Designing a Flat Slab-Worked Example

Flat slabs are reinforced concrete slabs supported directly on columns, without beams. They are sometimes thickened at the columns to form column drops. These drops are very effective when the negative moments at supports are quite high or where punching stresses are very critical and punching reinforcement is not desired. It is also not unusual to increase the size of columns at the slab-column interface to produce a column head. Column heads can reduce the effective span of the slab, hence, the internal forces and deflection are reduced.

The benefits of a flat slab over other slab types are numerous: They are suitable for irregular column layout. The absence of drop beams means that their formwork is simplified, thus increasing the pace of construction. They also offer reduced storey heights, level soffits, flexible arrangement and distribution of services est.

Analysis and Design

Analysis of Flat Slabs

There are several methods available for analyzing a flat slab. This includes the use of simplified coefficients, equivalent frame method, yield line analysis, grillage analysis and the finite element method. The use of simplified coefficients is subject to certain conditions (please see Annex I of BS EN 1992-1- 1). The yield line and finite element methods are particularly more suited for computer programs. Hence, this post will focus only on the equivalent frame method of analysing flat slabs.

In the equivalent frame method, we divide the slab in the two orthogonal directions into frames consisting of column strips and middle strips as shown in figure 1. The vertical load on the frame is taken based on the width between the centrelines of panels. Each of this frame will be divided into a series of subframes, consisting of the slab at one level and the columns above and below it.

Figure 1: Division of Flat Slab into Sub-frames

In analyzing the subframes, the stiffness’s of edge columns are usually multiplied by 0.7. This is because the equivalent frame method overestimates the restraints provided by edge and corner columns. This is based on a research conducted by the concrete society on flat slabs.

Analysis of subframes have been presented in a previous post, therefore, in this post, only the bending moment diagrams might be presented.

Transfer Moments

The maximum moment, that can be transferred to an edge or corner column is given as:

{ M }_{ t,max }=0.17{ b }_{ e }{ d }^{ 2 }{ f }_{ ck }

Where:

  • Is the effective width of the flat slab (see figure 2)
  • effective depth of the top reinforcement in the column strip
Figure 2: Effective width of a flat slab

Where the moment at an end support exceeds the maximum transfer moment, moment redistribution must then be carried out.

Design of Flat Slabs

Haven determined the moments. The positive moment in the spans, as well as the negative moment at supports will be divided into a column and a middle strip, see figure 1. The moment is then apportioned to the column and middle strip using table 1.

Table 1: Distribution of Moment in Flat Slab

Once the moment has been determined, the design will proceed in a similar fashion for concrete slabs. The critical areas include flexural design, punching shear and deflection.

An additional requirement in flat slabs is to place at least 50% of the hogging reinforcement over the middle quatre of the column strip.

Worked Example

Figure 3 shows the layout of a shopping mall building. The floors are required to have level soffits, hence drops and column heads are not permitted. All columns are 400mm2 and the floor-floor distance =4m. The slab is expected to be constructed from C20/25 concrete and concrete steel bars having a yield strength of 410Mpa.

Figure 3: Flat Slab Worked Example

Designing the whole slab will make this post unnecessarily long. Therefore to illustrate the process only sub-frame 3-3 will be designed. You could complete the design in your free time to gain confidence.

Actions

Note that the slab thickness has been sized conservatively i.e. 275mm thick. Guidance on sizing a flat slab can be obtained from the concrete centre publication: Economic Concrete Frame Elements.

Permanent Actions
a.\quad self\quad weight\quad of\quad slab\quad =0.275\times 25=6.875kN/{ m }^{ 2 }
b.\quad finishes\quad \& \quad services\quad =1.5kN/{ m }^{ 2 }
Permanent\quad Actions\quad { g }_{ k }=6.875+1.5=8.375kN/{ m }^{ 2 }
Variable Actions
a.\quad Imposed\quad Load\quad (shopping\quad mall)=4kN/{ m }^{ 2 }
b.\quad movable\quad partitions\quad =1kN/{ m }^{ 2 }
{ Variable\quad Actions\quad q }_{ k }=4+1=5kN/{ m }^{ 2 }
Design Values of Actions

By inspection the permanent actions are less than 4.5 times the variable actions, therefore equation 6.10b of BS-EN 1991-1-1 can be used.

