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

Piled foundationsIn 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

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

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

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

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)

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.

Strip Foundations

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.

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

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

[Assume the soil]

Red soil

Safe bearing capacity of the soil =

200KN/m2

Area of the footing req. = 210.66/200 = 1.053

m2

Assuming Square footing = B = A =

1.053

= 1.026m

Say = 1.2m

Hence provide size of footing = 1.2 x 1.2m

Up word pressure intensity (P)

P =load on the column/area of footing

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

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

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

d = 288.566 mm

Let provide 10mm bars

In the form of mesh with a clear 40mm

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

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

Check for shear force:

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

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

Breadth of the footing at the top of the

vertical section is:

485 – 305 = 180mm

Cover = 20mm

b1 = 230 – 20 – 20 = 190mm

Shear Areas q1 =

q1= 0.486

Shear stress is less than 1.0

So section is safe.

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

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]

Ast = 265.47mm2

Provide 10mm bars

No. of bars = Ast / A

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

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

So Ks = 1.0

Permissible Shear Stress = qc1 = Ks x qc

= 1.0 x 0.44 = 0.44 N/mm2

qc1 = 0.44 N/mm2

shear stress at critical section

qv = S1/b1 x d1 = 165.64 x 103 /2140 x 254.69

qv = 0.30 N/mm2

qc > qv

0.44 > 0.30

Hence the section is safe.

DESIGN OF FOOTING