Design of Pad Footings

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Basic Steps
01. Calculate the size of the footing considering allowable bearing pressure and service load.
02. Calculate the bearing pressure for ultimate loads
03. Check the vertical line shear (shear at face of the column)
04. Check for punching shear
05. Calculate reinforcement for bending
06. Check shear at critical section

Design Example

• Live load 400kN
• Dead load 900kN
• Allowable bearing pressure 175kN
• fck 30N/mm2
• fyk 500N/mm2
• Size of column 400mm
• Assume 150kN as footing weight

Calculation of Base Area

Design service load             = 1.0Gk + 1.0Qk
                                            =  900 +150 + 400
                                            =  1450kN

Required footing area          = 1450/175
                                            =  8.3m2
Hence, provide 2.9m square footing (area 8.41m2)

Calculate Ultimate Loads

Axial load                           =  1.35Gk + 1.5Gk
                                          =  1.35*900 + 1.5*400
                                          =  1815kN

Ultimate pressure                = 1815/(2.9*2.9)
                                            = 216kN/m2

Check for adequacy of footing thickness

• Check for Maximum Shear

Assume, footing thickness 500mm, 16mm diameter bars at base and cover to the reinforcement as 40mm

d                                       = 500-40-16/2
                                         = 452mm
Max. Shear resistance
                         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
 Hence,             VRd,max  >  VEd = 1815kN

• Punching Shear 

Critical section is considered at 2d from the face of the column

Critical Perimeter               = Column Perimeter + 4πd
                                          = 4*400 + 4π*452
                                          = 7280mm
Area with the perimeter     = (400-4d)^2 -(4-π)(2d)^2
                                          = (400+4*452)^2 -(4-π)(2*452)^2      
                                          = 4.17*10^6 mm
Punching Shear Force       = 216*(2.9^2-4.17)
                                          = 915.84kN
Shear Stress                      = VEd/(perimeter * d)
                                          = 915.84*E3/ (7280* 452)
                                         = 0.28
Shear stress is not that large so 500mm thickness can be used
                                
Bending Reinforcements
Consider critical section (at column face)

MEd                                 = 216*2.9*(2.9/2-0.4/2)*(2.9/2-0.4/2)/2
                                       = 490kNm
K                                    = M/[b*(d^2)*fck]
K                                    = 490*E6/[1000*(452^2)*30]
K                                    = 0.08

Kbal = 0.167 (balance codition is taken based on x = 0.45d)
Therefore
K<Kbal
Compression reinforcement is not 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                                     = 417.5mm  (Z/d = 0.92<0.95)

As                                  = M/(0.87*fyk*z)
As                                  = 490*E6/(0.87*500*417.5
                                      = 2698mm2
Provide T25@175mm spacing (As Provided = 2804mm2)
OR you may increase the depth of the footing in order to reduce the reinforcement area.

Check for Punching Shear
As/(bd)                           = 2698/(1000*452)
                                      = 0.006
                                      = 0.6% < 2%
Therefore,
Shear stress                   = 0.4N/mm2

VRd,c                                        = 0.4*7280*542
                                     = 1316.2kN > 915.84kN
Hence, punching shear is ok.

Check for Maximum Shear
 Consider 1.0d form face of the column

Design Shear Force      = 216*2.9*0.798
                                      = 499.9kN
 As above
VRd,c @ 1.0d                  = 0.4*1000*452
                                    = 180.8kN < 499.9kN
Hence,
Shear reinforcement is required.
Normally, Shear reinforcements are not provided for pad footings. Therefor, the thickness of the footing may be increased and re-do the design as done above.