{ n }_{ s,max }=1.35\xi { g }_{ k }+1.5{ q }_{ k }
=(1.35\times 0.925\times 8.375)+(1.5\times 5)=17.96kN/{ m }^{ 2 }
{ n }_{ s,min }=1.35\xi { g }_{ k }
=(1.35\times 0.925\times 8.375)=10.46kN/{ m }^{ 2 }

Analysis of Sub-Frame 3

Analysis of Sub-frames have been presented in a previous post, therefore only bending moment diagram will be shown here. If you would like to review it click: How to Analyse Elements in Braced Frames

Figure 4- Sub Frame 3
Load transferred to subframe 3
tributary\quad width\quad =\frac { 7.5 }{ 2 } +\frac { 7.5 }{ 2 } =7.5m
{ w }_{ max }={ n }_{ s,max }\cdot width=17.96\times 7.5=134.7kN/m
{ w }_{ min }={ n }_{ s,max }\cdot width=10.46\times 7.5=78.45kN/m

Analysis of this subframe will be carried out for three loadcases:

  • All spans loaded with the maximum actions, wmax
  • Alternate spans loaded with the maximum actions while the other spans are loaded with the minimum actions.

The analysis will then be carried out by moment distribution. Limited moment redistribution of the support moment into the spans can be carried out if need be.

Figure 5: Bending Moment Envelope for Frame 3

Flexural Design

The moment is apportioned between the column and middle strips within the specified limits in table 1. For this frame, the negative moment at supports will be shared in the proportion of 70% to the column strip and 30% to the middle strip while the positive moment in the spans will be shared in the proportion of 50-50%.

assuming 16mm bars in both direction and a concrete cover of 25mm

{ d }_{ y }=h-(c+\phi /2)=275-(25+16/2)=242mm
{ d }_{ z }=h-(c+\phi /2+\Phi )=275-(25+16/2+16)=226mm
{ d }_{ min }=\frac { { d }_{ y }+{ d }_{ z } }{ 2 } =\frac { 242+226 }{ 2 } =234mm

End Supports

The width of the end support, be=cx +y =(600+200)+400=1200mm

{ M }_{ t,max }=0.17{ b }_{ e }{ d }^{ 2 }{ f }_{ ck }
=0.17\times 1200\times { 234 }^{ 2 }\times 20=223.4kN.m
{ M }_{ Ed }=159kN.m(<223.4kN.m)
k=\frac { { M }_{ Ed } }{ b{ d }^{ 2 }{ f }_{ ck } } =\frac { 159\times { 10 }^{ 6 } }{ 1200\times { 234 }^{ 2 }\times 20 } =0.121
z=d\left[ 0.5+\sqrt { 0.25-0.882k } \right] \le 0.95d
=0.88d=0.88\times 234=205.92mm
{ A }_{ s }=\frac { { M }_{ Ed } }{ 0.87{ f }_{ yk }z } =\frac { 159\times { 10 }^{ 6 } }{ 0.87\times 410\times 205.92 } =2164.7{ mm }^{ 2 }2
Try\quad 12Y16\quad @\quad 100mm\quad centres\quad ({ A }_{ s,prov }=2412{ mm }^{ 2 })

