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Design Guide: Concrete Slabs to AS3600's banner
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Design Guide: Concrete Slabs to AS3600

A one-way and two-way slab is designed in the same way. The difference is how the design actions are determined.
Different slab behaviour


One-way vs Two-way Slabs

Although not specified in the standards, it is typical to define a two-way slab as:

Ly/Lx≤2L_y/L_x \leq 2
Where:
  1. 
    
    is the length of the shorter slab span
  1. 
    
    is the length of the longer slab span
Otherwise, the slab is considered a one-way slab.


Simplified Slab Analysis

For continuous slabs over two or more spans, design actions,

can be determined from a finite element model of your slab.
Otherwise, Clause 6.10 of the code allows you to calculate

by hand using simplified elastic analysis
A summary of the simplified elastic analysis permitted by the code, is provided below for one-way and two-way slabs.

One-way Slab Analysis

As per Clause 6.10.2, the design moments of a one-way slab using simplified elastic analysis is given by:

M∗=coefficient×FdLn2M^*=\text{coefficient} \times F_d L_n^2
Where:
  1. coefficient depends on the location of the
    
    under consideration, see figures below
  2. 
    
    is the factored ULS design load per unit length
    
    
  3. 
    
    is the clear span length
Simplified analysis of one-way slabs - two spans

Simplified analysis of one-way slabs - more than two spans

As per Clause 6.10.2, the simplified method is valid provided the conditions in the below toggles are met.

One-way slab requirements for simplified analysis

For one-way slabs, as per Clause 6.10.2, the simplified method is valid provided:
  1. the ratio of the longer to the shorter length of any two adjacent spans does not exceed
    
    
  1. loads are uniformly distributed
  2. 
    
    , where Q = live load & G = dead load
  1. slab has a constant cross-section
  1. reinforcement is arranged in accordance with Clause 9.1.3.2, which describes how much and the extent of top & bottom reinforcement with the slab span.
  2. 
    
    at supports are only caused by loads applied to the beam or slab, not from the columns


Two-way slab analysis

As per Clause 6.10.3, the positive design moments (i.e. sagging moments) at midspan of a two-way slab supported on four sides, using simplified elastic analysis is given by:

+Mx∗=βx×FdLx2+My∗=βy×FdLx2+M^*_x=\beta_x \times F_d L_x^2 \\+M^*_y=\beta_y \times F_d L_x^2
Where:
  1. 
    
    are moment coefficients, taken from Table 6.10.3.2(A)
  2. 
    
    is the factored ULS design load per unit area
    
    
  3. 
    
    is the shorter span length
AS3600-2018, Table 4.10.3.2(A)

👉 AS2600-2018 commentary provide equations for

and

that derives Table 4.10.3.2(A), and these equations are used in this calculator.

The code talks about edges of a two-way slabs supported on four sides by beams or walls as either continuous or discontinuous:
  1. continuous edge means the slab is able to transfer load to the adjacent slab through that edge
  2. discontinuous edge means the slab is unable to transfer load to the adjacent slab through that edge, either because it is not detailed as continuous or it is the outside edge of a slab
Edge conditions of a two-way slab

As per Clause 6.10.3, the negative design moments (i.e. hogging moments) of a two-way slab using simplified elastic analysis is given by:
  1. at a continuous edge:

−Mx∗=1.33×(+Mx∗)−My∗=1.33×(+My∗)-M^*_x=1.33 \times (+M^*_x) \\-M^*_y=1.33 \times (+M^*_y)
  1. at a discontinuous edge:

−Mx∗=0.5×(+Mx∗)−My∗=0.5×(+My∗)-M^*_x=0.5 \times (+M^*_x) \\-M^*_y=0.5 \times (+M^*_y)
The code says if the

on one side of the support is different from that on the other side, the unbalanced moment can be redistributed or design the slab on both sides with the larger support moment.
As per Clause 6.10.3, the simplified method is valid provided the conditions in the below toggle are met.

