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Concrete Rectangular Footing Designer to AS3600

Verified by the CalcTree engineering team on August 2, 2024

This calculator allows the user to design an isolated reinforced concrete footing supporting a single load-carrying column, that can be eccentric. It includes the section design to AS3600-2018 and checks for overturning, sliding, uplift and soil bearing at the four corners of the footing.
Detailed explanation of the behaviour of footings and required checks can be found in CalcTree's Design Guide: Concrete Footing to AS3600.
All calculations are performed in accordance with AS3600-2018.

Results Summary

Summary 
Design Check
Action
Resistance
Utilisation
Status
Bearing
q,max = 85.8 kPa
q,allow = 150.0 kPa
0.57
🟢
Overturning
ΣMo = 17.0 kNm
ΣMr = 66.0 kNm
0.26
🟢
Uplift
ΣPu = 0.0 kN
ΣPz = 131.4 kN
0
🟢
Sliding
Ff = 53.0 kN
ΣFr = 68.0 kN
0.78
🟢
Flexural
M* = 31.0 kNm
ϕMu = 212.0 kNm
0.15
🟢
Shear
V1* = 44.0 kN
ϕVu = 409.0 kN
0.11
🟢
Punching shear
V2* = 137.0 kN
ϕVuo = 1275.0 kN
0.11
🟢

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Calculation

Assumptions

  1. Applicable for a single column only
  1. The values
    
    assume to be the plan dimensions of a concrete column on the slab, or the end plate dimensions for a steel column on the slab

Inputs

Material Properties

Concrete


f'c
:{"mathjs":"Unit","value":40,"unit":"MPa","fixPrefix":false}​



γc
:{"mathjs":"Unit","value":25,"unit":"kN / m3","fixPrefix":false}​



Ec
:32,800MPa​



f'ct.f
:3.79 MPa​



f'ct
:2.28 MPa​

Reinforcement


fsy
:{"mathjs":"Unit","value":500,"unit":"MPa","fixPrefix":false}​



Es
:200 GPa​

Soil


qa
:150 kPa​



γs
:18 kN / m3​



μ
:0.40​



θ
:{"mathjs":"Unit","value":35,"unit":"deg","fixPrefix":false}​



Kp
:4​



Ka
:0​




Footing Geometry and Reinforcement



L
:2.0 m​



B
:1.0 m​



T
:0.40 m​



D
:0.3 m​


Longitudinal reinforcement (X-dir)


(x) Cover
:{"mathjs":"Unit","value":50,"unit":"mm","fixPrefix":false}​



(x) Reinforcement size, dst
:{"mathjs":"Unit","value":16,"unit":"mm","fixPrefix":false}​



(x) Number of bars, n
:15​



(x) Spacing
:63 mm​

Transverse reinforcement (Y-dir)


(y) Cover
:{"mathjs":"Unit","value":66,"unit":"mm","fixPrefix":false}​



(y) Reinforcement size, dst
:{"mathjs":"Unit","value":16,"unit":"mm","fixPrefix":false}​



(y) Number of bars, n
:30​



(y) Spacing
:64 mm​



Column Geometry



Lc
:0.3 m​



Bc
:0.3 m​



Hc
:0.5 m​



Loads



Q
:0 kPa​

Positive surcharge is defined as downwards load.


N*
:-100 kN​



ex
:{"mathjs":"Unit","value":0,"unit":"m","fixPrefix":false}​



ey
:0.00 m​



F*x
:0 kN​



F*y
:0 kN​



M*ox
:17 kN m​



M*oy
:0 kN m​

A positive axial load

is defined as an upwards load. Therefore:
For a

:
  1. 
    
    produces a
    
    
  1. 
    
    produces a
    
    
For a

:
  1. 
    
    produces a
    
    
  1. 
    
    produces a
    
    

Output

Resultant Forces and Eccentricities



Column weight
:1kN​



Footing weight
:20kN​



Soil weight
:10kN​



Surcharge
:0kN​



ΣP
:131.439kN​


Resultant axial load,



Resultant axial load is the sum of all column loads, the surcharge and the concrete self-weight:

∑Pz=∑(Ncol∗+Wcol)+Wslab+QsurchargeBL\sum P_z= \sum (N^*_{\text{col}}+ W_{\text{col}})+W_{\text{slab}}+Q_{\text{surcharge}}BL