End Spans

width\quad of\quad column\quad strip\quad =\frac { 6000 }{ 4 } \times 2=3000m
width\quad of\quad middle\quad strip\quad =7500-3000=4500mm
M=313kN.m
Column Strip
{ M }_{ Ed }=0.5\times \frac { 313 }{ 3.0 } =52.17kN.m/m
k=\frac { { M }_{ Ed } }{ b{ d }^{ 2 }{ f }_{ ck } } =\frac { 52.17\times { 10 }^{ 6 } }{ { 10 }^{ 3 }\times { 234 }^{ 2 }\times 20 } =0.048
z=d\left[ 0.5+\sqrt { 0.25-0.882k } \right] \le 0.95d
=0.95d=0.95\times 234=222.3mm
{ A }_{ s }=\frac { { M }_{ Ed } }{ 0.87{ f }_{ yk }z } =\frac { 52.17\times { 10 }^{ 6 } }{ 0.87\times 410\times 222\times .3 } =657.92{ mm }^{ 2 }/m
Try\quad 20Y12-150mm\quad centres\quad ({ A }_{ s,prov }=753{ mm }^{ 2 }/m)
Middle Strip
{ M }_{ Ed }=0.5\times \frac { 313 }{ 4.5 } =34.78kN.m/m
k=\frac { { M }_{ Ed } }{ b{ d }^{ 2 }{ f }_{ ck } } =\frac { 34.78\times { 10 }^{ 6 } }{ { 10 }^{ 3 }\times { 234 }^{ 2 }\times 20 } =0.032
z=d\left[ 0.5+\sqrt { 0.25-0.882k } \right] \le 0.95d
=0.95d=0.95\times 234=222.3mm
{ A }_{ s }=\frac { { M }_{ Ed } }{ 0.87{ f }_{ yk }z } =\frac { 34.78\times { 10 }^{ 6 } }{ 0.87\times 410\times 222.3 } =438.62{ mm }^{ 2 }/m
Try\quad 23Y12-200mm\quad centres\quad ({ A }_{ s,prov }=565{ mm }^{ 2 }/m)

Interior Supports

width\quad of\quad column\quad strip\quad =\frac { 7500 }{ 4 } \times 2=3750m
width\quad of\quad middle\quad strip\quad =7500-3750=3750mm
M=602kN.m
Column Strip
{ M }_{ Ed }=0.7\times \frac { 602 }{ 3.75 } =112.37kN.m/m
k=\frac { { M }_{ Ed } }{ b{ d }^{ 2 }{ f }_{ ck } } =\frac { 112.37\times { 10 }^{ 6 } }{ { 10 }^{ 3 }\times { 234 }^{ 2 }\times 20 } =0.103
z=d\left[ 0.5+\sqrt { 0.25-0.882k } \right] \le 0.95d
=0.90d=0.90\times 234=210.6mm
{ A }_{ s }=\frac { { M }_{ Ed } }{ 0.87{ f }_{ yk }z } =\frac { 112.37\times { 10 }^{ 6 } }{ 0.87\times 410\times 222.3 } =1495.6{ mm }^{ 2 }/m
Try\quad 30Y16-125mm\quad centres\quad ({ A }_{ s,prov }=1608{ mm }^{ 2 }/m)
Middle Strip
{ M }_{ Ed }=0.3\times \frac { 602 }{ 3.75 } =48.16kN.m/m
k=\frac { { M }_{ Ed } }{ b{ d }^{ 2 }{ f }_{ ck } } =\frac { 48.16\times { 10 }^{ 6 } }{ { 10 }^{ 3 }\times { 234 }^{ 2 }\times 20 } =0.044
z=d\left[ 0.5+\sqrt { 0.25-0.882k } \right] \le 0.95d
=0.95d=0.95\times 234=222.3mm
{ A }_{ s }=\frac { { M }_{ Ed } }{ 0.87{ f }_{ yk }z } =\frac { 48.16\times { 10 }^{ 6 } }{ 0.87\times 410\times 222.3 } =607.4{ mm }^{ 2 }/m
Try\quad 25Y12-150mm\quad centres\quad ({ A }_{ s,prov }=753{ mm }^{ 2 }/m)

