construction Requires Rugged Steel Forms.

Publication #C960738 The Aberdeen Group

[6] Karadelis J (2009): Concrete Grandstands. Part 1: Experimental investigations. Proceedingsto the Institution of Civil Engineers – Engineering and Computational mechanics. Volume162,Issue 1 ISSN 1755-0777

[7] Salyards K.A., Honagan L.M (2005): Evaluation of a finite element model for dynamiccharacteristic prediction of stadium facility.

Documents

Analysis and Design of a Continuous R. C. Raker Beam using Eurocode 2