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The University of LeedsSchool of Civil Engineering Reference

Sheet No: 1 of 13 Date:16 -April -2012 Job title : Reinforced Concrete Column Design to EC2 Calculation and Drawing Results

1. Actions and given data: It is required to design a symmetrically reinforced concrete column to carry the following design ultimate load effects: Axial Load = 1950 kN End moment at top of column = 105 kNm End moment at base of column = 0 Bending is about one axis only. Therefore, the first order moments are shown in figure(1)

Figure (1): First order moments. It is assumed that the column does not experience any sway effects, then the design will be for Braced column. The effective heights of the column is assumed to be taken as 7.5m about both axes (i.e. ley = lez = 7.5m) The creep coefficient, ef = 0.9. 1. Initial assumptions : It is required to make initial reasonable assumptions about materials and the cross-sectional dimensions of the column and the area of reinforcement are to be used. It is essential to taking into account (fire/exposure conditions/cover) and all requirements to produce a design that is economical, buildable, durable and safe (under all credible situations). 2.1 General assumption: The design working life of the structure is 50 years.Sheet No: 2 of 13

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The University of LeedsSchool of Civil Engineering Reference

Date:16 -April -2012 Job title : Reinforced Concrete Column Design to EC2 Calculation and Drawing Results

Table 4.1 : BS EN 1992-1-1:2004 EN 1992-1 -1 12004 (E)

Clause 3.2.2(3) BS EN 1992-1-1:2004 EN 1992-1 -1 12004 (E)

Normal quality control is put in place. The max aggregate size is: dg = 20 mm (< 32 mm). Assuming the exposure condition due to Corrosion induced by carbonation is XC1 (dry) The recommended Structural Class (for design working life of 50 years) is S4 Assuming 1 hour fire resistance required for.

Table 3.3[Structural elements design manual working with Eurocodes]

2.2 Materials assumptions: Eurocode 2 permits to use a various steel yielding grades ranging from 400 MPa to 600 MPa. Class H is the normally specified Class. So, the characteristic fyk can be taken as 500 MPa. Ii is originally assumed the concrete is with strength class C25/30. In order to obtain an adequate concrete durability, the reference (min.) concrete strength class for exposure class XC1 is C20/25; the resistance class adopted (C25/30) is suitable as it is higher than the reference strength class. It is recommended to do modifications to the structural class. Then, form table 4.3N for concrete strength class 30/37 the class of structural will reduce by 1. Therefore, the structural class is

Table 3.10 [Structural elements design manual working with Eurocodes]

[Clause 4.4.1.2(5) and Table 4.3N ] BS EN 1992-1-1:2004 EN 1992-1 -1 12004 (E)

structural class is S3

fck = MPa,

25

Table 3.8 [Structural elements design manual working with Eurocodes]

3

becoming S3. As the design calculations are always based on fck which is the cylinder compressive strength and it is the lower of the two values in the grade designation: Therefore; fck = 25 MPa, The partial factor m for concrete and steel used in reinforced concrete design are : At ULS : c for Concrete = 1.5 ; S for Steel =1.15 At SLS : c for Concrete = 1 ; S for Steel =1 Sheet No: 3 of 13 The University of LeedsSchool of Civil Engineering Reference Date:16 -April -2012 Job title : Reinforced Concrete Column Design to EC2 Calculation and Drawing Results

Therefore: fcd = 0.85(25/1.5) = 16.67MPa fyk = 500 MPa ; fyd= (500/1.15)= 434.8 MPa

fcd =16.67MPa fyd =434.8 MPa

Section 3.11 [Structural elements design manual working with Eurocodes]

2.3 The Cross-Sectional dimensions of the column: Initial dimensions are normally determined by taking into account requirements for durability and fire resistance and it is not practical to cast vertical columns smaller than 200X200 mm. Hence, Assuming : Large dimension =450 mm; Small Ac =157500mm2 dimension =350 mm. Therefore, (gross section of the column)=Ac =157500mm2As = 4825 mm2

