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​≤2
Where:


Lx​
is the length of the shorter slab span


Ly​
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,

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

M∗
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×Fd​Ln2​
Where:
coefficient depends on the location of the

M∗
under consideration, see figures below


Fd​
is the factored ULS design load per unit length

(kN/m)



Ln​
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:
the ratio of the longer to the shorter length of any two adjacent spans does not exceed

1.2

loads are uniformly distributed


Q<2G
, where Q = live load & G = dead load
slab has a constant cross-section
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.


M∗
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:
are moment coefficients, taken from Table 6.10.3.2(A)


Fd​
is the factored ULS design load per unit area

(kN/m2)



Lx​
is the shorter span length

AS3600-2018, Table 4.10.3.2(A)
👉 AS2600-2018 commentary provide equations for

βx​
and

βy​
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:
continuous edge means the slab is able to transfer load to the adjacent slab through that edge
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:
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:
loads are uniformly distributed
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.


M∗
at supports are only caused by loads applied to the beam or slab, not from the columns
openings in the slab are not big enough to adversely affect strength or stiffness
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​=8Fd​Lt​Lo2​​
Where:


Fd​
is the uniformly distributed design load per unit area (

kPa)



Lt​
is the width of the design strip, which for an interior design strip is taken as

Lt​=Lx​
in the longer span and

Lt​=Ly​
in the shorter span. For edge design strips, it is taken as

Lt​=Lx​/2
in the longer span and

Lt​=Ly​/2
in the shorter span.


Lo​
is the effective span, which is taken as

Lo​=Ly​
in the longer span and

Lo​=Lx​
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×Mo​
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

Ly​


Flat slab analysis for the shorter span

Lx​

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

Mo​=408 kNm
in the longer span of the interior design strip. The sagging moment in the middle of the span is calculated as

+My∗​=0.35×Mo​=142.8 kNm
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

0.5×(+My∗​)=71.4 kNm
is to be distributed to the column strip, and therefore leaving

71.4/2=35.7Â kNm
to each of the half middle strips.

Example of flat slab analysis with

Mo​=408 kNm
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:
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
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
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
2x

Mtotal∗​={Mx∗​,My∗​}
for each direction (shorter and longer span direction), making 2x20 unique moments
That's 40 unique

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

Ly​
and

Lx​
are constant!
The code says if the

−M∗
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
For two-way slabs supported by columns (i.e. flat slabs), as per Clause 6.10.4, the simplified method is valid provided:
loads are uniformly distributed


Q<2G
, where Q = live load & G = dead load
at least two continuous spans in each direction
support grid is roughly rectangular
successive span lengths in each direction do not differ by more then 1/3 of the longer span, and the end span cannot be longer then the adjacent interior span
lateral forces are resisted by shear walls or braced frames
reinforcement with ductility class L is not used for flexural reinforcement
reinforcement is arranged in accordance with Clause 9.1.3.4, which describes how much and the extent of top & bottom reinforcement with the slab span.


M∗
at supports are only caused by loads applied to the beam or slab, not from the columns
Design Checks
Flexure (ULS)
Three main flexural checks are required by Australian Standards:
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​≤0.36
Where

kuo​=ku​
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

Ast​
satisfies the following:

Ast​≥[αb​(D/d)2fct.f′​/fsy​]bw​d
Where:


αb​=0.2
for one-way slabs


αb​=0.24
for two-way slabs supported by columns at their corners


αb​=0.19
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: