Series of concrete pad footings (Heaton Manufacturing)

Foundations are un-sung heros of structures, rarely seen, but critical for structural integrity. They are often aesthetically underwhelming compared to more flashy structural components above ground, but foundations are responsible for the structure’s stability and the safety of its occupants. This guide specifically outlines procedures for the design of a very common type of foundations, pad footings.

The primary role of a foundation is to transfer loads from the superstructures above to the ground below. Different types of foundations have been developed to accommodate different ground and loading conditions, and are generally categorised as either deep or shallow. Shallow foundations transfer loads to the upper soil layers which are relatively shallow compared to the effective size of the bearing  area. If the upper soil layer is not strong enough, deep foundations can be used to reach stronger soil layers deeper in the ground, much deeper in comparison to the pad’s width. Their behaviour, and therefore the method of design for shallow foundations, is different from deep foundations.

Figure 2 - Different types of foundations

The design of a foundation varies from project to project, and depends on various factors such as the design of the structure above, applied loads, site accessibility, proximity to nearby structures, soil conditions, budget, etc. Pad foundations are commonly used in structures with smaller heights which have relatively small deck areas supported by each columns, like residential dwellings or single-story industrial buildings.

Design Procedure

The design of pad footings follow AS 3600 [1] and AS 2159 [2]. The table below shows import parameters and their definitions that are used constantly throughout this guide:

Table 1 - Commonly used variables in this guide and their definitions

Step 1. Calculate the ultimate soil bearing capacity

The first step in designing a pad footing is calculating the ultimate bearing capacity of the surrounding soil (qu), a nominal pressure at the foundation base at which the soil fails. The actual pressure beneath a foundation is often not uniform, but the nominal pressure takes the overall distribution into account, provided that a compliant set of equations are used end to end of the design procedure. The most relevant equations for calculating qu are using those proposed by Terzaghi and Hansen. For cohesionless soils, Terzaghi and Hansen’s equations are equal; however for cohesive soils, Terzaghi’s equations are more conservative than Hensen’s. Generally, Terzaghi’s equations are only applicable to shallow footings of certain shapes (i.e. circular, strip and square), but Hansen’s equation is applicable to all. 

\begin{align*} q_u& =cN_cs_cd_ci_cg_cb_c + q_oN_qs_qd_qi_qg_qb_q + 0.5B'N_γs_γd_γi_γg_γb_γ \end{align*}
\begin{align*} s = shape\ factor\; accounts\ for\ the\ shape\ of\ the\ footing \\ d = depth\ factor\; accounts\ for\ the\ depth\ to\ the\ footing\ base \\ i = inclination\ factor\; accounts\ for\ cases\ where\ loads\ act\ other\ than\ normal \\ g = ground\ factor\; accounts\ for\ the\ slope\ of\ the\ ground \\ b = base\ factor\; accounts\ for\ the\ inclined\ surface\ of\ the\ footing \\ \end{align*}

Figure 3: Visualisation of variables in Table 1

Table 2: Hansen’s bearing capacity factors #1
Table 3 - Hansen’s bearing capacity factors #2

Step 2. Calculate the required area of the footing

Generally, the ultimate bearing capacity is reduced by a ‘factor of safety’ to obtain the allowable bearing capacity qa. The allowable bearing capacity is used as the limit for the design of the footing instead of the ultimate bearing capacity for safety. This factor of safety generally varies between 2 ~ 4, depending on the type of the structure.

The required footing area can be calculated by rearranging the ‘pressure = force / area’ equation. P represents the column axial force, which includes the loads transferred from the superstructure and the column’s self-weight. For footings, the shorter side is referred to as width (denoted B), and the longer side as length (denoted L). 

\begin{align*} A = BL = {\frac{P}{factor \ of \ safety\ *\ q_u}} \end{align*}
Figure 4: Visualisation of applied load P and bearing capacity qu

You should note that some of the bearing capacity coefficients depend on the pad geometry. Therefore, you often need to do a bearing capacity ‘check’ after first design iteration.