BS EN 1992-1-1:2004 EN 1992-1 -1 12004 (E) (EC2 clause 4.4) Clause 4.4.1.1 (2) Exp. (4.1) Clause 4.4.1.2

2.4Reinforcement in the column : Assuming the longitudinal bars are 6 no. H32; and the links are 8 at 180 mm. Therefor ; As = 4825 mm2 3. Concrete cover : The nominal value for

concrete cover can be determined from : Nominal cover: Cnom = Cmin + Cdev

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Exp. (4.2)

Clause 4.4.1.3 Clause 4.4.1.2(3)& Table 4.2

Clause 4.4.1.2 (6) Clause 4.4.1.2 (7) Clause 4.4.1.2 (8)

Where, Cmin is: The minimum concrete cover, which shall be provided in order to ensure: The safe transmission of bond forces. The protection of the steel against corrosion (durability). An adequate fire resistance [according to (EN 1992-1-2). Cmin can be determined from: Cmin = max {C min,b; C min,dur + c dur, , - c dur,st c dur,add ; c dev } For the above equation : The c dev (allowance in design for deviation)= 10 mm C min,b = minimum cover due to bond requirement ,the minimum cover C min,b = diameter of bar. c dur, , The recommended value is 0 mm. c dur,st ,. The recommended value is 0 mm. c dur,add ,The recommended value is 0 mm.Sheet No: 4 of 13 Date:16 -April -2012 Job title : Reinforced Concrete Column Design to EC2 Calculation and Drawing Results

The University of LeedsSchool of Civil Engineering ReferenceClause 4.4.1.2 (5)

3.1 Design the cover for environmental condition and durability c dur,: The amount of cover necessary to

Table 4.4N: -EC2

Table 4.2:EC2

Table 4.4N: EC2

protect reinforcement against corrosion depends on the exposure conditions and quality of the concrete used. C min,dur = minimum cover due to environmental conditions. In this example, the environmental condition is assumed to be class XC1. Then, the minimum concrete cover for XC1 is equal to 10 mm for modified Structural Class S3, cmin,dur=10 Concrete cover for the links The minimum cover, cmin,b due to bond = diameter of bar. cmin,b = 8 mm cmin,dur = 10 mm Hence: cmin = max (cmin,b; cmin,dur + cdur, - cdur,st -

Exp.(4.2): EC2

Exp.(4.1): EC2

5

Table 4.2- EC2 Table 4.4N - EC2

cdur,add; 10 mm) max (8 ; 10 + 0 0 0; 10 mm) = 10 mm then : cnom = cmin + cdev = 10 + 10 = 20 mm . Concrete cover for longitudinal reinforcement bars: c min,b = Diameter of bar = 32mm. c min,dur = 10 mm . Hence: From Exp.(4.2) c min = max (32; 10 + 0 0 0; 10 mm) =32mm . From Exp.(4.1) cnom = cmin + cdev = 32 + 10 = 42 mm . For the ordinary reinforcement bars, the diameter for links is reduced The concrete cover for links is 42 - 8 = 34 mm. say C,nom =35mm to links . C,=35mm

nom

The University of LeedsSchool of Civil Engineering ReferenceBS EN 1992-1-2:2004 EN 1992-1-2:2004 (E) (EC2 -clause 5.3) clause 5.3.2 (2) & clause 5.3.3(3)

Sheet No: 5 of 13 Date:16 -April -2012 Job title : Reinforced Concrete Column Design to EC2 Calculation and Drawing Results

3.2 Design cover for fire resistance requirement :"For

Table 5.1 Concise EC2

braced building structures where the required Standard fire exposure is higher than 30 minutes, the effective length l o,fi may be taken as 0,5 l for intermediate floors and 0,5 l l 0,fi 0,7l for the upper floor, where l is the actual length of the column (centre to centre)" From the given data: Effective heights of the column=7.5m. lo = factor clear height = 7500mm Conservatively, choose to use tabular method. For critical direction condition 2 at top and condition 3 at bottom (pinned support). Then, the factor is 0.95 Therefore , the clear height = 7.5/0.95 =7.89m The effective length of column in fire lo,fi = 0.5 clear height Therefore: for fire resistance the effective

