this is a simple design,which is very useful to mini projects ,most of the studunts has bother about projects

1. Foundations:
 The foundations of houses must carry
the dead loads (weight of the
structure) of the walls, roof and floors
etc
 Introduction to foundation types:
strip foundation:
 The strip foundation is basically a
strip, or ribbon, of in situ concrete running
under all the load bearing walls.

The foundation size can be determined by
referring to the Building Regulations

2. Piled foundations
In clay and other cohesive soils piles can be
used to distribute the loads into the ground
through the friction forces along the length of the
pile sides
Piles are usually made from insitu or
precast concrete but can also be steel and
timber
Raft foundations
Rafts are an expensive form of
construction, probably the most expensive of
the three, and are used where only a very
low load can be applied, for example, on soft
or variable ground

3. Concrete:
House foundations are invariably
formed in concrete. It is available in a
range of strengths and is usually brought
onto site ready-mixed as, and when,
required.
It is a mixture of several constituents
which behaves as a single material. In its
simplest form concrete comprises cement,
aggregate and water

4.  The major constituent by weight in
concrete is aggregate - stone with a range
of particle size from 40mm down to
0.1mm. The aggregate is a mixture of:
 coarse aggregate - naturally occurring
gravel or crushed rock
 fine aggregate - sand or crushed rock.
 The aggregate is bound together by
cement paste, a mixture of cement and
water

5.  Properties
 The properties of the cement paste are
extremely important and largely
determine the properties of the concrete:
 it must be fluid enough for some time
after mixing to allow the concrete to be
placed and

6.  it must then set and gain strength so that it
binds the aggregates together to make a
strong material
 The mechanism by which cement sets and
hardens depends on the type of cement,
usually due to a chemical reaction
between the cement and the mixing water
(eg Portland cement)

8. Reinforced concrete contains steel
reinforcing rods, usually 20-30mm in
diameter. These rods are positioned where
the principal tensile stresses will occur in
the structure, and then the concrete is
poured and compacted around the
reinforcement. Reinforced concrete is
therefore a composite material, where the
concrete takes the compressive forces and
the reinforcing steel takes the tensile forces.

9.  Strip Foundations

11.  Strip foundations, the most common
form, can either be 'traditional' or trench-fill
(see below). They are usually 500 to
700mm wide and as deep as necessary
for the type of ground
 In weaker ground the foundation has to
be wider than 700mm to spread the
building load over an adequate area of
ground.

12.  Reasons for choosing traditional strip
foundations:
 Proven method, most builders are familiar with
traditional strip foundations
 Mistakes (eg, setting out) are not too
expensive to rectify once concrete is poured
 Builder may want work to keep bricklayer
occupied
 Services will mostly cross the wall above the
concrete - so not an immediate problem
 Cheaper than trench fill for wider foundations

13. 4.1DESIGN OF FOOTING FOR FOUNDATION
 Load on the column ()= 171.975 KN
= 171.975 x 103 N
 Self wt. of footing = 15% of column load
= 15/100 x 171.975
= 25.79KN
Factored load = 25.79 x 1.5
= 38.685 KN
 Total load =171.975+38.685
=210.66KN

14. [Assume the soil]
 Red soil
Safe bearing capacity of the soil =
200KN/m2
Area of the footing req. = 210.66/200 = 1.053
m2

15. Assuming Square footing = B = A =
1.053
= 1.026m
Say = 1.2m
Hence provide size of footing = 1.2 x 1.2m

16. Up word pressure intensity (P)

17. P =load on the column/area of footing

18.  Depth of footing from b.m consideration:
 Critical section B.M. = B-b/2
 Here l = 1.2m
 B = 0.23m
 B.M = 1.2 – 0.23/2 = 0.485 m (or) 485mm

20. B.m = 119.42 x 103 x 1.2 x 0.4852/2
= 16.854 x 103 N-m
= 16.854 x 106 N-m

21. Using working stress method:
Qbd2 = B.m
Use m20 grade concrete = Q = 0.88
Fe 415 grade steel = st = 230 N/mm2
B = 0.23m = 230 mm

22.  d = 288.566 mm
 Let provide 10mm  bars
 In the form of mesh with a clear 40mm

24. Effective cover = 40 + 10 + 10/2 = 55mm
 Overall depth of footing:
 = Effective depth + effective cover
 = 288.566 + 55 = 343.566mm
 Hence provided overall depth of footing =
360mm
 Actual effective depth = 360 – 55 = 305
mm
 d = overall depth – effective cover
 d = 305 mm

25.  Check for shear:
 Depth for punching shear consideration
 Punching load = column load – upward
pressure intensity on column area
 = 171.975 x 103 – 119.42 x 103 x (0.23 x
0.23)
 = 171975 – 6317.318
 = 165.65 x 103 N
 Ast = 2654.47 N
 = 265.47/78.53 = 3.380 No. Say 4 No.
  provide 4 No. of 10 mm  bars in both
directions

26. Check for shear force:

27.  Check for one way shear:
 The critical section for one way shear is
taken at a distance equal to the effective
depth from the column face (i.e. let depth
of footing at edge = 200mm)
 The overall depth at critical section
= 278.247

29.  The effective depth at critical section
 d1 = 278.247 – 55 (effective cover)
 d1 = 223.247mm
 Shear force at critical section = S1
 S1 = Upward pressure x length of
footing x shear constant
 (Where shear constant = 0.144 as per
code)
 S1 = 119.42 x 103 x 1.2 x 0.144
 S1 = 20.635 x 103 N

31. Breadth of the footing at the top of the
vertical section is:

32.  485 – 305 = 180mm
 Cover = 20mm
 b1 = 230 – 20 – 20 = 190mm

33. Shear Areas q1 =
q1= 0.486
Shear stress is less than 1.0
So section is safe.

34.  Check for two way shear:
 The critical section for two ways Shear at the
distance of half of the effective depth from the
force of the column all around overall depth of the
footing at a distance d/2.
 d/2 = 305/2 = 152.5 mm from the column face.
 Permissible punching shear stress is 1 N/mm2
 Equating the punching resistance to the punching
load
4 x 230 x D x l = 165.65 7 x 103N

35. Hence provide overall depth of 360mm
B.M Consideration for steel required:
Where B.m = 16.854 x 106 N-mm
J = 0.905
St = 230 N/mm2
d=305mm [Actual effective depth]

36. Ast = 265.47mm2
Provide 10mm  bars
No. of bars = Ast / A 

38. l1 = d/2 + 180 = 305/2 + 180 = 332.5mm
= 332.5/2 x 160 = 109.69mm
x=x1 + 200 = 109.69 + 200 = 309.69mm
effective depth of the critical section
d1 = x – effective cover
d1 = 309.69 – 55 = 254.69mm

39. Critical perimeter:
= 4 x (dimension of column + effective depth)
b1=4 x (230 + 305)
b1=2140 mm
shear force at critical section = S1
S1= upward pressure x (l2 – dimension of column2)
S1= 119.42 x 103 x (1.22 – 0.232)
S1= 165.64 x 103N
Ks = 0.5 + Bc = 0.5 + 1 = 1.5
Ks = should not be taken greater than 1.0

40. So  Ks = 1.0
1 = Ks x qc
Permissible Shear Stress = qc
= 1.0 x 0.44 = 0.44 N/mm2
q1 = 0.44 N/mm2
c
shear stress at critical section
q= S1/b1 x d1 = 165.64 x 103 /2140 x 254.69
v q= 0.30 N/mm2
v q> qc v
0.44 > 0.30
Hence the section is safe.

41. DESIGN OF FOOTING