**Solution 1:**

Plain concrete footing is given in secs.11.28.2(A)1 and 11.28.5(b).

**Step 1: Transfer of axial force at the base of column**

It is essential that the total factored loads must be transferred at the base of column without any reinforcement. For that the bearing resistance should be greater than the total factored load Pu.

Here, the factored load Pu = 400(1.5) = 600 kN.

The bearing stress, as per cl.34.4 of IS 456 and given in Eqs.11.7 and 8 of sec.11.28.5(g) of Lesson 28, is brσ = 0.45 fck (A1/A2)1/2 (11.7)

with a condition that

(A1/A2)1/2 ≤ 2.0 (11.8)

Since the bearing stress brσ at the column-footing interface will be governed by the column face, we have A1 = A2 = 400(400) = 160000 mm2. Using A1 = A2, in Eq.11.7, we have

Pbr = Bearing force = 0.45 fck A1 = 0.45(20)(160000)(10-3) = 1440 kN > Pu (= 600 kN).

Thus, the full transfer of load Pu is possible without any reinforcement.

**Step 2: Size of the footing**

Then, the base area required = 400(1.15)/300 = 1.533 m2. Provide 1250 x 1250 mm (= 1.5625 m2) as shown in Fig.11.29.1. The bearing pressure qa = 400(1.15)/(1.25)(1.25) = 294.4 kN/m2.

**Step 3: Thickness of footing**

The thickness of the footing h is governed by Eq.11.3 of sec.11.28.5 of Lesson 28. From Eq.11.3, we have

tan ≤α0.9{(100qa/fck) + 1}1/2 …. (11.3)

≤ 0.9[{100(0.2944)/20} + 1]1/2

≤ 1.415

We have from Fig.11.29.1a:h = {(1250 - 400)/2}(tanα) = 601.375 mm

Provide 1250 x 1250 x 670 mm block of plain concrete.

**Step 4: Minimum reinforcement**

The plain concrete block 1250 x 1250 x 670 shall be provided with the minimum reinforcement 0.12 per cent for temperature, shrinkage and tie action.

Minimum Ast = 0.0012(1250)(670) = 1005.0 mm2.

Provide 9 bars of 12 mm diameter (= 1018 mm2) both ways as shown in Fig.11.29.1b. The spacing of bars = (1250 - 50 - 12)/8 = 148.5 mm c/c. Provide the bars @ 140 mm c/c.

**Step 5: Check for the gross base pressure**

Assuming unit weights of concrete and soil as 24 kN/m3 and 20 kN/m3

Service load = 400.00 kN

Weight of footing = (0.67)(1.25)(1.25)(24) = 25.125 kN

Weight of soil = (0.33)(1.25)(1.25)(20) = 10.3125 kN

Total = 435.4375 kN

qa = 435.4375/(1.25)(1.25) = 278.68 kN/m2 < 300 kN/m2

Hence, o.k.

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