Design a reinforced concrete combined footing (rectangular) for two columns lo

Design a reinforced concrete combined footing (rectangular)
for two columns located 3.60 metres apart. The overall
sizes of the columns are 40cm x 40cm and 69cm x 60cm and
the loads on them are 1000 kn and 1500 kn respectively. The
space available for the width of the footing is restricted
to 180cm. The S.B.C. of the soil is 280 kn/mm2 use M15
concrete.

Question Posted / rajamrizwan

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Design a reinforced concrete combined footing (rectangular) for two columns located 3.60 metres ap..

Answer / Maneesha Meena

First, calculate the total load: Load_total = 1000 kn + 1500 kn = 2500 kNn Next, determine the total area needed to support this load (A_total): A_total = Load_total / S.B.C.n Let L be the length of each column (L = 360 cm). Since the space available for the width of the footing is restricted to 180 cm, let W be the width of the footing. The total area of the combined footing is: A_total = 2 * (W * L)n Solving for W: W = sqrt((A_total / 2) / L) = sqrt((2500 kN / 280 kn/mm2 / 2) / 360 cm) = 2.74 mn To determine the required reinforcement, let d be the depth of the footing (d > L), b_c be the effective breadth of the column at the base, and fy_c be the yield strength of M15 concrete.n The ultimate load that each column can carry is: Ultimate_Load = fy_c * b_c * dn Since each column has an area of 40cm x 40cm and 69cm x 60cm, let's assume the effective breadth of the columns at the base (b_c) to be a value between these two. Let's try b_c = 50 cm.n Ultimate_Load = fy_c * 50 cm * dn The required footing depth can be determined by solving:n Ultimate_Load > Load_total n Solving for d: d > (Load_total / (fy_c * 50 cm))

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