Slab Design to BS 8110

Slab design is comparatively easy when compared with the design of other elements. The first stage of the design is finding the bending moment of the slab panels.

Depending on the boundary condition and the properties of the slabs, methods of finding bending moment is expressed in the BS 8110 Part 01 as follows.

One way spanning slabs

the following table can be referred.

There is two conditions that should be satisfied to use the above table.

1. The ratio of the characteristic imposed load to characteristic dead load does not exceed 1.25
2. Characteristic imposed load does not exceed 5 N/mm² excluding partitions

Simply supported slabs – Two way spanning

The bending moment can be found using bending moment coefficients given in the code. There are two equations that can be used to find the bending moments.

m_sx= α_sxnl_x²m_sy= α_synl_x²
Here m_{sx  and} m_sy are the bending moments at each direction, α_sx  and α_syy are the bending moment coefficients and l_x is the shorter span of the slab panel.

The following table can be referred to find the bending moment coefficients.

• 

Restrain slabs – Two way spanning

There is a table to find the bending moment coefficients depending on the boundary conditions and the ratio of the two spans which is a large span to a small span.

Using the above-explained method, binding moments can be found in each case. In addition, the span behind moments at supports also can be found from the above Table. Notation of the bending moments at span and supports are as follows.

Design Example

The ration of shorter span to a longer span = 1.3
Characteristic strength of concrete                 = 25 N/mm²
Cover to the reinforcement                             = 25 mm

Characteristic strength of reinforcement       = 460 N/mm²

Shorter Span                                                = 3m
Slab thickness                                             = 150mm
Live Load                                                       = 2.5 kN/mm²
Finishes and partitions                           = 1 kN/mm²

Consider a slab panel with two adjacent edgers discontinues

From table 3.14 (see above), found bending moment coefficients are as follows.

At Short Span
Negative moment at continues edge           =  0.069
Positive moment at mid-span                        =  0.051

At Longer Span
Negative moment at continues edge           =  0.045
Positive moment at mid-span                        =  0.035

Find Desing Loads

Dead Load              = 0.15x 24
                                = 3.6 kN/mm²

Total Dead Load = 3.6 + 1

= 4.6 kN/mm²

Live load                 = 2.5  kN/mm²

Design Load            = 1.4 Dead Load + 1.6 Live Load
                                = 1.4×4.6 + 1.6×2.5
                                = 10.44 kN/mm²

Find the bending moments

Positive moment at mid span (shorter span)

m_sx                                   = 0.051×10.44×3²
                                 = 4.8 kNm

Positive moment at mid span (longer span)

m_sy                                   = 0.035×10.44×3²
                                 = 3.3 kNm

Similarly, bending moments at supports also can be found from the above coefficients.

Design of shorter span
Effective depth            = 150-25-10/2
                                    = 120 mm

                     K            = M/bd²fcu
                                    = 4.8×10⁶/(1000×120²x25)
                                    = 0.028

                     K            < K’ = 0.156

Hence, section is singly reinforced.
                     Z             = d[0.5+{0.25-K/0.9}^1/2]
                                    = 120[0.5+{0.25-0.028/0.9}^1/2]
                                    = 116.174 mm

                     As           = M / (0.95fyZ)
                                    = 4.8×10⁶ / (0.95x460x116.174)
                                    = 197 mm²

Provide T10 bars at a spacing of 200mm.
Provided reinforcement 393 mm²/m