1

Introduction

Three reinforced concrete elements have been designed according to Eurocode 2 ; a simply supported slab, a simply supported beam and a column.

The slab is a two-way spanning slab so the moment values are determined by the coefficients given in the British Standards and the dimensions were assumed according to structural class and environmental conditions.

The moment and the axial load acting on the column were slightly low, so the column has been designed with smaller dimensions.

Hence, the beam had to have smaller dimensions, but had to carry the moment on it due to relative larger opening and the uniformly distributed load acting on it. This has been made possible by designing the beam with a singly reinforced section, to take the advantage of larger lever arm. This also made the need of reinforcement less on the tension side, and none on the compression. Also, with the less reinforcement required, the concrete space between the reinforcement bars were at acceptable values.

2

Simply Supported Slab Design to Eurocode 2

Properties of the Slab

ly = 6.75 m

lx = 4.50 m

ly / lx = 1.50

Bending Moment Coefficients

According to BS8110

αsx = 0.104

αsy = 0.046

Assumptions of Height of Slab and Reinforcement Coverage to Eurocode

It is assumed that Structural Class is S3 and exposure class related to environmental conditions is XC1. From Eurocode 2 Table 4.4 the minimum cover is 20 mm.

Assumed height of the slab = 200 mm

Assumed reinforcement coverage = 20 mm

Ultimate Load

Ultimate Load given in Eurocode = 1.35 g × 1.50 q

1.35 × (0.20 × 24) + 1.50 × 3.5 = 11.73 kN / m2

Bending Moment for Short Span

Msx = αsx × n × lx2

Msx = 0.104 × 11.73 × 4.52 = 24.70 kN m

Determining Ko

K0,sx = Moment / fck / b / d2

K0,sx = 24.70 × 106 / 30 / 1000 / 1802

K0,sx = 0.0254 < 0.167

Singly Reinforced Section

Determining Lever – Arm for Short Span

zsx = d × [ 0.5 + ( 0.25 – ( K0,sx / 1.134 ) )0.5 ]

zsx = 180 × [ 0.5 + ( 0.25 – ( 0.0254 / 1.134 ) )0.5 ]

zsx = 175.87 mm

3

Reinforcement Calculation

Asx = Msx / 0.87 / fyd / zsx

Asx = 24.70 × 106 / 0.87 / 460 / 175.87

Asx = 350.92 mm2/m

Minimum Reinforcement

According to Eurocode Clause 9.2.1

The total amount of longitudinal reinforcement should not be less than Asmin

As,min = (0.26 × fctm × bt × d) / (fyk) but not less than 0.0013 × bt × d

As,min = (0.26 × 2.9 × 1000 × 180) / (460) = 295.00 mm2/m

As,min = 0.0013 × 1000 × 180 = 234.00 mm2/m

Calculated Reinforcement is greater than the minimum reinforcement clauses. Calculated Reinforcement will be used

Provide T8 at 125 mm (=402mm2/m)

ρ = As / b / d [ Reinforcement Ratio ]

ρ = 402 / 1000 / 180 = 0.0022

ς = 310 × 460 × 350.92 / 500 / 402 [ Design Service Stress ]

ς = 248.96 kN/m2

300 / 248.96 = 1.2050 [ Modification Factor ]

20 × 1.2050 = 24.10 [ Modified span/depth Ratio ]

d = span / (span / depth ratio) [ Minimum Effective Depth ]

d = 4500 / 24.10 = 186.72 > 180

Deflections are too much. Increase the reinforcement to prevent this situation.

Provide T8 at 100 mm (=503mm2/m)

ρ = As / b / d [ Reinforcement Ratio ]

ρ = 503 / 1000 / 180 = 0.0028

ς = 310 × 460 × 350.92 / 500 / 503 [ Design Service Stress ]

ς = 198.97 kN / m2

300 / 198.97 = 1.5580 [ Modification Factor ]

20 × 1.55 = 31.16 [ Modified Span /Depth Ratio ]

d = span / (span/depth ratio) [ Minimum Effective Depth ]

d = 4500 / 31.16 = 144.42 < 180

With the increased reinforcement area, deflections are in the limits.

4

Bending Moment for Long Span

Msy = αsy × n × lx2

Msy = 0.046 × 11.73 × 4.52 = 10.93 kN m

Determining Lever – Arm for Long Span

zsy = 175.8780 – 8 = 167.8780 mm

Reinforcement Calculation

Asy = Msy / 0.87 / fyd / zsx

Asy = Msy / 0.87 / 460 / 167.8780

Asy = 162.69 mm2/m

As,min = (0.26 × 2.9 × 1000 × 180) / 460 = 295 mm2 /m

Minimum reinforcement is greater than the necessary reinforcement area. Minimum reinforcement will be used.