BS EN 1992-1-2:2004 EN 1992-1-2:2004 (E)

6

clause 5.3.2

heights of the column is: lo,fi=0.57.89= 3.95m. a) Check validity of using Method A :

BS EN 1992-1-2:2004 EN 1992-1-2:2004 (E) clause 5.3.3

Check if lo,fi 3.0 m

As lo,fi > 3 m Method A is not adequate. b) Check validity of using Method B.

BS EN 1992-1-2:2004 EN 1992-1-2:2004 (E) Exp. (5.8a)

Ac =157500mm2 As = 4825 mm2

Determine parameters n, , e and check fi. n = N0Ed,fi / (0.7(Acfcd + Asfyd)) n = 0.7 1950 / 0.7 (Ac cc fck / C + As fyk / S) BS EN 1992-1-2:2004 EN 1992-1-2:2004 (E) =0.71950 103 / 0.7 (157500 0.85 25 / 5.3.3(2) 1.5 + 4825 500 / 1.15) BS EN 1992-1-2:2004 =1365/ 0.7 (2231.25+ 2097.83) EN 1992-1-2:2004 (E) Exp. (5.8b) n= 0.45 = Asfyd / Acfcd 1.0 = 4825434.8 / 15750016.67 = 0.799 0.8 < 1 OK e = M0Ed,fi / N0Ed,fi = [105 / 0.70 1950 ] 1000=76.9 77 Sheet No: 6 of 13 The University of LeedsSchool of Civil Engineering Referenceclause 4.6.2 Concise EC2

n= 0.45fcd =16.67MPa fyd =434.8 MPa =0.8 e=77

Date:16 -April -2012 Job title : Reinforced Concrete Column Design to EC2 Calculation and Drawing

Results

The limiting values of e:

BS EN 1992-1-2:2004 EN 1992-1-2:2004 (E) Exp. (5.8c)

BS EN 1992-1-2:2004 EN 1992-1-2:2004 (E)

=0.15b=0.15350=52.5 Or , emax 0.25b=0.25350=87.5 When, 0.15b < e < 0.25b, Method B is satisfying to be used. fi = lo,fi / i where lo,fi = 3.95m. i = radius of gyration = h / 3.46 for a rectangular section = 3.95m 1000/ (350 / 3.46)about yfi fi=39 y axis=39 fi = 3.95m 1000/ (450/ 3.46) about z-z axis=30.11 About y-y axis is critical, = 39 > 30 this is proved that the column is slender. Then, BS EN 199212 Annex C Tables C1 C9 are to be used. From table C5 ( = 0.5, e = 0.25b): Assume column exposed to fire on more than one sideemax

7

and Standard fire resistance is R60. For = 30 interpolate between n=0.3 and n=0.5 n=0.3 b,min = 150 , a = 35 (minimum values are used) n=0.45 b,min? , a =? mmm n=0.5 b,min = 250, a = 35 (minimum values are used) b min =225mm, a = 35mm for R60 at n=0.45 For = 40 interpolate between n=0,3 and n=0.5 n=0.3 b,min = 200, a = 30 (minimum values are used) n=0.45 bmin=? , a = ? n=0.5 bmin = 300, a= 35(minimum values are used )b min = 275 mm, a=33.75 for R60 at n=0.45 Hence ,interpolate between = 30 and = 40 = 30 b,min = 225mm , a = 35 mm = 39 b,min = ? , a = ? mm = 40 b,min =275 mm , a = 33.75 mm b,min = 270 mm , a = 33.9 mm for = 39 and =0.5 From table C8 ( =1, e = 0.25b) For = 30 interpolate between n=0.3 and n=0.5 n=0.3 b,min =150, a=30 mm (minimum values are used) n=0.45 b,min ? , a = ? mm n=0.5 b,min=200, a= 40mm (minimum values are used) b min = 178.5mm, a 37.5 mm for R60 at n=0.45 For = 40 interpolate between n=0,3 and n=0.5 n=0.3 b,min=150,a=40mm(minimum values are used) n=0.45 b,min ? , a = ? mm n=0.5 b,min = 250, a =25(minimum values are used) b min =225mm , a =28.75 mm for R60 at n= n=0.45 The University of LeedsSchool of Civil Engineering Reference Sheet No: 7 of 13 Date:16 -April -2012 Job title : Reinforced Concrete Column Design to EC2 Calculation and Drawing Results

