Isolated footing design

Design of an Isolated RCC Column footing

1. FOR 5
TH
SEMESTER DIPLOMA IN CIVIL ENGINEERING
CONCEPTs AND PROBLEMS
DESIGN OF RCC COLUMN FOOTING

2. Learning Outcomes:
1.Concept of column footings.
2.Design criteria.
3.Design steps for column footing.
4.Reinforcement detailing.

3. Column footing:
RCC columns are supported by the foundation structures which are
located below the ground level are called footings.
Purpose of Footings:
1)To support the upper structure.
2)To transfer the Loads and moments safely to subsoil.
3)Footings are designed to resist the bending moment and shear
forces developed due to soil reaction.
S.J.(Govt.) POLYTECHNIC BENGALURU

4. Types of footing:

5. Isolated footing.
The footings which are provided below the column
independently are called as isolated footings.
This type of footing may be square ,
rectangular or circular in section
Isolated footing consists of thick slab which
may beflat or slopedor stepped.
S.J.(Govt.) POLYTECHNIC BENGALURU
P

6. The structural design of the footing includes the design of
1)Depth of footing
2)Reinforcement requirement
3)Check on serviceability
S.J.(Govt.) POLYTECHNIC BENGALURU

7. S.J.(Govt.) POLYTECHNIC BENGALURU

8. S.J.(Govt.) POLYTECHNIC BENGALURU

9. Design Considerations:
Minimum reinforcement : (As per IS456:2000, clause 26.5.2.1&2
The mild steel reinforcement in either direction in slabs shall not be
less than 0.15 percentof the total cross sectionalarea. However,
this value can be reduced to 0.12 percentwhen high strength
deformed bars or welded wire fabric are used
The diameter of reinforcing bars shall not exceed one eight of the
total thickness of the slab.
S.J.(Govt.) POLYTECHNIC BENGALURU

10. Design Considerations:
S.J.(Govt.) POLYTECHNIC BENGALURU

11. S.J.(Govt.) POLYTECHNIC BENGALURU

12. ..\IS456:2000.pdf
When the depth required for the above development length or the
other causes is very large, it is more economical to adopt a stepped or
sloped footing so as to reduce the amount of concrete that should go
into the footing.
SHEAR:
1.One way shear(Wide beam Shear):
One way shear is similar to Bending shear in slabs considering the
footing as a wide beam. Shear is taken along the vertical plane
extending the full width of the base
Lowest value of allowable shear in Table 13 of IS 456:2000
Is 0.35N/mm
2
is recommended.

13. S.J.(Govt.) POLYTECHNIC BENGALURU

14. S.J.(Govt.) POLYTECHNIC BENGALURU

15. S
H
E
R
A
F
A
I
L
U
R
E

16. 3. Bending Moment for Design:
Consider the entire footing as
cantilever beam from the face of
The column and calculate the
BM.
Calculate span for the
cantilever portion (Hashed portion)
= plx
??????
2
=
????????????
2
2
Substitute l=[(B-D)/2]
M
xx= p (
�−�
2
)
2
x
1
2
This is BM for 1m width of the beam

17. DESIGN STEPS:
1. Assume self weight of footing =0.1p
Total load w= P+0.1P
2. Area of footing required, A =
�
??????��
For square footing, Size of footing = ??????
For Rectangular footing, assume L
LxB=A
Provide L x B square footing,
Total Area = _ _ _ _ _ m
2

18. 3. Bending Moment
Pressure P =
??????
??????
�
_ _ _kN/m
2
4. Factored moment/m
M
u= p (
�−�
2
)
2
x
1
2

19. 5. Effective depth :
d
required =
M
u
0.138xf
ck
xb
increase depth for 1.75 to 2 times more than calculated value for shear
considerations.
6. Area of tension reinforcement :
M
u= 0.87f
y A
std(1-
A
st
fy
bdf
ck
)
This is a quadratic equation, calculate the value for A
stand consider the
minimum of values
Area of steel per m = A
st/span = _ _ _mm
2

20. Assume diameter bars
Area of one bar a
st=
πx????????????�
2
4
=___mm
2
Spacing of reinforcement , S =
1000a
st
A
st
7) Check for one way shear :
The critical section is taken at a distance “d” away from the face of the
column y-y axis.
Shear force per m,
V
u= p x B x [(
L−D
2
)-d]

21. Nominal Shear stress,??????
??????=
??????
??????
�??????
=_ _ _N/mm
2
Percentage steel =
100Ast
�??????
=
pt?
Refer table No. 19 of IS 456:2000 for ??????
??????
??????
??????= _ _ _N/mm
2
??????
vshould be less than ??????
??????,
design is safe against one way shear.
??????
v< ??????
??????

22. 8) Check for two way shear :
The critical section is taken at a distance
“d/2” away from the faces of the column
Shear force per m,
V
u= p x [A-(0.4+0.5)
2
]
Nominal Shear stress, ??????
??????=
V
u
b
0
d
b
0 = perimeter = 4( D + d)
Maximum shear stress permitted
??????
??????=0.25??????
????????????
??????
??????should be greater than ??????
v, Then design is safe against Punching shear /
two way shear.