Two-way slab requirements for simplified analysis

For two-way slabs supported on all four sides by walls or beams, as per Clause 6.10.3, the simplified method is valid provided:
  1. loads are uniformly distributed
  2. reinforcement is arranged in accordance with Clause 9.1.3.3, which describes how much and the extent of top & bottom reinforcement with the slab span.
  3. 
    
    at supports are only caused by loads applied to the beam or slab, not from the columns
  4. openings in the slab are not big enough to adversely affect strength or stiffness
  5. if using reinforcement with ductility class L, slabs are supported by walls (not beams)


Flat slab analysis

Flat slab is defined as a two-way slab supported by columns. Clause 6.10.4 presents a simplified elastic method where the slab is analysed one panel (unit strip) at a time. A total static moment in each direction in each panel is calculated as:

Mo=FdLtLo28M_o=\dfrac{F_dL_tL_o^2}{8}
Where:
  1. 
    
    is the uniformly distributed design load per unit area (
    
    
  2. 
    
    is the width of the design strip, which for an interior design strip is taken as
    
    in the longer span and
    
    in the shorter span. For edge design strips, it is taken as
    
    in the longer span and
    
    in the shorter span.
  3. 
    
    is the effective span, which is taken as
    
    in the longer span and
    
    in the shorter span, for simplicity
The total static moment is then shared between the supports (negative moments) and the mid-span (positive moments) as shown below.
AS3600-2018 Commentary, Figure C6.10.4.2

At any critical section the design moment is determined by:

M∗=factor×MoM^*=\text{factor}\times M_o
Where the factor is given in Tables 6.10.4.3(A) & (B).
AS3600-2018, Table 6.10.4.2(A) & (B)

The positive and negative design moments are then distributed to the column and middle strips using the column strip moment factor in Table 6.9.5.3.
AS3600-2018, Table 6.9.5.3

Flat slab analysis for the longer span

Flat slab analysis for the shorter span

Let's break this down, for example: you could calculate an

in the longer span of the interior design strip. The sagging moment in the middle of the span is calculated as

using the factor in Table 6.10.4.3(B). This sagging moment then needs to be distributed transversely to the column and middle strips. Using Table 6.9.5.3, we calculate

is to be distributed to the column strip, and therefore leaving

to each of the half middle strips.
Example of flat slab analysis with for the sagging moment in an interior design strip, distributed to the column strip and middle strip

So essentially, we end up the following unique design moments:
  1. 3x
    
    for the end spans in the interior design strip, which are then each distributed to the column strip and half of a middle strip, making 6x unique moments
  2. 2x
    
    for the interior spans in the interior design strip, which are then each distributed to the column strip and two halves of a middle strip, making 4x unique moments
  3. 3x
    
    for the end spans in the edge design strip, which are then each distributed to the column strip and half of a middle strip, making 6x unique moments
  4. 2x
    
    for the interior spans in the edge design strip, which are then each distributed to the column strip and two halves of a middle strip, making 4x unique moments
  5. 2x
    
    for each direction (shorter and longer span direction), making 2x20 unique moments
That's 40 unique

for a flat-slab system, and even that's assuming a consistent grid i.e.

and

are constant!
The code says if the

on one side of an interior support is different from that on the other side, the unbalanced moment can be redistributed or design the slab on both sides with the larger negative moment.
As per Clause 6.10.4, the simplified method is valid provided the conditions in the below toggle are met.

Flat slab requirements for simplified analysis

Design Checks

Flexure (ULS)

Three main flexural checks are required by Australian Standards:
  1. moment capacity check
  2. ductility check
  3. minimum reinforcement check (i.e. minimum strength check)

Moment Capacity
Slab moment capacity is calculated the same way as a beam.
The strain and equivalent rectangular stress block of a typical RC slab section, without top reinforcement

The strain and equivalent rectangular stress block of a typical RC slab section, with top reinforcement

Let's use the example of a cross-section with top and bottom reinforcement.
Using the Rectangular Stress Block as per Cl. 8.1.2, we can determine the moment capacity of a section by taking moments about the

force:

Mu=ϕ[Ts×lever arm1+Cs×lever arm2]Mu=ϕ[Astfsy×(dst−γkud2)+Ascσsc×(γkud2−dsc)]M_u=\phi [T_s\times \text{lever arm}_1 + C_s \times \text{lever arm}_2]\\M_u=\phi [A_{st}f_{sy}\times (d_{st}-\frac{\gamma k_ud}{2}) + A_{sc} \sigma_{sc} \times (\frac{\gamma k_ud}{2}-d_{sc})]
Where:
  1. 
    
    is the capacity reduction factor, taken from Table 2.2.2.
  2. 
    
    factor is the depth ratio of the neutral axis from the extreme compressive fibre to the bottom reinforcement, at ultimate strength. Therefore
    
    represents the depth to the neutral axis.
Factor

is not defined explicitly in the standards and must be computed via force equilibrium.

Cc+Cs=Tsα2fc′γkudb+Ascσsc=AstfsyC_c+C_s=T_s\\ \alpha_{2}f'_c\gamma k_u d b + A_{sc}\sigma_{sc} = A_{st}f_{sy}
Where:
  1. 
    
    
  2. 
    
