Design Guide: Pad Footings

IntroductionFoundations are unsung 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 foundation, pad footings.
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Series of concrete pad footings (Heaton Manufacturing)
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 categorized 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 behavior, and therefore the method of design for shallow foundations, is different from deep foundations.
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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 column, like residential dwellings or single-story industrial buildings.
Design Proceduread foundation design follows AS 3600 [1] and AS 2159 [2]. The table below shows import parameters and their definitions that are used constantly throughout this guide:
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Table 1: Commonly used variables in this guide and their definitions
Step 1. Calculate the ultimate soil-bearing capacityThe 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 ultimate bearing capacity formula are 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 only apply to shallow footings of certain shapes (i.e. circular, strip, and square), but Hansen’s equation applies to all. 
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𝑞𝑢=𝑐𝑁𝑐𝑠𝑐𝑑𝑐𝑖𝑐𝑔𝑐𝑏𝑐+𝑞𝑜𝑁𝑞𝑠𝑞𝑑𝑞𝑖𝑞𝑔𝑞𝑏𝑞+0.5𝐵′𝑁γ𝑠γ𝑑γ𝑖γ𝑔γ𝑏γ𝑞_𝑢=𝑐𝑁_𝑐𝑠𝑐𝑑_𝑐𝑖_𝑐𝑔_𝑐𝑏_𝑐+𝑞_𝑜𝑁_𝑞𝑠_𝑞𝑑_𝑞𝑖_𝑞𝑔_𝑞𝑏_𝑞+0.5𝐵'𝑁_γ𝑠_γ𝑑_γ𝑖_γ𝑔_γ𝑏_γ qu​=cNc​scdc​ic​gc​bc​+qo​Nq​sq​dq​iq​gq​bq​+0.5B′Nγ​sγ​dγ​iγ​gγ​bγ​
s = shape factor accounts for the shape of the footing
d = depth factor accounts for the depth of 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
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Figure 3: Visualisation of variables in Table 1
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Table 2: Hansen’s bearing capacity factors #1
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Table 3: Hansen’s bearing capacity factors #2
Step 2. Calculate the required area of the footingGenerally, 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 ~ and 4, depending on the type of 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). 
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A=BL=Pfactor of safety ∗ quA = BL = {\frac{P}{factor \ of \ safety\ *\ q_u}}A=BL=factor of safety ∗ qu​P​
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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 the first design iteration.
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Step 3. Calculate the minimum reinforcement and spacingA footing is responsible for the transfer of loads to the 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 2.1.3.1 :
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Ast,min=0.19(Dd)2fct,f′fsyA_{st,min} = 0.19 ({\frac{D}{d}})^2{\frac{f'_{ct,f}}{f_{sy}}}Ast,min​=0.19(dD​)2fsy​fct,f′​​
where,
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D=depth of the footingd=depth from the compressive fibre to the centroid of the tensile reinforcementf′ct,f=characteristic tensile flexural strength of concretelyfsy=yield strength of steel reinforcement ( 500MPa)D = \text{depth of the footing}\\ d= \text{depth from the compressive fibre to the centroid of the tensile reinforcement}\\ f′_{ct,f} = \text{characteristic tensile flexural strength of concretely}\\  f_{sy}=\text{yield strength of steel reinforcement ( 500MPa)}\\D=depth of the footingd=depth from the compressive fibre to the centroid of the tensile reinforcementf′ct,f​=characteristic tensile flexural strength of concretelyfsy​=yield strength of steel reinforcement ( 500MPa)
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Figure 5: Example cross-section of reinforced concrete footing
Then, select the appropriate bar size and the number of bars required, considering the 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 below.
AS3600 Clause 8.1:
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Mu=fsyAstd(1−γku2)M_u = f_{sy}A_{st}d(1-{\frac{{\gamma}k_u}{2}})Mu​=fsy​Ast​d(1−2γku​​)
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ku=the ratio between depth to the neutral axis from the extreme compressive fibre to tensile reinforcementk_u = the\ ratio\ between\ depth\ to\ the\ neutral\ axis\ from\ the\ extreme\ compressive\ fibre\ to\ tensile\ reinforcement \\ku​=the ratio between depth to the neutral axis from the extreme compressive fibre to tensile reinforcement
Clause 1.7:
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ku=Astfsya2fc′Bdk_u = {\frac{A_{st}f_{sy}}{a_2f'_cBd}}ku​=a2​fc′​BdAst​fsy​​
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γ=the ratio of depth of the assumed rectangular compressive stress block to kud\gamma = the\ ratio\ of\ depth\ of\ the\ assumed\ rectangular\ compressive\ stress\ block\ to\ k_{ud} \\γ=the ratio of depth of the assumed rectangular compressive stress block to kud​
Clause 8.1.3:
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y=0.97 −0.0025fc′y = 0.97\ - 0.0025f'_cy=0.97 −0.0025fc′​
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Figure 6: Example of a rectangular stress block
Step 5. Find the ultimate shear strengthCritical 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 below.
AS3600 Clause 8.2.4.1:
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Vuc=kvbvdvfc′V_{uc} = k_vb_vd_v \sqrt {f'_c}Vuc​=kv​bv​dv​fc′​​
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kv=concrete shear factorbv=effective width for shear =bdv=effective depth for shear =max(0.72D,0.9d)k_v = concrete\ shear\ factor \\                                b_v = effective\ width\ for\ shear\ = b \\                                 d_v = effective\ depth\ for\ shear\ = max(0.72D, 0.9d) \\kv​=concrete shear factorbv​=effective width for shear =bdv​=effective depth for shear =max(0.72D,0.9d)
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 8.2.4.3 may be used:
No prestressed elements
No axial tension or torsion
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f’c<65MPafsy<500MPaf’_c < 65 MPa \\ f_sy < 500 MPaf’c​<65MPafs​y<500MPa
The aggregate particle size of the concrete > 10mm
In such cases, the following can be assumed as per Clause 8.2.4.3:
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θv=angle of the compression reinforcement to the longitudinal axis = 0.36kv = min[0.15, 200(1000+1.3dv)]\theta_v = angle\ of\ the\ compression\ reinforcement\ to\ the\ longitudinal\ axis\ =\ 0.36 \\                                  k_v\ =\ min [0.15,\ {\frac{200} {(1000+1.3d_v)}}]θv​=angle of the compression reinforcement to the longitudinal axis = 0.36kv​ = min[0.15, (1000+1.3dv​)200​]
If the above conditions are not met, then Clauses 8.2.4.2.1 ~ 3 must be followed.
Step 6. Perform checksThe calculated flexural and shear strength must satisfy the following:
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AS3600 Clause 8.1: M∗ < ΦMu AS3600 Clause 8.1: V∗ < ΦVucAS3600\ Clause\ 8.1:\ M*\ <\ ΦM_u \ AS3600\ Clause\ 8.1:\ V*\ <\ ΦV_{uc} AS3600 Clause 8.1: M∗ < ΦMu​ AS3600 Clause 8.1: V∗ < ΦVuc​
📚 Sources[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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