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Concrete Rectangular Footing Designer to AS3600's banner
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Concrete Rectangular Footing Designer to AS3600

This calculator allows the user to design an isolated reinforced concrete footing supporting a single load-carrying column. It includes the section design to AS3600-2018 and checks for overturning, sliding, uplift and soil bearing at the four corners of the footing.

Note

❗ This calculation has been written in accordance with AS3600-2018

Calculation

Inputs

Geometry nomenclature

Geometry and load nomenclature



Material Properties

Concrete


f'c
:40MPa​



γc
:24kN/m3​



Ec
:32,800MPa​



f'ct.f
:3.79 MPa​



f'ct
:2.28 MPa​

Reinforcement


fsy
:500MPa​



Es
:200,000MPa​

Soil


qmax
:150kPa​



γs
:18kN/m3​



μ
:0​



θ
:35​



Kp
:4​



Ka
:0​



Footing Geometry



L
:5m​



B
:7m​



T
:0m​



D
:2m​

Section X-X reinforcement


(x) Cover
:50mm​



(x) Reinforcement size, dst
:16mm​



(x) Number of bars, n
:17​



(x) Spacing
:305 mm​

Section Y-Y reinforcement


(y) Cover
:60mm​



(y) Reinforcement size, dst
:16mm​



(y) Number of bars, n
:43​



(y) Spacing
:163 mm​



Column Geometry



Lc
:1m​



Bc
:1m​



Hc
:1m​



Distance to centroid of tensile reinforcement in column
:0mm​

↑ put in '0' if unknown

Loads



N*
:-200kN​



Fx*
:50kN​



Fy*
:-120kN​



Mx*
:-150kN m​



My*
:213kN m​



FoS and Capacity Reduction Factors



Factor of Safety
:1.5​

ϕ shall be selected as per AS3600 Table 2.2.2.


ϕ (bending)
:0.85​



ϕ (shear)
:0.7​


Output

Resultant Forces and Eccentricities



Column weight
:24kN​



Footing weight
:252kN​



Soil weight
:1242kN​



Surcharge
:0.0kN​



Total vertical force, ΣPz
:1718.0kN​



ex
:0.162 m​



ey
:-0.004 m​



Bearing Check

Different cases of biaxial bearing pressure

Bearing corner pressures


4​
58.76​
58.47​
39.41​
39.7​
5​
7​
100​


Maximum bearing pressure, qu
:17.36 kPa​



Bearing check
:PASS​



Overturning Check


Me = Overturning moment due to applied external shear and momentMN =Overturning moment due to eccentric axial loadΣMo = Me +MN = Total overturning momentΣMr = Σ(soil, column, footing self-weight and surcharge)=Total resisting moment \small{M_{e}\ =\ \text{Overturning\ moment\ due\ to\ applied\ external\ shear\ and\ moment}}\\\small{M_N\ =\text{Overturning\ moment\ due\ to\ eccentric\ axial\ load}}\\\small{ΣM_o}\ =\ M_e\ + M_N\ =\ \text{Total\ overturning\ moment}\\\small{ΣM_r\ =\ Σ(\text{soil,\ column,\ footing\ self-weight\ and\ surcharge})=\text{Total\ resisting\ moment}}\
Section x-x


(x) Me
:6 kNm​



(x) MN
:0 kNm​



(x) ΣMo
:6 kNm​



(x) ΣMr
:6013 kNm​

Section y-y


(y) Me
:278 kNm​



(y) MN
:0 kNm​



(y) ΣMo
:278 kNm​



(y) ΣMr
:4295 kNm​



Overturning check
:PASS​



Uplift Check



ΣPz
:1718 kN​



ΣPu
:0 kN​



Uplift check
:PASS​



Sliding Check


Fkp = Passive earth pressure resistanceFf = Frictional resistanceΣFr = Fp + Ff = Total resisting forceF∗ = Pushing force = applied shear in the direction being considered\small{F_{kp}\ =\ \text{Passive\ earth\ pressure\ resistance}}\\F_f\ =\ \text{Frictional\ resistance}\\ΣF_r\ =\ F_p\ +\ F_f\ =\ \text{Total\ resisting\ force}\\F^*\ =\ \text{Pushing\ force}\ =\ \text{applied\ shear\ in\ the\ direction\ being\ considered}
Section x-x