Provide T8 at 150 mm (=335mm2/m)

Check For Bar Spacing:

Main Reinforcement Spacing = 100mm < 400 √

Secondary Reinforcement Spacing = 150mm < 450 √

Section of short – span. T8 bars at 100 mm

Section of long – span. T8 bars at 150 mm

5

Maximum Reinforcement

According to Eurocode Clause 9.2.1

The cross-sectional area of tension or compression reinforcement should not exceed As,max outside lap locations.

As,max = ( 0.04 × Ac )

As,max = ( 0.04 × 1000 × 200 )

As,max = 8000 mm2 / m

Provided reinforcements are in within range with the minimum and maximum reinforcement clauses given in Eurocode 2.

6

Simply Supported Beam Design to Eurocode 2

Loading on the Beam

Total ultimate load on the slab is 11.70 kN / m2. The loading on the beam will be trapezoidal, with a maximum of value of 26.39 kN /m starting at 2.25 meters and ending at 4.50 meters as shown in the figure.

Support Reactions

Support Reactions = Total Load on the Beam / 2

Support Reactions = (26.39 × 2.25 × ½ × 2 + 26.39 × 2.25) / 2 = 59.38 kN

Maximum Moment

Maximum Moment will be in the midpoint of the beam. Taking moment about that point:

Moment = 59.38 × 6.75 / 2 – [ 2.25 × 26.39 × ½ × ( 2.25/2 + 2.25 /3 ) ] – 2.25/2 × 26.39 × 2.25/4

Design Moment = 128.04 kN m

The system is symmetrical so the maximum moment and the support reactions will be doubled.

Considered Loads Designing the Beam

Reaction at the Support = 59.39 × 2 = 118.76 kN

Design Moment = 128.04 × 2 = 256.08 kN m

Estimate The Depth Of The Beam To Design It Singly Reinforced

The necessary dimensions will be checked for if the beam can be designed singly reinforced.

0.167 > Moment / bw / d2 / fck [ Boundary Value For Single Reinforced Beam ]

Assume bw = 250 mm

0.167 > 256.08 × 106 / 250 / d2 / 30

d > 452.1617 mm

7

Properties of the Beam

Assume Concrete cover to reinforcement = 25 mm

Links ø = 10 mm, Reinforcement ø = 25, h = 500 mm

d = h – d’ – links – reinforcement / 2 [ Effective Depth Of Reinforcement ]

d = 500 – 25 – 10 – 25 / 2 = 452.50

452.50 > 452.1617 so 250 × 500 is enough for singly reinforced cross section. The dimensions are reasonable so design the beam to 250 × 500 mm.

Determining K0

K0 =Design Moment / fck / b / d2

K0 = 256.08 × 106 / 30 / 250 / 452.502

K0 = 0.166 < 0.167

Singly Reinforced Section

Determining Lever – Arm

z = d × [ 0.5 + ( 0.25 - K0 / 1.134 )0.5 ]

z = 452.50 × [ 0.5 + ( 0.25 - 0.166 / 1.134 )0.5 ]

z = 371.90 mm

Reinforcement Calculation

As1 = Design Moment / 0.87 / fyd / zsx

As1 = 256.08 × 106 / 0.87 / 460 / 371.90

As1 = 1720.57 mm2

π × 252/4 = 491 mm2 [ Area of single T25 bar ]

1720.57 / 491 = 3.50 [ T25 bars needed ]

4T25 bars selected for the longitudinal bars (Total area = 1964.00 mm2)

Check for Minimum And Maximum Reinforcement Limits

ρ min = 0.0013

As,min = 0.0013 × 250 × 452.50 = 147.06 mm2

As,min = (0.26 × fctm × bt × d) / (fyk)

As,min = (0.26 × 2.9 × 250 × 452.50) / (460) = 185.43 mm2

ρ max = 0.04

As,max = 0.04 × 300 × 452.50 = 5430.00 mm2

4T25 bars are in the limits of the conditions

8

Design for Shear

VEd = 118760 Newton [ Design Shear Force ]