23. 9) Development Length
Ld =
f
sxdiaofbar
4??????
&#3627408463;??????
=
0.87x415xdiaofbar
4&#3627408485;2.4
=37.6∅
For Fe 415 steel and M20 concrete the values substituted to the above
equation and Ld= 37.6∅
Taken to be , Ld= 40∅
available Ld=( L-D)/2 = _ _ _mm
This is alright

24. PLAN OF FOOTING
10) Reinforcement details:

25. 10) Reinforcement details:

26. PROBLEM 1:
Designasquarefootingtocarryacolumnloadof1100kN
froma400mmsquarecolumn.Thebearingcapacityofsoilis
100kN/mm
2
.UseM20concreteandFe415steel.

27. 1. Assume self weight of footing =0.1p= 0.1x 1100 =110kN
Total load w= P+0.1P = 1100+110= 1210kN
2. Area of footing required, A =
&#3627408484;
??????&#3627408437;&#3627408438;
=
1210
100
= 12.1m
2
Size of footing = ??????= 12.1= 3.478
Provide 3.5m x 3.5msquare footing,
Total Area = 12.25m
2

28. 3. Bending Moment
Pressure P =
??????
??????
&#3627408436;
=
1.5x1210
12.25
=148.16kN/m
2
4. Factored moment/m

29. BM about axis x-x passing through face of the
Column as shown in fig.
M
u= p x B x [
L−D
2
]
2
X
1
2
=148.16x3.5x[
3.5−0.4
2
]
2
X
1
2
=622.92kN−m
L = B for square footing
D = Size of column = 400mm = 0.4m
M
u=622.92kN−m

30. 5. Effective depth :
d
required =
M
u
0.138xf
ck
xb
=
622.92x106
0.138x20x3500
= 253.93mm
Adopt 500mm effective depth and overall depth 550mm. (increase
depth for 1.75 to 2 times more than calculated value for shear
considerations)
6. Area of tension reinforcement :
M
u= 0.87f
y A
std(1-
A
st
fy
bdf
ck
)
622.92 x 10
6
= 0.87 x 415 xA
stx500(1-
A
st
x415
3500x500x20
)
622.92 x 10
6
= 180525A
st-2.14 A
st
2

31. 622.92 x 10
6
= 180525A
st-2.14 A
st
2
2.14 A
st
2
-180525A
st+ 622.92 x 10
6
= 0
This is a quadratic equation, calculate the value for A
st and consider the
minimum of values
There fore, A
st= 3604.62mm
2
Area of steel per m = 3604.5/3.5 = 1029.85mm
2
Provide 12mm diameter bars
Area of one bar a
st=
πx122
4
=113.09mm
2

32. Spacing of reinforcement , S =
1000a
st
A
st
=
1000x113.09
1029.85
= 109.81mm
Providing 12mm diabars @ 100mm c/c.
7) Check for one way shear :
The critical section is taken at a distance “d” away from the face of the
column y-y axis.
Shear force per m,
V
u= p x B x [(
L−D
2
)-d] = 148.16x 1 x [(
3.5−0.40
2
)-0.50] = 155.57kN
Nominal Shear stress,??????
??????=
??????
??????
&#3627408463;??????
=
155.57x103
1000x500
= 0.31N/mm
2

34. Percentage steel =
100Ast
&#3627408437;??????
=
100x3604.62
3500x500
= 0.20
Refer table No. 19 of IS 456:2000 for ??????
??????
Since ,
%
steel ??????
??????
0.15 0.28
0.20 x
0.25 0.36
For ??????
??????at 0.2 = 0.28 +
(0.36−0.28)
(0.25−0.15)
x(0.2-0.15)
??????
??????= 0.32N/mm
2

35. ??????
vis less than ??????
??????, design is safe against one way shear.
8) Check for two way shear :
The critical section is taken at a distance
“d/2” away from the faces of the column
Shear force per m,
V
u= p x [A-(0.4+0.5)
2
]
= 148.16 x [12.25-(0.4+0.5)
2
] = 148.16x 11.44
=1695kN

36. b
0 = perimeter = 4( D + d)
= 4(400+500)
=3600mm
Nominal Shear stress, ??????
??????=
V
u
b
0
d
=
1695x103
3600x500
??????
??????= 0.941N/mm
2
Maximum shear stress permitted
??????
??????=0.25??????
????????????=0.2520= 1.11N/mm
2
since,??????
??????>??????
v, design is safe against Punching / two way shear.

37. 9) Development Length
Ld =
f
sxdiaofbar
4??????
&#3627408463;??????
=
0.87x415xdiaofbar
4&#3627408485;2.4
=37.6∅
For Fe 415 steel and M20 concrete the values substituted to the above
equation and Ld= 37.6∅
Taken to be , Ld= 40∅=40 x 12 = 480mm
available Ld=( 3500-400)/2 = 1550mm
This is alright

38. 10) Details of Reinforcement :

39. ANY QUESTIONS
?????
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