This article presents a worked example on the design of flat slab to EN 1992-1-1-2004. The plan of the flat slab to be designed is shown below. It is a 250mm thick office slab of 18 x 18m, which is meant to bear an impose load of 4.0KN/m2 and finishes of 1.0KN/m2. The slab is supported on 300 x 300mm columns, and the concrete strength is for slabs and columns is C25/30.

Steps in Designing the Flat Slab

  Below are the steps taken to design the flat slab Steps taken in Designing the Flat Slab

• Estimate the slab load and analyse the slab for internal forces using equivalent frame analyses
• Redistribute 20% of the support moment gotten from elastic analysis
• Check maximum transferable moment to edge column
• Divide Slab into middle strip and column strip
• Distribute design moment into column strip and middle strip
• Design the strips
• Design for punching Shear

  Load Estimation and Analysis

In order not to make this article excessively long, we shall only reproduce the bending moment diagram after analysis in this article as shown below. The detailed analysis of the flat slab using equivalent frame method has already been published here (check it out!).

  Redistribution of moment

The bending moments diagram above is from linear elastic analysis. The moments in the flat slab have to be redistributed to get the design moment. This 20% redistribution of support moment from elastic analysis is necessary because maximum ultimate load is used throughout the continuous span of the slab rather than pattern loading. For this case of maximum ultimate load, redistributing the 20% of support moments to adjacent span is in accordance with UK national annex to BS EN 1992-1-1-2004 and also BS 8110-1-1997.

The redistribution using basic static principles is performed below and the redistributed bending moment diagram is plotted  

 

Redistribution of support moment of span BE

Support moment = 336.66

Reduce support moment by 20% = 0.8 x 336.66 = 269.328

Calculate the shear force at B using the formular below

V_BE = wl/2 + (M_BE – M_EB)/L

V_BE= 94.74 x 6/2 + (79.506 – 269.328)/6

V_BE= 252.583KN

Considering that summation of upward forces equals zero

V_BE + V_EB  = 94.74 x 6

V_EB = 568.44 – 252.583

V_EB = 315.857KN

Calculate the Span moment after redistribution

Position of maximum moment (x) = 252.583/94.74 = 2.67m

Span moment = 0.5 x 252.583 x 2.67 = 257.195KNm  

 

Redistribution of support moment of span EH

Support moment 1= 315.92KNm

Support moment 2= 315.06KNm

Reduce support moment 1 by 20% = 0.8 x 315.92 = 252.736

Reduce support moment 2 by 20% = 0.8 x 315.06 = 252.048

Calculate the shear force at E using the formular below

V_EH = wl/2 + (M_EH – M_EB)/L

V_EH = 94.74 x 6/2 + (252.048 – 252.736)/6

V_EH = 284.1KN

Considering that summation of upward forces equals zero

V_EH + V_HE= 94.74 x 6

V_HE = 568.44 – 284.1

V_HE = 284.33KN

Calculate the Span moment after redistribution

Position of maximum moment (x) = 284.1/94.74 = 3m

Span moment = 0.5 x 284.1 x 3 = 173.93KNm  

Redistribution of support moment of span HK

Support moment = 336.04KNm

Reduce support moment by 20% = 0.8 x 336.66 = 270.43

Calculate Shear force at H using the formular below

V_HK = wl/2 + (M_HK – M_KH)/L

V_HK = 94.74 x 6/2 + (270.43 – 81.061)/6 = 252.658KN  

Considering that summation of upward forces equals zero

V_HK + V_KH = 94.74 x 6

V_KH = 568.44 – 252.658

V_KH = 315.782KN

Calculate the Span moment after redistribution

Position of maximum moment (x) = 252.658/94.74 = 2.67m

Span moment = 0.5 x 252.658 x 2.67 = 255.8KNm  

The bending moment diagram for the redistributed moment is shown below:

  Check the moment transferable to end/edge columns

We shall check whether the moment transferred to edge/end column is beyond the acceptable limit which is given by 0.17b_ed²f_ck

b_e for edge columns = c_z+ y (fig 9.9 of BS EN 1992-1-1-2004)

b_e = 300 + 150 = 450mm

f_ck = 25N/mm²

Effective depth = (d_x +d_y)/2

d_x = h-c-ϕ/2 = 250 -25-12/2 = 219mm

d_y = h-c- ϕ- ϕ/2 = 250 -25-12-12/2 = 207mm

d = (207 + 219)/2 = 213mm

M = 0.17b_ed²f_ck

M = 0.17 x 450 x 213^2 x 25

M =86KNm

The moment transferred (81KNm) to edge column is lesser than the allowed moment (86KNm). Check is OK.  

