Page 1 Module 11 Foundations - Theory and Design Version 2 CE IIT, Kharagpur Page 2 Module 11 Foundations - Theory and Design Version 2 CE IIT, Kharagpur Lesson 29 Design of Foundations Version 2 CE IIT, Kharagpur Page 3 Module 11 Foundations - Theory and Design Version 2 CE IIT, Kharagpur Lesson 29 Design of Foundations Version 2 CE IIT, Kharagpur Instructional Objectives: At the end of this lesson, the student should be able to: â€¢ understand and apply the design considerations to satisfy the major and other requirements of the design of foundations, â€¢ 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 Page 4 Module 11 Foundations - Theory and Design Version 2 CE IIT, Kharagpur Lesson 29 Design of Foundations Version 2 CE IIT, Kharagpur Instructional Objectives: At the end of this lesson, the student should be able to: â€¢ understand and apply the design considerations to satisfy the major and other requirements of the design of foundations, â€¢ 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 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 s = 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 s 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 Page 5 Module 11 Foundations - Theory and Design Version 2 CE IIT, Kharagpur Lesson 29 Design of Foundations Version 2 CE IIT, Kharagpur Instructional Objectives: At the end of this lesson, the student should be able to: â€¢ understand and apply the design considerations to satisfy the major and other requirements of the design of foundations, â€¢ 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 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 s = 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 s 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 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: 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 = a 0.9{(100q a /f ck ) + 1} 1/2 â€¦. (11.3) = 0.9[{100(0.2944)/20} + 1] 1/2 = 1.415 We have from Fig.11.29.1a: h = {(1250 - 400)/2}(tan a ) = 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 A st = 0.0012(1250)(670) = 1005.0 mm 2 . Provide 9 bars of 12 mm diameter (= 1018 mm 2 ) 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/m 3 and 20 kN/m 3 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 q a = 435.4375/(1.25)(1.25) = 278.68 kN/m 2 < 300 kN/m 2 Version 2 CE IIT, KharagpurRead More

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