Desing of Combined Footings

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• Footing has two columns
• 300mm square column has 100kN and 500kN live and dead loads respectively
• 400mm square column has 200kN and 800kN live and dead loads respectively
• Spacing of columns 3m
• fck = 30N/mm2
• fyk = 500N/mm2
• Assumed footing thickness 500mm
• Allowable bearing pressure 200N/mm2
• Assume effective depth 440mm for initial calculations

Calculation of required footing area

Net bearing pressure   = 200-25*.500
                                     = 187.5N/mm2
Service load                 = 100 + 500 + 200 + 800
                                     = 1600kN
Required footing area  = 1600/187.5
                                     = 8.53m2
Select footing dimensions as 4.3m X 2m
(550mm offset from 300mm square column and 750mm offset from 400mm square column)

Calculation of Load Center

Take moment about center of 300mm square column
                       x          =  (200+800)*3/(100+500+200+800)
                                   = 1.875m

Calculation of Bearing 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 Punching Shear

Max. Shear Resistance   = 0.5ud[0.6(1-fck/250)](fck/1.5)

At 300mm column face
                                     = 0.5*300*4*440*0.6*(1-30/250)*(30/1.5)
                                     = 2788kN > (1.35*500+1.5*100 = 825kN)

At 400mm column face
                                     = 0.5*400*4*440*0.6*(1-30/250)*(30/1.5)
                                     = 3717kN > (1.35*800+1.5*200 = 1380kN)
Hence, Punching Shear is OK

Maximum bending was calculated
It was 209kNm, and it occurred at1.05m from 300mm square column

Design for Bending
Longitudinal reinforcements

MEd                                 = 209kNm                                      
K                                    = M/[b*(d^2)*fck]
K                                    = 209*E6/[1000*(440^2)*30]
K                                    = 0.04

Kbal = 0.167 (balance condition 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                                     = 440[0.5+(0.25-0.04/1.134)^0.5]
z                                     = 424mm  (Z/d = 0.96>0.95)
Therefore
Z                                    = 0.95*440
                                      = 418mm

As                                  = 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,
Proved T16@150mm spacing

Transverse reinforcements

Bending Moment             = 257*(1^2)/2                          
MEd                                 = 128.5kNm                             
K                                    = M/[b*(d^2)*fck]
K                                    = 128.5*E6/[1000*(440^2)*30]
K                                    = 0.022

Kbal = 0.167 (balance condition 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                                     = 440[0.5+(0.25-0.022/1.134)^0.5]
z                                     = 431mm  (Z/d = 0.97>0.95)
Therefore
Z                                    = 0.95*431
                                      = 409.5mm
As                                  = 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
Provide T12@150mm spacing