Shear Resistance of Concrete Alone

The design value for the shear resistance without any shear reinforcement is

VRd,c = [CRdC × k × (100 × ρ1 × fck )1/3

+ k1 × ςcp ] ×b × d given in Eurocode clause 6.2 where k = 1 + (200 / d)0.5

k1 = 1 + (200 / d)0.5

k1 = 1.66

ρ1 = 1473 / 250 / 452.50 = 0.013

ρ < 0.02; ρ = 0.02

NED = 0; ςcp = 0 [ Normal Force On Section ]

CRd,C = 0.18 / γC = 0.12 and k1 = 0.15

Vmin = 0.035 × k3/2 × fck

0.5 = 0.035 × 1.66483/2 × 300.5

Vmin = 0.4118 N / mm2

VRd,c = [ 0.12 × 1.6648 × ( 100 × 0.02 × 30 )1/3 + 1.66 × 0 ] × 250 × 452.50

VRd,c = 88475 N < VED = 118760 N

Shear reinforcement required.

According to Eurocode clause 6.2.2 , the shear force VEd, calculated without reduction, should always satisfy the condition VEd < 0.5 × v × fcd × bw × d where v is a strength reduction factor for concrete cracked in shear. v = 0.6 × [ 1 – (fck / 250) ]

v = 0.6 × [ 1 – ( 30 / 250 ) ]

v = 0.48

VEd < 0.5 × 0.48 × 30 / 1.5 × 250 × 452.50

118760 < 543000 √ [ Satisfied ]

According to Eurocode 2 clause 6.2.3, element 3, for members with vertical shear reinforcement, the shear resistance VRd is the smaller value of VRd,s = Asw / s × z × fywd × cot ϴ and VRd,max = ( αcw × bw × z × v1 × fcd ) / (cot ϴ + tan ϴ)

Calculation Of VRd,max

Assuming to use vertical shear reinforcement: α = 90°; ϴ = 45°

VRd,max = ( 1 × 250 × 387.7473 × 0.48 × 30 / 1.5 ) / (1 + 1)

VRd,max = 465296.67

VRd,c < VEd < VRd,max

9

Shear Resistance with Shear Reinforcement Bars

VRd = Asw / s × z × fywd × cot ϴ

VRd = Asw / s × 0.9 × 387.7473 × 250 × 1

87243.1425 × Asw / s > VED = 118760

Asw / s > 1.3613

Minimum Shear Reinforcement Required

According to Eurocode ρ w,min = 0.08 × fck0.5 / fywk

ρ w,min = 0.08 × 300.5 / 250 / 1.15

ρ w,min = 0.0015

0.0015 = Asw / s / bw / sin α

Asw / s > 0.3125

Calculated Shear Reinforcement > Minimum Shear Reinforcement

Necessary Shear Reinforcement

Asw / s > 1.3613

Area of cross section = 78.50 (for ø = 10 mm)

2 × Area = 157.00 mm2

157 / s > 1.3613

s < 115.4412 mm

Chosen value for s = 115.00mm

Maximum Allowed Shear Reinforcement

According to Eurocode slmax = 0.75 × d × (1+cot α)

slmax = 339.37 mm √

T10 at 115 is in allowed range, provide ø = 10 mm links at 115 mm for whole length of the bar.

Four ø25 bars for the tension zone, having 27.5 millimetres space between them and ø10 bars at 115 millimetres. (Drawing not to scale.)

10

Column Design to Eurocode 2

Normal Force Acting On The Column

N1 = [ (6.75 / 2 + 4.5 / 2) × (4.5 / 2 + 4.5 / 2) ] × 11.73 = 296.92 kN

NEd = 296.92 × 3 = 890.76 kN [ 3 Stories in total ]

Slenderness Ratio Of Column

Height of Column = 4.5 m

Effective Height of the Column = 0.7 × lcolumn for the given restraint conditions according to Eurocode 2 Clause 5.8

lo = 0.7 × lcolumn = 0.7 × 4500 = 3150 mm

The beam has a width of 250 mms so assume the column to be 250 × 250 mm

i = ( I / A )0.5 [ Radius Of Gyration ]

i = ( b × h3 / 12 / b / h )0.5

i = ( 250 × 2503 / 12 / 250 / 250 )0.5

i = 72.17 mm

λ = lo / i [ Slenderness Ratio ]

λ = 3150 / 72.17 = 43.65

Classification of Column

Eurocode 2 places an upper limit on the slenderness ratio of a single member below which second order effects may be ignored. The limit is given by : λlim = 20 × A × B × C / (n)0.5