ΣM*x
:17​



ΣM*y
:0​


Resultant moments,



Resultant moment about the x-axis:

∑Mx∗=N∗ex+Mox∗+Fx∗(Hc+T)\sum M^*_x=N^*e_x+M^*_{ox}+F^*_x (H_c+T)
Resultant moment about the y-axis:

∑My∗=N∗ey+Moy∗+Fy∗(Hc+T)\sum M^*_y=N^*e_y+M^*_{oy}+F^*_y (H_c+T)


ex, resultant
:0.129 m​



ey, resultant
:0 m​


Resultant eccentricities

Resultant eccentricity from the x-axis:

ey=ΣMyΣP e_y = \dfrac{\Sigma M_y}{\Sigma P}
Resultant eccentricity from the y-axis:

ex=ΣMxΣPe_x = \dfrac{\Sigma M_x}{\Sigma P} \\


Geotechnical Checks (Stability)



Factor of Safety
:3​


Bearing Check


2​
91.2​
91.2​
91.2​
91.2​
2​
1​
100​


Maximum bearing pressure, qu
:85.82 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
:17 kNm​



(x) MN
:0 kNm​



(x) ΣMo
:17 kNm​



(x) ΣMr
:66 kNm​

Section y-y


(y) Me
:0 kNm​



(y) MN
:0 kNm​



(y) ΣMo
:0 kNm​



(y) ΣMr
:131 kNm​



Overturning check
:PASS​


Uplift Check



(z) ΣPz
:131.44 kN​



(z) 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*
:0 kN​



(x) Fkp
:15 kN​



(x) Ff
:53 kN​



(x) ΣFr
:68 kN​

Section y-y


(y) F*
:0 kN​



(y) Fkp
:28 kN​



(y) Ff
:53 kN​



(y) ΣFr
:81 kN​



Sliding check
:PASS​


Structural Checks (ULS)

Flexure



ϕ (bending)
:0.85​



α2
:0.79​



γ
:0.87​

ϕ as per AS3600 Table 2.2.2

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


(x) M*
:31 kNm​



(x) ku
:0.0801915620388398​



(x) do
:342 mm​



(x) ϕMu
:212 kNm​



(x) M* / ϕMu
:0.14655744064367338​

Section y-y


(y) M*
:5 kNm​



(y) ku
:0.3365093768991806​



(y) do
:326 mm​



(y) ϕMu
:713 kNm​



(y) M* / ϕMu
:0.0073683570580787075​


One-Way Shear



ϕ (shear)
:0.7​

ϕ as per AS3600 Table 2.2.2

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


(x) V1*
:44 kN​



(x) ϕVu
:409 kN​



(x) V1* / ϕVu
:0.10664286064672857​




(x) kv
:0.15​



(x) bw
:2000 mm​



(x) dv
:307.8 mm​



(x) √f'c
:6.32 MPa​

Section y-y


(y) V1*
:4 kN​



(y) ϕVu
:195 kN​



(x) V1* / ϕVu
:0.021142084927982828​




(y) kv
:0.15​



(y) bw
:1000 mm​



(y) dv
:293.4 mm​



(y) √f'c
:6.32 MPa​


Two-Way Shear (Punching Shear)



dom
:334 mm​



u
:2536 mm​



V2*
:137 kN​

If shear head is not provided:

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


ϕVuo1
:1275 kN​



V2* / ϕVuo1
:0.10756526315413556​

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
:1875 kN​



V2* / ϕVuo2
:0.07314437894481254​


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.

Note

Typically, footing designs can be done by hand as the required calculations are relatively simple. However, it should be noted that the most accurate method of structural analysis is finite element modelling (FEA). FEA allows the engineer to capture information that would otherwise be near-impossible to incorporate into hand calculations, such as soil spring stiffness, two-way load distribution and other boundary conditions.

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. Design Guide: Concrete Footing to AS3600-2018
  2. Concrete Slab-on-Grade Designer to AS3600
  1. Foundation Bearing Failure Modes and Capacities
  2. Rectangular Spread Footing Design to ACI-384
  3. Concrete Slab-on-grade Calculator to ACI 360R-10


(y) Depth to reinforcement centroid, dst
:326.0​