Interior Span

width\quad of\quad column\quad strip\quad =\frac { 7500 }{ 4 } \times 2=3750m
width\quad of\quad middle\quad strip\quad =7500-3750=3750mm
M=398kN.m
Column Strip
{ M }_{ Ed }=0.5\times \frac { 398 }{ 3.75 } =53.07kN.m/m
k=\frac { { M }_{ Ed } }{ b{ d }^{ 2 }{ f }_{ ck } } =\frac { 53.07\times { 10 }^{ 6 } }{ { 10 }^{ 3 }\times { 234 }^{ 2 }\times 20 } =0.048
z=d\left[ 0.5+\sqrt { 0.25-0.882k } \right] \le 0.95d
=0.95d=0.95\times 234=222.3mm
{ A }_{ s }=\frac { { M }_{ Ed } }{ 0.87{ f }_{ yk }z } =\frac { 53.07\times { 10 }^{ 6 } }{ 0.87\times 410\times 222.3 } =669.28{ mm }^{ 2 }/m
Try\quad 25Y12-150mm\quad centres\quad ({ A }_{ s,prov }=753{ mm }^{ 2 }/m)
Middle Strip
{ M }_{ Ed }=0.5\times \frac { 398 }{ 3.75 } =53.07kN.m/m
k=\frac { { M }_{ Ed } }{ b{ d }^{ 2 }{ f }_{ ck } } =\frac { 53.07\times { 10 }^{ 6 } }{ { 10 }^{ 3 }\times { 234 }^{ 2 }\times 20 } =0.048
z=d\left[ 0.5+\sqrt { 0.25-0.882k } \right] \le 0.95d
=0.95d=0.95\times 234=222.3mm
{ A }_{ s }=\frac { { M }_{ Ed } }{ 0.87{ f }_{ yk }z } =\frac { 53.07\times { 10 }^{ 6 } }{ 0.87\times 410\times 222.3 } =669.28{ mm }^{ 2 }/m
Try\quad 25Y12-150mm\quad centres\quad ({ A }_{ s,prov }=753{ mm }^{ 2 }/m)

Deflection Verification

The deflection requirement is more critical in the interior span, therefore we can carry out the check this span alone.

Limiting L/d ratio
{ \left[ \frac { l }{ d } \right] }_{ limit }=N\cdot k\cdot F1\cdot F2\cdot F3
\rho =\frac { { A }_{ s,req } }{ bd } =\frac { 669.28 }{ { 10 }^{ 3 }\times 234 } =0.0029
{ \rho }_{ o }={ 10 }^{ -3 }\sqrt { { f }_{ ck } } ={ 10 }^{ -3 }\times \sqrt { 25 } =0.00447
N=\left[ 11+\frac { 1.5\sqrt { { f }_{ ck } } { \rho }_{ o } }{ \rho } +3.2\sqrt { { f }_{ ck } } \left( \frac { { \rho }_{ o } }{ \rho } -1 \right) ^{ \frac { 3 }{ 2 } } \right]
=\left[ 11+\frac { 1.5\times \sqrt { 20 } \times 4.47 }{ 2.9 } +3.2\sqrt { 20 } \left( \frac { 4.47 }{ 2.9 } -1 \right) ^{ \frac { 3 }{ 2 } } \right]
=27.04
F1=1.0;\quad k=1.5;\quad F2=1.0
F3=\frac { 310 }{ { \sigma }_{ S } } \le 1.5
{ \sigma }_{ S }=\frac { { f }_{ yk } }{ { \gamma }_{ s } } \left[ \frac { { g }_{ k }+{ \psi q }_{ k } }{ { n }_{ s,max } } \right] \left( \frac { { A }_{ s,req } }{ { A }_{ s,pro } } \right) \cdot \frac { 1 }{ \delta }
=\frac { 410 }{ 1.15 } \left[ \frac { 8.375+(0.6\times 5) }{ 17.96 } \right] \times \frac { 669.28 }{ 753 } \times \frac { 1 }{ 0.8 } =236.11
F3=\frac { 310 }{ 236.11 } =1.31
{ \left[ \frac { l }{ d } \right] }_{ limit }=27.04\times 1.5\times 1\times 1\times 1.31\quad =53.2
Actual Deflection
{ \left[ \frac { l }{ d } \right] }_{ actual }=\frac { 7500 }{ 234 } =32.05

Since actual span-effective depth ratio is less than the limiting value, the deflection requirement is okay.

This concludes the design of subframe 3. The frame in the other directions must also be analyzed and designed accordingly.

Punching verification must be carried out and if shear reinforcement is required, it should be provided. This was also the subject of a previous post, see: Designing for Punching Shear in Concrete Slab Other detailing checks must also be carried out before detailing the slab.

THANK YOU!!!

Omotoriogun Victor
About Omotoriogun Victor 66 Articles
A dedicated, passion-driven and highly skilled engineer with extensive knowledge in research, construction and structural design of civil engineering structures to several codes of practices

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