Division of Flat Slab into column and middle strip

Width of column strip = 2(ly) = 2(1.5) = 3m

Width of middle strip =  ly/2 = 3m

  Check figure 3.12 of BS 8110-1-1997 or figure I.1 of EN 1992-1-1-2004 for further information on division of flat slabs into column and middle strips  

Lateral distribution of slab moment between column strip and middle strip

The design moment shall be distributed between column strips and middle strips according to the table below. This table corresponds to the average percentage in table I.1 of BS EN 1992-1-1-2004.

We shall pick the largest positive moment and negative moment in the bending diagram above and use it as the moment throughout the span and internal supports respectively. As for the edge columns, we can either use the larger negative edge moment or use the ultimate moment of resistance 0.17b_ed²f_ck

Using the larger span moment for all spans and larger support moment for all internal support is more practical, saves time, allows for easier detailing as well as reinforcement placement on site.

The positive moment and negative moment are therefore 257.195KNm and 270.43KNm respectively

Span Moment for column strip

span moment = Percentage distribution x moment/width of strip

span moment = 0.6 x 257.195/3 = 51.44KNm/m

Span Moment for middle strip

span moment = Percentage distribution x moment/width of strip

span moment = 0.4 x 257.195/3 = 34.29KNm/m  

Support Moment for column strip

support moment= Percentage distribution x moment/width of strip

support moment = 0.7 x 270.43/3 = 63.1KNm/m

Support Moment for middle strip

support moment = Percentage distribution x moment/width of strip

support moment = 0.3 x 270.43/3 = 27.04KNm/m  

 

Design of Strips

Design of Middle Strip

Flexural design of Middle Strip Span

1. Effective depth is 213 as calculated previously

2. Check whether section is to be designed as singly or doubly reinforced beam

$ K\,=\,\,\frac{M}{bd^2f_{ck}} $

$ K\,=\,\,\frac{34.29×10^6}{1000×213^2×25} $

 

= 0.03

Since K (0.03) < K’ (0.168); design as singly reinforced.

 

3) Calculate the lever arm (Z)

$ Z\,\,=\,\,d\left( 0.5+\sqrt{\text{0.25}-\,\,\frac{K}{1.134}} \right) $

 

$ Z\,\,=\,\,213\left( 0.5+\sqrt{\text{0.25}-\,\,\frac{0.03}{1.134}} \right) $

Z = 207.16mm

Since 207.16 > 0.95d (202.35): use Z = 202.35mm

 

4. Calculate the area of steel

$ A_{st\,\,=\,\,\frac{M_{Ed}}{0.87f_{yk}Z}} $

 

$ A_{st\,\,=\,\,\frac{34.29×10^6}{0.87x500x202.35}} $

 

Ast = 389.5mm²

Provide Y12@250mmc/c (449.86mm²/m)

5) Check Whether area of tensile steel provided satisfies minimum area requirement

               A_smin = 0.26 (f_ctm/f_yk )b_t d ≥ 0.0013bh

fctm = 0.3 x f_ck ^2/3 fctm = 0.3 x 25^2/3 = 2.56

              A_smin  = 0.26 (2.56/500) 1000 x 213

      A_smin      =   284mm²

         Since Ast > Asmin, minimum area requirement is satisfied

 

Deflection Check of Middle Strip Span

1. Calculate the actual span-effective depth ratio

Span/depth ratio = 6000/213   = 28.16

(it should be noted that in this article the clear span is taken as the effective span)

2. Calculate the limiting Span-effective depth ratio

ρ = As_required/b x d

ρ = 389.5/1000 x 213

= 0.0018

ρ^o = 10^-3√fck

ρ^o = 10^-3√25

= 0.005

K = 1.3 (for end spans)

Since ρ < ρ^o , then we will use the below expression:

l/d = K[11 + 1.5√f_ck ρ0/ρ + 3.2√f_ck (ρ^o/ρ – 1)^3/2]

l/d = K[11 + 1.5√25 x 0.005/0.0018 + 3.2√25 (0.005/0.0018 – 1)^3/2]

= 88.4                   

Since actual span-effective depth ratio is less than the limiting span-effective depth ratio, the slab passes deflection check.

