Worked example according to Eurocode 2: Design of individual foundations
Fundamentals
01. Calculate the size of the foundation taking into account the allowable support pressure and the operational load.
02. Calculate bearing pressure for maximum loads
03. Check the shear of the vertical line (shear in front of the pillar).
04. Punching test
05. Calculate reinforcement for bending
06. Check the shear in the critical area
Project example
- Payload 400kN
- Dead load 900kN
- Allowable bearing pressure 175kN
- fuck 30N/mm2
- fyk 500N/mm2
- Column size 400mm
- Assume 150 kN as the weight of the foundation
Calculation of useful area
Design operating load = 1.0 Gk + 1.0 Qk
= 900 +150 + 400
= 1450 kN
Required foundation area = 1450/175
= 8.3 m²
Therefore, provide 2.9 m² of usable area (area 8.41 m²).
Calculate maximum loads
Axial load = 1.35 Gk + 1.5 Gk
= 1.35*900 + 1.5*400
= 1815kN
Final pressure = 1815/(2.9*2.9)
= 216kN/m2
Check whether the foundation thickness is sufficient
Check maximum shear
Suppose the foundation thickness is 500 mm, bars with a diameter of 16 mm at the base and reinforcement cover of 40 mm
d = 500-40-16/2
= 452mm
Maximum shear strength
VRD,max = 0.5ud(0.6(1-fck/250))(fck/1.5)
= 0.5*(4*400)*452*0.6*(1-30/250)*(30/1.5)*10^-3
= 3818.5kN
Therefore VRd,max > VEd = 1815kN
- piercing scissors
The critical section is seen in 2d from the front of the column.
Critical perimeter = column perimeter + 4πd
= 4*400 + 4π*452
= 7280mm
Area with perimeter = (400-4d)^2 -(4-π)(2d)^2
= (400+4*452)^2 -(4-π)(2*452)^2
= 4.17*10^6mm
Punching force = 216*(2.9^2-4.17)
= 915.84 kN
Shear stress = VEd/(perimeter * d)
= 915.84*E3/(7280*452)
= 0.28
The shear stress is not that large, so a thickness of 500 mm can be used
Flexion reinforcements
Consider the critical section (on the column surface)
MEd = 216*2.9*(2.9/2-0.4/2)*(2.9/2-0.4/2)/2
= 490 kNm
K = M/(b*(d^2)*fck)
K = 490*E6/(1000*(452^2)*30)
K = 0.08
Kbal = 0.167 (balance value is based on x = 0.45d)
therefore
K
No compressive reinforcement required
z = d(0.5+(0.25-K/1.134)^0.5)
z = 452(0.5+(0.25-0.08/1.134)^0.5)
z = 417.5 mm (Z/d = 0.92 < 0.95)
Als = M/(0.87*fyk*z)
As = 490*E6/(0.87*500*417.5
= 2698mm2
Provide clearance of T25 to 175 mm (as intended = 2804 mm2).
OR You can increase the depth of the foundation to reduce the area of reinforcement.
Check for perforation
As/(bd) = 2698/(1000*452)
= 0.006
= 0.6% <2%
Therefore,
Shear stress = 0.4N/mm2
VRd,c = 0.4*7280*542
= 1316.2 kN > 915.84 kN
So drilling is ok.
Check maximum shear
Consider the 1.0d shape surface of the column
Design shear force = 216*2.9*0.798
= 499.9kN
As above
VRd,c@1.0d = 0.4*1000*452
= 180.8 kN < 499.9 kN
That's why,
Shear reinforcement is required.
Shear reinforcement is not normally provided for individual foundations. Therefore, the thickness of the foundation can be increased and the structure rebuilt as described above.