Step 3. Calculate the minimum reinforcement and spacing

A footing is responsible for the transfer of loads to surrounding stratum and soil. Much like a beam or a slab, footings are subject to flexure and shear forces. Therefore, reinforcements are necessary.

Find the minimum requirement steel reinforcement:

AS3600 Clause :

\begin{align*} A_{st,min} = 0.19 ({\frac{D}{d}})^2{\frac{f'_{ct,f}}{f_{sy}}} \end{align*}
\begin{align*} D = depth\ of\ the\ footing \\ d = depth\ from\ the\ compressive\ fibre\ to\ the\ centroid\ of\ the\ tensile\ reinforcement \\ f’_{ct,f} = characteristic\ tensile\ flexural\ strength\ of\ concrete \\ f_{sy} = yield\ strength\ of\ steel\ reinforcement\ (~500MPa) \\ \end{align*}
Figure 5: Example cross-section of reinforced concrete footing

Then, select the appropriate bar size and the number of bars required, considering cover on each side.

Step 4. Find the ultimate flexural strength 

The critical flexural failure (or two-way shear failure) occurs at the face of the column. The flexural strength of the footing can be calculated as below.

AS3600 Clause 8.1:

\begin{align*} M_u = f_{sy}A_{st}d(1-{\frac{{\gamma}k_u}{2}}) \end{align*}
\begin{align*} k_u = the\ ratio\ between\ depth\ to\ the\ neutral\ axis\ from\ the\ extreme\ compressive\ fibre\ to\ tensile\ reinforcement \\ \end{align*}

Clause 1.7:

\begin{align*} k_u = {\frac{A_{st}f_{sy}}{a_2f'_cBd}} \end{align*}
\begin{align*} \gamma = the\ ratio\ of\ depth\ of\ the\ assumed\ rectangular\ compressive\ stress\ block\ to\ k_{ud} \\ \end{align*}

Clause 8.1.3:

\begin{align*} y = 0.97\ - 0.0025f'_c \end{align*}
Figure 6: Example of a rectangular stress block

Step 5. Find the ultimate shear strength

The critical shear failure (or one-way shear failure) occurs at some distance away from the face of the column. The shear strength of the footing (specifically the concrete) can be calculated as below.

AS3600 Clause

\begin{align*} V_{uc} = k_vb_vd_v \sqrt {f'_c} \end{align*}
\begin{align*} k_v = concrete\ shear\ factor \\ b_v = effective\ width\ for\ shear\ = b \\  d_v = effective\ depth\ for\ shear\ = max(0.72D, 0.9d) \\ \end{align*}

This section of the standard presents two methods for estimating the ultimate shear strength of concrete. In cases where the following are satisfied, a simplified method outlined in Clause may be used:

No prestressed elements

No axial tension or torsion

\begin{align*} f’_c < 65 MPa \\ f_sy < 500 MPa \end{align*}

Aggregate particle size of the concrete > 10mm

In such cases, the following can be assumed as per Clause

\begin{align*} \theta_v = angle\ of\ the\ compression\ reinforcement\ to\ the\ longitudinal\ axis\ =\ 0.36 \\ k_v\ =\ min [0.15,\ {\frac{200} {(1000+1.3d_v)}}] \end{align*}

If the above conditions are not met, then Clauses ~ 3 must be followed.

Step 6. Perform checks

The calculated flexural and shear strength must satisfy the following:

\begin{align*} AS3600\ Clause\ 8.1:\ M*\ <\ ΦM_u \ AS3600\ Clause\ 8.1:\ V*\ <\ ΦV_{uc} \end{align*}

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[1] Standards Australia, AS 3600-2018: Concrete Structures. Standards Australia, 2018.

[2] Standards Australia, AS 2159-2009: Piling - Design and Installation. Standards Australia, 2009.

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