interpolate between = 30 and = 40 = 30 b,min =178.5mm, a = 33.75 = 39 b,min = ? , a =? mm = 40 b,min =225mm, a = 28.75 mm mm

8

b,min = 220.3 mm , a =29.25mm for = 39 and =1 Then interpolate between =0.5 and =1 =0.5 b,min= 270 mm, a = a = 33.9 mm =0.8 b,min= ? , a=? =1 b,min= 220.3mm, a=29.25mm b,min =240.18 mm and a = 31.11 mm . Therefore , the small dimension shall not be less than 240.18 mm and the cover for fire resistance also shall not less than 31.11 mm Thus, the assumption for b=350mm is satisfy. And, the cover to be sufficient for all requirements is 35 mm. 3. Structural design: The second order effects may be ignored if the column slenderness is below a certain value lim . Limiting slenderness ratio, lim ,can be calculated from the next expression : lim = 20 ABC / n0.5 =0.8 where: A = 1 / (1 + 0.2ef) = 1 / (1 + 0.2x0.9)=0.847 B = (1 + 2 Asfyd / Acfcd) 0.5 = (1 + 2)0.5 = 0.8 as before B = (1 + 2 0.8) 0.5= 1.61 C = 1.7 rm where rm = Mo1 / Mo2, Mo1 = 0, then rm = 0 C = 1.7 BS EN 1992-1-1:2004 n = NEd / Acfcd EN 1992-1 -1 12004 (E) =1950 103 / (350 450 0.85 25 / 1.5)=0.87 Clause 5.8.3.2 lim = 20 0.847 1.61 1.7 / (0.87) 0.5 Exp. (5.14) lim= 49.71 Check Actual slenderness ratios: The effective length= lo=7.5m Slenderness ratio, = l0 / i Where : i = radius of gyration Sheet No: 8 of 13 The University of LeedsSchool of Civil Engineering Reference Date:16 -April -2012 Job title : Reinforced Concrete Column Design to EC2 Calculation and Drawing Results

BS EN 1992-1-1:2004 EN 1992-1 -1 12004 (E)

clause 5.8.3.1(1) Exp. (5.13N)

slenderness ratios are : y =(7.5/0.35) 3.46=74.14 >49.71 z=(7.5/0.45) 3.46=57.7 > 49.71

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Nominal curvature method based on [Reinforced Concrete design to Eurocode 2 Book ]

As both > lim the column is slender, and y is critical. The column is considered as slender column. Hence, the nominal curvature method will be used to design the column, where the second order moment is added to the first order moment to give the total column design moment. Mt = nominal 2nd order moment = Ned etot Where , etot= ee + ea+ e2 Since: ee is Equivalent eccentricity ea is the accidental eccentricity e2 is the second order eccentricity The top and bottom eccentricities are : eo2= M o2 N =

eo2=54mm eo1=0mm

Ed

105 x 10 3 =53.84 m m 1950

M o1 eo1= N =0 mm Ed

Equivalent eccentricity = (0.6 eo2 + 0.4 eo1) 0.4 eo2 (0.6 x 5 3 .8 4+ 0.4(0) 0.4( 53.8 4 (0.6 x 5 3 .8 4+ 0.4(0)=32.304

ee= 32mm ea= 1 9mm

0.4( 53.8 4 =21.536Therefore the equivalent eccentricity is ee= 32.304mm By taking v as 2 0 0 the accidental eccentricity is :ey ea= v x 2 = 200 x 2 = 18.75 m m

1

l

1

7500

The second order eccentricity is:

e2= K

2 x 1 035001

K 2 l2 f o

yk

d

e2= 9 5 mm 25 74.14

k1=1+(0.35+ 200 150 ) ef=1+(0.35+ 200 150 )x 0.9 k1= 0.98 1. Then , use k1=1 d ' = c,nom + link + / 2 = 35+8+32/2 =59mm d =350-59 =291mm With k2=1 for initial value.1 x 1 x 7500 2 x 500 =94.6 m m 2 x103500 x 291

f ck

e2=

10

etot= ee + ea+ e2 =32+19+95 =146The University of LeedsSchool of Civil Engineering Reference Sheet No: 9 of 13 Date:16 -April -2012 Job title : Reinforced Concrete Column Design to EC2 Calculation and Drawing Results

etot = 146mm Design total moment Mt = nominal 2nd order moment = NEd etot Mt =195014610 -3 =284.7 KNm As d ' = 59mm d ' /h =59 /350= 0.168 Then, the chart for d ' /h=0.15 can be used as shown in theConcise EC2

etot=146mm

figure (2).

Figure (2) design column chart for d ' /h=0.2 For :N Ed 1950x 103 = h b f c k 450 x 350x 25 =0.49Mt hb f ck2

=

284.7 x106 =0.206 2 450 x 350 x 25ck As f y k bh f

Then,

=0.68

And K2= 0.71 The new value K2is used to calculate e2 and Mt for the second iteration. The design chart is again used to

11

determine

ck As f y k bh f

and a new value of K2 as shown in table

(1). The iterations are continued until the values of K2 in the last column are becoming very close which gives a reasonable agreement .In this design ,it occurs after two iterations Sheet No: 10of 13 The University of LeedsSchool of Civil Engineering Reference Table (1) K2 iteration Trail No. 1 2 Date:16 -April -2012 Job title : Reinforced Concrete Column Design to EC2 Calculation and Drawing Results

K21 0.71

e29567.18

etot146 118.2

Mt

Mt hb f ck2

ck As f y k bh f

K20.71 0.69

284.7 230.5

0.20 0.17

0.68 0.56

So, the steel area required is

As =

0.56b h f c k = 4410 mm2 f ykAc =157500mm2 As = 4825 mm2

Then the provided area of steel As = 4825 mm2 is satisfying . And k2=0.69 As a check on the final value of k2 interpolated from the deign chart:

N bal = 0.29 f c k AcN bal = =0.29 2515750010 -3 = 1141.875 KN

N ud =0.567 f c k Ac+0.87 f yk AsN ud =(0.56725157500+0.875004825 ) 10 -3= 4331.4375 KN = K2= NN u d N b al ud N E D

=

4 3 .4 31 4 3 .4 31

1 4 .8 5 11 7 1 5 90

= 0.74

K2 from design chart can be accepted as it close to the final

value obtained from equation above. The small different (0.05) between the values of K2 calculated from the equation and the design chart is because the chart used was for d ' /h =0.15 but, in fact the value of d ' /h

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is 0.168.

The University of LeedsSchool of Civil Engineering Reference

Sheet No: 11 of 13 Date:16 -April -2012 Job title : Reinforced Concrete Column Design to EC2 Calculation and Drawing Results

Design summary

Drawing should be here:

6H 32 H8 links @180cc 35 cover to links Check Reinforcement requirements :IStructE EC2 (Concrete) Design Manual

Minimum area of reinforcement is given by the greater of: 0.18N or 0.3% of the gross cross-sectional area of the concrete . Ok satisfy. Longitudinal bars should not be less than 12mm diameter. (The used bars are H32) . Columns should be provided with links whose diameter should not be less than one-quarter the diameter of the largest longitudinal bar nor less than 6mm.(the used diameter for links 32/4=8mm) The maximum spacing of links should be the lesser of: 12 times the diameter of the smallest compression bar =332 =384mm the least dimension of the column=350mm 300mm. The spacing used is 180 mm which is less than 300mm and satisfy all conditions.

