m11l29 - Module 11 Foundations - Theory and Design Version...

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Module 11 Foundations - Theory and Design Version 2 CE IIT, Kharagpur
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Lesson 29 Design of Foundations Version 2 CE IIT, Kharagpur
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Instructional Objectives: At the end of this lesson, the student should be able to: understand and apply the design considerations to satisfy the major and other requirements of the design of foundations, design the plain concrete footings, isolated footings for square and rectangular columns subjected to axial loads with or without the moments, wall footings and combined footings, as per the stipulations of IS code. 11.29.1 Introduction The two major and some other requirements of foundation structures are explained in Lesson 28. Different types of shallow and deep foundations are illustrated in that lesson. The design considerations and different codal provisions of foundation structures are also explained. However, designs of all types of foundations are beyond the scope of this course. Only shallow footings are taken up for the design in this lesson. Several numerical problems are illustrated applying the theoretical considerations discussed in Lesson 28. Problems are solved explaining the different steps of the design. 11.29.2 Numerical Problems Problem 1: 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/m 2 at a depth of 1 m below the ground level. Use M 20 and Fe 415 for the design. 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 P u . Here, the factored load P u = 400(1.5) = 600 kN. Version 2 CE IIT, Kharagpur
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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 br σ = 0.45 f ck ( A 1 / A 2 ) 1/2 (11.7) with a condition that ( A 1 / A 2 ) 1/2 2 . 0 (11.8) Since the bearing stress br at the column-footing interface will be governed by the column face, we have A 1 = A 2 = 400(400) = 160000 mm 2 . Using A 1 = A 2 , in Eq.11.7, we have P br = Bearing force = 0.45 f ck A 1 = 0.45(20)(160000)(10 -3 ) = 1440 kN > P u (= 600 kN). Thus, the full transfer of load P u is possible without any reinforcement. Step 2: Size of the footing Version 2 CE IIT, Kharagpur
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Let us assume the weight of footing and back fill soil as 15 per cent of P u . Then, the base area required = 400(1.15)/300 = 1.533 m 2 . Provide 1250 x 1250 mm (= 1.5625 m 2 ) as shown in Fig.11.29.1. The bearing pressure q a = 400(1.15)/(1.25)(1.25) = 294.4 kN/m 2 . Step 3:
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m11l29 - Module 11 Foundations - Theory and Design Version...

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