Answer: (Assuming each column has same sectional area) Total vertical load W=50x3x40x40=240000kN, Each column take W'=0.5W/32=3750kN Mmax=1x50x3x40x75=450000kNm Average column load by Mmax=(3/4450000/(40x8)=1054.7kN According“Shear lag'”, the load in corner column =1.5x1054.7+3750=5332.05kN, Ac=5332050/315=16927mm2 the load in middle column=0.5x1054.7+3750=4277.35kN Am=4277350/315=13579mm2Answer: (Assuming each column has same sectional area) Total vertical load W=50x3x40x40=240000kN, Each column take W’=0.5W/32=3750kN Mmax=1x50x3x40x75=450000kNm Average column load by Mmax=(3/4)450000/(40x8)=1054.7kN According “Shear lag”, the load in corner column =1.5x1054.7+3750=5332.05kN, Ac=5332050/315=16927mm² the load in middle column=0.5x1054.7+3750=4277.35kN Am=4277350/315=13579mm²