Practical example of combined foundation design according to Eurocode
- The base has two columns
- The 300 mm square column has a payload of 100 kN or 500 kN and a dead load of 100 kN.
- The 400 mm square column has a payload of 200 kN or 800 kN and a dead load of 300 kN.
- Distance between supports 3m
- fck = 30N/mm2
- fyk = 500N/mm2
- Assumed foundation thickness 500 mm
- Allowable bearing pressure 200N/mm2
- For initial calculations, assume an effective depth of 440 mm
Calculation of the required foundation area
Net bearing pressure = 200-25*0.500
= 187.5 N/mm²
Operating load = 100 + 500 + 200 + 800
= 1600kN
Required foundation area = 1600/187.5
= 8.53 m²
Choose the dimensions of the foundation 4.3 mx 2 m
(550mm offset from 300mm square column and 750mm offset from 400mm square column)
Load center calculation
Seize the moment around the center of the 300mm square column
x = (200+800)*3/(100+500+200+800)
= 1.875m
Calculation of support pressure
Design load = (1.35*800+1.5*200) + 1.35*500+1.5*100)
= 2205kN
Pressure = 2205/(4.3*2)
= 257N/mm2
Check for perforation
Maximum shear strength = 0.5ud(0.6(1-fck/250))(fck/1.5)
With support area of 300 mm
= 0.5*300*4*440*0.6*(1-30/250)*(30/1.5)
= 2788kN > (1.35*500+1.5*100 = 825kN)
With support area of 400 mm
= 0.5*400*4*440*0.6*(1-30/250)*(30/1.5)
= 3717kN > (1.35*800+1.5*200 = 1380kN)
So drilling is OK
The maximum curvature was calculated
It was 209 kNm and occurred at a distance of 1.05 m from a 300 mm square pillar.
Flexible construction
Longitudinal reinforcements
MEd = 209kNm
K = M/(b*(d^2)*fck)
K = 209*E6/(1000*(440^2)*30)
K = 0.04
Kbal = 0.167 (equilibrium state is assumed based on x = 0.45d)
therefore
K
No compressive reinforcement required
z = d(0.5+(0.25-K/1.134)^0.5)
z = 440(0.5+(0.25-0.04/1.134)^0.5)
z = 424 mm (Z/d = 0.96>0.95)
therefore
Z = 0.95*440
= 418mm
Als = M/(0.87*fyk*z)
As = 209*E6/(0.87*500*418)
= 1150mm2
As min. = 0.15*b*d/100
= 0.15*1000*440/100
= 660mm2
Therefore,
Proven T16@150mm spacing
Transverse reinforcements
Bending moment = 257*(1^2)/2
MEd = 128.5 kNm
K = M/(b*(d^2)*fck)
K = 128.5*E6/(1000*(440^2)*30)
K = 0.022
Kbal = 0.167 (equilibrium state is assumed based on x = 0.45d)
therefore
K
No compressive reinforcement required
z = d(0.5+(0.25-K/1.134)^0.5)
z = 440(0.5+(0.25-0.022/1.134)^0.5)
z = 431 mm (Z/d = 0.97>0.95)
therefore
Z = 0.95*431
= 409.5mm
Als = M/(0.87*fyk*z)
As = 128.5*E6/(0.87*500*409.5)
= 722mm2
As min. = 0.15*b*d/100
= 0.15*1000*440/100
= 660mm2
Ensure a distance of T12@150mm