(x) F*
:50 kN​



(x) Fkp
:499 kN​



(x) Ff
:687 kN​



(x) ΣFr
:1186 kN​

Section y-y


(y) F*
:120 kN​



(y) Fkp
:413 kN​



(y) Ff
:687 kN​



(y) ΣFr
:1101 kN​



Sliding check
:PASS​



ULS Capacity Checks

One-Way Shear


Vu = kvbwdvfc′V_u\ =\ k_vb_wd_v\sqrt{f'_c}
Section x-x


(x) V1*
:214 kN​



(x) ϕVu
:723 kN​



(x) V1* / ϕVu
:0.295464868736963​




(x) kv
:0.15​



(x) bw
:5000 mm​



(x) dv
:217.8 mm​



(x) √f'c
:6.32 MPa​

Section y-y


(y) V1*
:240 kN​



(y) ϕVu
:1004 kN​



(x) V1* / ϕVu
:0.23933219170171746​




(y) kv
:0.15​



(y) bw
:7000 mm​



(y) dv
:216 mm​



(y) √f'c
:6.32 MPa​



Two-Way Shear (Punching Shear)



dom
:237 mm​



u
:4948 mm​



V2*
:581 kN​

If shear head is not provided:

Vuo = udom(fcv+0.3σcp) \small{V_{uo}\ =\ ud_{om}(f_{cv}+0.3\sigma_{cp})}\\\


ϕVuo1
:1765 kN​



V2* / ϕVuo1
:0.3292339734812499​

If shear head is provided:

Vuo = udom(0.5fc′+0.3σcp)≤0.2udomfc′\small{V_{uo}\ =\ ud_{om}(0.5f'_c+0.3\sigma_{cp})}\\\small{\leq0.2ud_{om}f'_c}


ϕVuo2
:2596 kN​



V2* / ϕVuo2
:0.2238791019672496​



Flexure


Mu = Astfsy(do−γkudo2)\large{M_u\ =\ A_{st}f_{sy}(d_o-\frac{{\gamma}k_{u}d_o}{2})}


α2
:0.79​



γ
:0.87​

Section x-x


(x) M*
:35 kNm​



(x) ku
:0.05137561891942855​



(x) do
:242 mm​



(x) ϕMu
:69 kNm​



(x) M* / ϕMu
:0.5052065024558157​

Section y-y


(y) M*
:78 kNm​



(y) ku
:0.0968224229345794​



(y) do
:232 mm​



(y) ϕMu
:117 kNm​



(y) M* / ϕMu
:0.6698218747386719​


Explanation

This section focuses on the limit state design principles of footing design to AS3600. Detailed explanation of the behaviour of footings and required checks can be found in CalcTree's Design Guide: Concrete Footing to AS3600.

Design Considerations

Footing design is an iterative process; it requires an initial judgement from the engineer on the required thickness and reinforcement, then repeating structural analysis until the desired strength is achieved.
The initial 'guess' of the footing size is governed by two things:
  1. Applied load and allowable bearing pressure - this determines bearing area i.e. length and width
  2. Required development length column reinforcement in the footing, and the concrete shear strength without shear reinforcement - this determines the depth
The required bearing area can be calculated by dividing the total applied load on the footing by the allowable bearing pressure:

A =FqallowableA\ =\large \frac{F}{q_{allowable}}
Although there are many combinations of lengths and widths that can achieve the same bearing area, squarer footings are better as they produce a more even pressure distribution than a rectangular one.
Longitudinal reinforcement from columns are required to continue into the footing to achieve sufficient development length for structural continuity and load transfer. Generally, at column-footing interface, these bars are cogged at 90° and extend parallel to the footing surface.

Columns are mostly in compression and hence the reinforcement is also in compression. AS3600 Cl. 13.1.5 provides the following formula for calculating the development length of bars in compression:

Lsy.cb=0.22fsyfc′db≥0.0435fsydbor 200mm, whichever is greaterL_{\mathrm{sy} . \mathrm{cb}}=\frac{0.22 f_{\mathrm{sy}}}{\sqrt{f_{\mathrm{c}}^{\prime}}} d_{\mathrm{b}} \geq 0.0435 f_{\mathrm{sy}} d_{\mathrm{b}}\\ \text{or\ 200mm,\ whichever\ is\ greater}
Considering that the minimum development length is 200mm, the required footing depth would generally be around 300mm or higher.
Isolated pad footing under construction (Source: TR Construction)


Related Resources

  1. Concrete Beam Design Calculator to AS3600-2018
  2. Concrete Column Design Calculator to AS3600-2018
  3. Concrete Slab-on-Grade Design to AS3600
  4. Design Guide: Concrete Footing to AS3600-2018
  5. Foundation Bearing Failure Modes and Capacities
  6. Rectangular Spread Footing Design to ACI-384
  1. Slab Thickness Calculator to ACI 360R-10