Design of Shallow Foundation (Isolated Footings)

DESIGN OF SHALLOW FOUNDATION ISOLATED FOOTING

Aim:

To design a square footing for a column of size 400x400 using some of the informations such as

• The compression axial load for the load combination of (1.5 DL + 1.5 LL) = 2000 KN.
• The safe soil bearing capacity = 150 KN/m2 at a depth of 2 metres below E.G.L.
• Grade of concrete = M25

Then to perform the following checks:

• If the gross bearing pressure under the footing is within safe bearing capacity
• Provide sufficient depth and longitudinal steel for resisting bending and shear
• Sufficient anchorage for column steel into the footing
• If the bearing pressure at column-footing joint is within 0.45 times the maximum concrete compressive strength

Introduction:

• Footings and other foundation units transfer loads from the structure to the soil or rock supporting the structure.
• The distribution of soil pressure under a footing is a function of the type of soil and the relative rigidity of the soil and the foundation pad.
• Most common types of footings used are:
  • Strip or wall footing
  • Spread footing
  • Stepped footing
  • Tapered footing
  • Pile cap
  • Combined footing
  • Mat or raft footing.

• This project is dealing with isolated footing for a column or spread footing.

Procedure:

- Given datas include:

• Column size = 400mmx400mm
• Compression axial load (1.5DL+1.5LL) = 2000kN - factored load
• Safe bearing capacity of soil at a depth of 2m below EGL = 150kN/`m^2`
• Assuming the self weight of footing and soil above it as 10% load.
• Grade of concrete = 25kN/`m^2`
• Assuming fy = 500N/`mm^2`

- Here a sloped footing is designed and the steps involved in the design are as follows:

1) Size of footing:

Unfactored or service load = `2000/1.5=1333.33kN`

Safe bearing capacity of soil , q_c = 150kN/`m^2`

Hence, the required base area = `(1333.33*1.1)/150=9.78m^2`

Hence, provide 3200mmx3200mm which gives 10.24`m^2` base area for footing. (This area later proves to be insufficient).

2) Thickness of footing slab based on one way shear.

Factored bearing pressure(Net soil pressure) = `2000/(3.2*3.2)=195.31`kN/`m^2`

Now, assuming 0.15% reinforcement in footing slab.

From table 19 of IS456:2000, design shear strength of concrete, `tau`_c= 0.29N/`mm^2` for M25 grade concrete.

From the condition, One way shear resistance`>=`One way shear force -->

0.29d x 3200 x d`>=`0.195x(1400-d)x3200

d`>=`562.88mm `~~`600mm

So, provide d = 600mm

Assuming 20mm dia bars and 70mm cover , total depth of the footing = 600+70+20+10 =700mm

3) Check for two way shear

Here, the critical section is ABCD, so shear resistance at critical depth = area of critical section x permissible shear stress

i.e, shear resistance = (400 + 600) x 4 x 600 x K_s`tau`_c =

K_s`tau`_c = 0.25`sqrtfck`= 1.25N/`mm^2`

Hence, shear resistance = 1000 x 4 x 600 x 1.25 = 3000kN

Now, the shear force = [3.2 x 3.2 - `(0.4+0.6)^2`] x 0.195 = 2944.5kN < 3000kN

Hence, safe.

4) Gross bearing capacity

Assuming unit weight of concrete and soil as 25kN/`m^3` and 18kN/`m^3`, we have,

Load on footing = 1333.3kN

Weight of footing = 3.2x3.2x0.7x25=179.2kN

Weight of soil = 3.2x3.2x(2-0.7)x18=239.61kN

Total load = 1751.81kN

Hence, gross bearing capacity = `1751.81/(3.2*3.2)=171.07`kN/`m^2` > 150kN/`m^2`. Hence, we need to increase the size of footing to 3600mmx3600mm.

So, now the weight of footing = 226.8kN

Weight of soil = 303.26kN

Total load = 1863.06kN

Now, gross bearing capacity = `1863.06/(3.6*3.6)= 143.75`kN/`m^2`<150kN/`m^2`. Hence safe.

So, the corrected net soil pressure = `2000/(3.6*3.6)=154.32`kN/`m^2`

5) Bending moment

We need to determine the area of steel in one direction as it is a square footing.

Consider effective depth = 600mm.

Bending moment at critical section is obtained as , Mu= 3600x16000.154x1600/2= 709.63kNm

`(Mu)/(bd^2)= (709.63*10^6)/(3600*600*600)=0.5475`N/`mm^2`

Now, from table 3 of SP16 for fck = 25N/`mm^2` and fy = 500N/`mm^2`, % of reinforcement = 0.1294%

Hence, area of steel = `0.1294/100*3600*600=2795.04``mm^2`.

So, no. of bars required is given by, `2795.04/314=8.9~~10`bars

Hence, provide 10 nos of 20mm dia bars both ways.

6) Development Length

For 20mm dia bars, Ld = `(0.87fyphi)/(4taubd)= 970.98mm~~1000mm.`

7) Providing slope in footing slab

Since the three critical sections are within a distance of 600mm from the face of the column , the full depth of footing slab is to be atleast 700mm, which is the provided depth of footing.

Hence, the one way shear force = 0.154(1600-600)3600=554400N=554.4kN

One way shear resistance is given by 0.29 x 3600 x 600= 626.4kN which is greater than the shear force. Hence safe.

So, we need to provide a depth of 700mm near the column and the depth can be reduced to about 400mm at the edges of the footing.

Result:

Sloped isolated footing for a square column has been designed with the given specification.