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The University of LeedsSchool of Civil Engineering Reference

Sheet No: 12 of 13 Date:16 -April -2012 Job title : Reinforced Concrete Column Design to EC2 Calculation and Drawing Results

Summary of the alternative attempting : The initially attempted was by using : b=400mm; h=350 mm. Ac = 140000 mm2(gross section of the column). And Assuming the longitudinal bars are 6 no. H32; the links are 8 at 180 mm. The value obtained for that solution are : n=0.47, =0.89, the cover =35mm ,lim= 49 y =(7.5/0.35)x3.46=74.14 >49 z=(7.5/0.4)x3.46=64.87 > 49 k1= 0.981 K2=1

d ' = c,nom + link + / 2 = 35+8+32/2 =59mm d =350-59 =291mme2=1 x 1 x 7500 2 x500 =95 m m 2 x 103500 x 291

etot= ee + ea+ e2 = 32.304+ 18 .7 5 9 5 =146mm. +Design total moment Mt = nominal 2nd order moment = NEd Mt =1950x 144.544x10 -3 =284.7 KNm

etot=

N Ed 1950x 103 = h b f c k 400 x350 x 25 =0.557 0.57Mt hb f ck2

=

284.7 x 106 =0.23 2 400 x 350 x 25

For d ' /h =59 /350= 0.168, the design chart for d ' /h =0.15 was used. The summary of results is shown in table (2)Table (2) results from design chart for 350400mm column. ck Mt Trail Mt K2 e2 etot 2 As f y k No. hb f bh fck

K2

14

1 2

1 0.71

9567

146 118

284.7 230.2

0.23 0.19

0.820.71

0.71 0.63

The solution was not satisfying because of the area of steel required is higher that area provided in the section .so, and the section is not adequate for resistance the applied forcesAs = 0.71 b h f c k =4970 m m 2 > 4825mm provided f ykSheet No: 13of 13 Date:16 -April -2012 Job title : Reinforced Concrete Column Design to EC2 Calculation and Drawing Results

The University of LeedsSchool of Civil Engineering Reference

Bigger section was also attempted, with 350 mm by 600mm, but it was not an economic solution. As the concrete area was adequate to sustain the loads, but the provided steel area used was big comparing with the required steel area.As = 0.3 b h f c k = 3150 m m 2 4825 mm2 provided. f yk

Conclusion : From all the attempted solution , the column of 450mm 350 can be consider as an economical, buildable, durable and safe design. If the steel is more expensive than concrete ,it is accepted to use large sections of concrete and reduce the provided steel area as possible but not less than the minimum area required by the code of design .Therefore, the column with 600350 dimensions can be also adequate but we need to reduce the amount of steel by using 2 bars of H25 (982mm2) and 4 bars of H32 (3217mm2) which together provided (4199mm2) . that could be better than to consuming (4825mm2) if 6H32 bars areused .

15

Reference:

BS EN 1992-1-1:2004 Eurocode 2: Design of concrete structures Part 1-1: Generalrules and rules for buildings.

BS EN 1992-1-2:2004 Eurocode 2: Design of concrete structures Part 1-2: Generalrules - Structural fire design. Concise Eurocode 2 for the design of in-situ concrete framed building to BS EN 1992-11:2004 and its National Annex: 2005, Narayanan R S and Goodchild CH , The concrete Centre ,2006. Draycott.T. Structural elements design manual working with Eurocodes2 ed. Great Britain .Elsevier Ltd.2009 EUROCODE 2 WORKED EXAMPLES. IStructE Manual for the design of reinforced concrete building structure to Eurocode 2, Institution of structural Engineers, The Institution of Civil Engineers 2000.

Mosley B et al . Reinforced Concrtet design to Eurocode 2,6 th , palgrave macmillan.