A = 0.7 B = 1.1 C = 2.2

n = 890.76 × 103 / 250 / 250 / ( 30 / 1.5 ) = 0.71

λlim = 20 × 0.7 × 1.1 × 2.2 / ( 0.71 )0.5

λlim = 40.15

λlim < λ

Column is Slender

The First Order Moment M0,Ed

ϴi ϴi = ϴ0 αh αm [ Imperfections ]

ϴ0 = 0.005, m = 4

αh = 2 / l0.5 = 0.94

αm = [ 0.5 × ( 1 + 1 / m) ]0.5 = 0.7905

ϴi = 0.005 × 0.9428 × 0.79 = 0.00372

ei = ϴi × lo / 2 = 5.85 × 10-3 [ Eccentricity ]

M0,ED = ei × NED = 5.85 × 10-3 × 890.76 = 5.21 kN m

11

Assumptions :

Cover To All Reinforcement = 20 mm

4 longitudinal bars with diameter ø = 25 mm are used at four corners of cross – section

Diameter of links ø’ = 8mm

d’ = 0.5 × ø + ø’ + c = 40.50 mm

The Nominal Stiffness of the Column

E × I = kc × Ecd × Ic + Ks × Es × Is

Ecd = Ecm / ɣce = 32 × 103 / 1.2 = 27.5 × 103

Is = As × ( 250 / 2 – 40.50 )2

As = 4ø25 = 1962.50 mm2

Is = 14.01 × 106 mm4

Ic = b × h3 / 12 – Ic = 269.99 × 106 mm4

Es = 200000 N / mm2

ρ = As / Ac

ρ = 1962.50 / 250 / 250 = 0.03140 > 0.002

Ks = 1

K1 = ( fck / 20 )0.5 = ( 30/ 20 )0.5 = 1.22

K2 = n × λ / 170 = 0.71 × 40.15 / 170 = 0.182 < 0.20

Kc = K1 × K2 ( 1 + φf ) = 1.22 × 0.182 / ( 1 + 0 ) = 0.22

E × I = 0.22 × 27.50 × 103 × 269.99 × 106 + 200000 × 14.01 × 106

E × I = 43.45 × 1011 Nmm2

NB = π2 × E × I / Io2 = π2 × 43.45 × 1011 / 31502 = 4317.45 × 103 N

MEd = M0,Ed / ( 1 – ( NEd / NB ) = 5.21 × 106 / [ 1 – ( 890.73 × 103 / 4317.45 / 103 ) ]

MEd = 6.56 × 106 kN m

50 > 6.56

Design Moment will be 50 kN m

Coefficients Needed for Determining the Reinforcement Area

MEd = 50 kN m

d’ / h = 40.50 / 250 = 0.1620

Round up to 0.20 and use the relevant chart to get the values needed for designing the column.

N / b / h / fcd = 890.76 × 103 / 250 / 250 / ( 30 / 1.5 ) = 0.7126

M / b / h2 / fcd = 50 × 106 / 250 / 2502 / ( 30 / 1.5 ) = 0.16

12

Reinforcement Required

As × fyd / b / h / fcd = 0.21 [ From the Relevant Chart ]

As,required = 656.25 mm2

As,provided = π × 252 / 4 × 4 = 1962.50 mm2 [Asprovided > Asrequired ]

Longitude Reinforcement Rules and Details for Columns According To Eurocode 2

The rules governing the minimum and maximum amounts of reinforcement in a load bearing column according to Eurocode 2 are

i )A minimum of four bars is required in a rectangular column (one bar in each corner). Bar diameter should not be less then 12mm.

4ø20 bars are provided at each corner.

ii) The minimum are of steel is given by; As,min = 0.10 × NED / fy

As,min = 0.10 × NED / fyd = 75 mm2

As,min = 0.002 × AC = 0.002 × 250 × 250 = 125 mm2

iii)The maximum area of steel allowed is given by; As,max = 0.08 × AC

As,max = 0.08 × 250 × 250 = 5000 mm2

As provided is in the acceptable range

Links Reinforcement Rules and Details for Columns According To Eurocode 2

i) Minimum size = ¼ × size of the compression bar but not less than 6mm

ii) Maximum spacing should not exceed the lesser of 20 × size of the smallest compression bar or the least lateral dimension of the column or 400mm

iii) Every longitudinal bar placed in corner should be held by transverse reinforcement.

Provide, R8 links at 250 mm centres.

Diameter of links is greater of 6 mm and ¼ ø = 6.25 mm. Spacing of links does not exceed the least of 20 × ø = 20 × 25 = 500mm and least dimension of column = 250.