 

Flexural design of Middle Strip Support

1. Effective depth is 213 as calculated previously

2. Check whether section is to be designed as singly or doubly reinforced beam

$ K\,=\,\,\frac{M}{bd^2f_{ck}} $

$ K\,=\,\,\frac{27.04×10^6}{1000×213^2×25} $

 

= 0.024

Since K (0.024) < K’ (0.168); design as singly reinforced.

 

3) Calculate the lever arm (Z)

$ Z\,\,=\,\,d\left( 0.5+\sqrt{\text{0.25}-\,\,\frac{K}{1.134}} \right) $

 

$ Z\,\,=\,\,213\left( 0.5+\sqrt{\text{0.25}-\,\,\frac{0.024}{1.134}} \right) $

Z = 208.42mm

Since 208.42 > 0.95d (202.35): use Z = 202.35mm

 

4. Calculate the area of steel

$ A_{st\,\,=\,\,\frac{M_{Ed}}{0.87f_{yk}Z}} $

 

$ A_{st\,\,=\,\,\frac{27.04×10^6}{0.87x500x202.35}} $

 

Ast = 307.19mm²

Provide Y12@250mmc/c (449.86mm²/m)

5) Check Whether area of tensile steel provided satisfies minimum area requirement

               A_smin = 0.26 (f_ctm/f_yk )b_t d ≥ 0.0013bh

fctm = 0.3 x f_ck ^2/3 fctm = 0.3 x 25^2/3 = 2.56

              A_smin  = 0.26 (2.56/500) 1000 x 213

      A_smin      =   284mm²

         Since Ast > Asmin, minimum area requirement is satisfied

Design of Column Strip

Flexural design of column strip span  

1. Effective depth is 213 as calculated previously

2. Check whether section is to be designed as singly or doubly reinforced beam

$ K\,=\,\,\frac{M}{bd^2f_{ck}} $

$ K\,=\,\,\frac{51.44×10^6}{1000×213^2×25} $

 

= 0.04

Since K (0.04) < K’ (0.168); design as singly reinforced.

 

3) Calculate the lever arm (Z)

$ Z\,\,=\,\,d\left( 0.5+\sqrt{\text{0.25}-\,\,\frac{K}{1.134}} \right) $

 

$ Z\,\,=\,\,213\left( 0.5+\sqrt{\text{0.25}-\,\,\frac{0.04}{1.134}} \right) $

Z = 204mm

Since 204 > 0.95d (202.35): use Z = 202.35mm

 

4. Calculate the area of steel

$ A_{st\,\,=\,\,\frac{M_{Ed}}{0.87f_{yk}Z}} $

 

$ A_{st\,\,=\,\,\frac{51.44×10^6}{0.87x500x202.35}} $

 

Ast = 584.4mm²

Provide Y12@150mmc/c (674.78mm²/m)

5) Check whether area of tensile steel provided satisfies minimum area requirement

               A_smin = 0.26 (f_ctm/f_yk )b_t d ≥ 0.0013bh

fctm = 0.3 x f_ck ^2/3 fctm = 0.3 x 25^2/3 = 2.56

              A_smin  = 0.26 (2.56/500) 1000 x 213

      A_smin      =   284mm²

    Since Ast > Asmin, minimum area requirement is satisfied

 

Deflection Check of Column Strip Span

1. Calculate the actual span-effective depth ratio

Span/depth ratio = 6000/213   = 28.16

(it should be noted that in this article the clear span is taken as the effective span)

2. Calculate the limiting Span-effective depth ratio

ρ = As_required/b x d

ρ = 584.4/1000 x 213

= 0.003

ρ^o = 10^-3√fck

ρ^o = 10^-3√25

= 0.005

K = 1.3 (for end spans)

Since ρ < ρ^o , then we will use the below expression:

l/d = K[11 + 1.5√f_ck ρ0/ρ + 3.2√f_ck (ρ^o/ρ – 1)^3/2]

l/d = K[11 + 1.5√25 x 0.005/0.003 + 3.2√25 (0.005/0.003 – 1)^3/2]

= 47.58                 

Since actual span-effective depth ratio is less than the limiting span-effective depth ratio, the slab passes deflection check.

