Design a plain concrete footing for a column of 400 mm x 400 mm carrying an axial load of 400 kN under service loads. Assume safe bearing capacity of soil as 300 kN/m2 at a depth of 1 m below the ground level. Use M 20 and Fe 415 for the design.
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 σbr 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.
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