    
  3. 
    
    is the stress in the reinforcement in the compressive zone
  4. 
    
    is the strain in the reinforcement in the compressive zone, and is derived from similar triangles in the figure above
  5. 
    
    is the maximum strain in the extreme compressive concrete fibre as per Clause 8.1.3

If

, then

is given by:

→ku=fsy(Ast−Asc)α2fc′γdb\\\rightarrow k_u = \dfrac{f_{sy}(A_{st}-A_{sc})}{\alpha_{2}f'_c\gamma db}
If

, then

is found by solving the following quadratic equation:

α2fc′γkudb+AscEsεcu(1−dsckud)=Astfsy\alpha_{2}f'_c\gamma k_u d b + A_{sc}E_{s}\varepsilon_{cu}(1-\dfrac{d_{sc}}{k_ud}) = A_{st}f_{sy}
We then substitute

in the moment capacity equation to arrive at our answer.

Ductility Check
It is universal practice to ensure your concrete section is ductile. A ductile section means the reinforcement will yield before the concrete crushes, which is deemed a less dangerous failure mechanism since the failure occurs relatively slow compared to a sudden brittle failure. As per Cl 8.1.5, the code ensures ductile failure by checking:

kuo≤0.36k_{uo} \le 0.36
Where

for a section in pure bending.

Minimum Reinforcement Check
Clause 8.1.6 specifies minimum strength requirements. Clause 9.1.1 specifies an adjustment to this for two-way slabs.
Therefore, for reinforced concrete sections, the minimum strength requirement is deemed to be satisfied if

satisfies the following:

Ast≥[αb(D/d)2fct.f′/fsy]bwdA_{st}\geq [\alpha_b(D/d)^2f'_{ct.f}/f_{sy}]b_wd
Where:
  1. 
    
    for one-way slabs
  1. 
    
    for two-way slabs supported by columns at their corners
  1. 
    
    for two-way slabs supported by beams or walls on four sides

Shear (ULS)

Since slab depths are typically small, shear is rarely a critical consideration. Except for flat slabs at column supports, punching shear must be checked. For punching shear calculation of a flat slab, check out our Concrete Punching Shear Calculator to AS3600.

Deflection (SLS)

Clause 9.4 provides guidance for the deflection of slabs. A popular approach because of it's simplicity, is to follow Clause 9.4.4 which describes a deemed to conform span-to-depth ratio for deflections in RC slabs.
As per Clause 9.4.4, slab deflections are deemed to conform to code requirements if the following is satisfied:

Lef/d≤k3k4[(Δ/Lef)1000EcFd.ef]1/3L_{ef} / d \leq k_3 k_4 \left[ \dfrac{(\Delta/L_{ef})1000 E_c}{F_{d.ef}}\right]^{1/3}
Where:
  1. 
    
    is the deflection limit, chosen by the user, as per Table 2.3.2
  2. 
    
    is the effective span, where
    
    in one-way slabs or two-way flat slabs, and
    
    in two-way slabs supported on four sides
  3. 
    
    is the slab effective depth
  4. 
    
    is the effective design service load
    

→(1.0+kcs)G+(ψs+kscψ1)Qfor total deflection→kcsG+(ψs+kscψ1)Qfor incremental deflection\hspace{1cm}\rightarrow (1.0+k_{cs})G+ (\psi_s+k_{sc} \psi_1)Q \hspace{0.5cm}\text{for total deflection} \\\hspace{1cm}\rightarrow k_{cs}G+ (\psi_s+k_{sc} \psi_1)Q \hspace{1.7cm}\text{for incremental deflection}
  1. 
    
    where
    
    is from Table 4.1 in AS1170.0 and
    
    is taken at midspan for simply supported and continuous slabs
  2. 
    
    for one-way slab and two-way slabs supported on four sides,
    
    for multi-span two-way flat slab without drop panels,
    
    for multi-span two-way flat slab with drop panels
  3. 
    
    for simply support slabs,
    
    for an end span of a continuous slab,
    
    for an interior span in a continuous slab
AS3600-2018, Table 2.3.2

To use the above deflection check, Clause 9.4.4 states conditions that must be satisfied:
  1. uniform slab depth
  2. fully propped during construction
  3. loads are uniformly distributed and
    
  4. for two-way flat slabs with drop panels, drop panels must extend at least
    
    in each direction on each side of the support centre-line, and have an overall depth not less then
    
    slab thickness beyond the drops
  5. for continuous one-way slabs or two-way flat slabs, in adjoining spans, the ratio of the longer span to the shorter span does not exceed
    
    , and no end span is longer than an interior span