 

Flexural design of Column Strip Support

1. Effective depth is 213 as calculated previously

2. Check whether section is to be designed as singly or doubly reinforced beam

$ K\,=\,\,\frac{M}{bd^2f_{ck}} $

$ K\,=\,\,\frac{63.1×10^6}{1000×213^2×25} $

 

= 0.06

Since K (0.06) < K’ (0.168); design as singly reinforced.

 

3) Calculate the lever arm (Z)

$ Z\,\,=\,\,d\left( 0.5+\sqrt{\text{0.25}-\,\,\frac{K}{1.134}} \right) $

 

$ Z\,\,=\,\,213\left( 0.5+\sqrt{\text{0.25}-\,\,\frac{0.06}{1.134}} \right) $

Z = 202mm

Since 202 = 0.95d (202.35): use Z = 202mm

 

4. Calculate the area of steel

$ A_{st\,\,=\,\,\frac{M_{Ed}}{0.87f_{yk}Z}} $

 

$ A_{st\,\,=\,\,\frac{63.1×10^6}{0.87x500x202}} $

 

Ast = 718.18mm²

Provide Y16@250mmc/c (799.7mm²/m)

5) Check Whether area of tensile steel provided satisfies minimum area requirement

               A_smin = 0.26 (f_ctm/f_yk )b_t d ≥ 0.0013bh

fctm = 0.3 x f_ck ^2/3 fctm = 0.3 x 25^2/3 = 2.56

              A_smin  = 0.26 (2.56/500) 1000 x 213

      A_smin      =   284mm²

         Since Ast > Asmin, minimum area requirement is satisfied

 

Design of Edge Columns (column Strip)

We shall design the edge column for moment of 0.17b_ed²f_ck, which has been evaluated to be 86KNm

1. Effective depth is 213 as calculated previously

2. Check whether section is to be designed as singly or doubly reinforced beam

$ K\,=\,\,\frac{M}{bd^2f_{ck}} $

$ K\,=\,\,\frac{86×10^6}{1000×213^2×25} $

= 0.168

Since K (0.06) is not greater than K’ (0.168); design as singly reinforced.

 

3) Calculate the lever arm (Z)

$ Z\,\,=\,\,d\left( 0.5+\sqrt{\text{0.25}-\,\,\frac{K}{1.134}} \right) $

 

$ Z\,\,=\,\,213\left( 0.5+\sqrt{\text{0.25}-\,\,\frac{0.168}{1.134}} \right) $

Z = 202mm

Since 202 = 0.95d (202.35): use Z = 202mm

 

4. Calculate the area of steel

$ A_{st\,\,=\,\,\frac{M_{Ed}}{0.87f_{yk}Z}} $

 

$ A_{st\,\,=\,\,\frac{63.1×10^6}{0.87x500x202}} $

 

Ast = 718.18mm²

Provide Y16@250mmc/c (799.7mm²/m)

 

Design of Free Edge of Middle Strip

We shall provide minimum reinforcement at the free end of the middle strip

A_smin = 0.26 (f_ctm/f_yk )b_t d ≥ 0.0013bh

fctm = 0.3 x f_ck ^2/3 fctm = 0.3 x 25^2/3 = 2.56

              A_smin  = 0.26 (2.56/500) 1000 x 213

      A_smin      =   284mm²

Provide Y12@250mmc/c (449mm²/m)

Design of Edge Frames

Since the equivalent end/edge frame is not analyzed separately as only the internal equivalent frame is analyzed, the reinforcement for internal column strip shall also be provided for the edge frame  

Design for Punching Shear 

A separate article has been written for the design of the slab for punching shear, click here to study the article !

Detailing  

The other orthogonal axis (y-axis) is also detailed just as shown.