Foundations - Theory and Design: Design of Foundations Notes | EduRev

: Foundations - Theory and Design: Design of Foundations Notes | EduRev

 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: 
 
• 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 
 
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: 
 
• 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 
 
 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: 
 
• 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 
 
 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
 ….  (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, Kharagpur 
 
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