Design of Shallow Foundation (Isolated Footings)

AIM: - To design a square footing for the column size of 400X400 mm. And to check whether the gross bearing pressure of the footing is within the safety or not. And to provide the sufficient depth and the longitudinal steel for the resisting bending and shear in footing.

INTRODUCTION: - The footing is a structural member that safely transmits the load of the upper building to the ground, and is the first construction member of the building. The basic method of footing design is to distribute the load so that the size per area of ​​the load transmitted from the upper part of the building is less than the strength that the ground can support, that is, the bearing capacity. With this function, the footing is a structure installed between the ground and the column or wall immediately above it, and the important point in the basic design is to reduce the total amount of settlement and prevent the occurrence of immobile settlement.

PROCEDURE: -

Design of Footing:

• Size of the column: 400mmX400mm.
• The factored load due to the column = 2000 KN/sq.m.
• Soil Bearing Capacity (SBC) = 150 KN/sq.m.
• Grade of concrete = M25.
• Grade of steel = Fe415.

CALCULATIONS: -

1. Find the size of the Footing:

To find the size of the footing, we have the expression from which we can find the size of the footing and the expression is given below:

P= F/A

where P = Soil Pressure.

F = Unfactored Load of column + self-weight of the footing and the surcharge pressure.

A = Area of Footing.

So, in order to find out the footing use the above expression P= F/A

But first of all, we need to find the unfactored load in addition to the self-weight of the footing.

So, Unfactored load due to column = Factored load/1.5.

Unfactored load due to column = 2000/1.5

Unfactored load due to column = 1333.33 KN.

Now assume the self-weight of the footing as 10%. of the axial load or the superimposed load.

∴ Self-weight of the Footing = 10% of the superimposed load.

where, Superimposed Load = 2000 KN.

Superimposed load = 10% X 2000.

Superimposed Load = (10/100) ×2000

Superimposed Load = 200 KN or WD= 200 KN

And now Total unfactored Load on footing (F) = WL+WD

F = 1333.33 +200

F = 1533.33 KN

Now, we have the value for F and P so now substitute these values in the expression P= F/A

Therefore, P= F/A

A= F/P

A=1533.33/150

A=10.22m^2

Area of footing = 10.22m^2

Breadth of the Footing (B) = √Area

B = √10.22

B = 3.19m = 3.20 m

Since we are adapting the square footing, so the width of the footing also will be the same as of the breadth of the footing.

So, we can say that B=L=3.20 m

Therefore, the size of the footing = 3.20 m X 3.20 m

2. To find the Upward Pressure:

Formula for the net upward pressure is as follows:

Upward Pressure (Po)= Total load (W) /Size of Footing.

Upward Pressure (P0) = 2200/3.20^2

Upward Pressure (PO) = 214.84 KN/sq.m.

3. To determine the effective depth of the footing based on the maximum Bending Moment:

The maximum bending moment for the footing can be found out by the below expression:

M=P0 B/8×(B−b) ^2

where, b= Column width = 0.4 m.

P0 =214.84 KN/sq.m.

B = 3.20m.

Now put these values

M=P0 B/8×(B−b) ^2

M = 214.84× (3.20/)8× (3.20−0.4) ^2

M = 214.84×0.4×7.84

M = 673.74673.74KNm

Now find the Ultimate Moment (Mu)

Ultimate moment can be found out by using the below expression:

Mu= 1.5×M

where M = 673.74 KNm

Mu= 1.5×673.74

Mu= 1010.61 KNm

The depth of the footing can be obtained by using the below formula:

D=d×50

We have to find the value of D, so in order to find the D value first we need to find the value of d.

And we can find the d value just by using the below expression:

Mu=0.138fck×Bd^2

where Mu = 1010.61 X 10^6 Nmm

fck = 25

B = 3200 mm

Now put these values in the equation:

Mu=0.138fck×Bd^2

1010.61 X 10^6 =0.138×25×3200×d^2

d^2= (1010.61×10^6)/ (0.138×25×3200)

d^2= (1010.61×10^6)/11040

d^2=91540.76

d=√91540.76

d=302mm

Our depth d = 302 mm and this much of depth is not sufficient according to the IS codes to carry the column confinement into the footing. Therefore, we need to go for the higher value for the effective depth.

So, we adopt 550mm as the effective depth value of the footing.

Therefore d= 550mm.

Assume 75mm cover, so the total depth of the footing D = 550 + 75 = 625mm.

4. Obtain the required Area of reinforcement:

To find the Ast we use below expression:

Ast=0.5fck/fy[1−√1−(4.6Mu/fckBd^2]x Bd

where, fck= 25

fy = 415

Mu = 1010.61×10^6 Nmm

B = 3200 mm

d = 550 mm

Now put all these values

ASt=(0.5×25)/415×[1−√1−(4.6×1010.61×10^6)/(25×3200×550^2]x3200x550

After solving the above equation, we get the below result:

Ast= 5280 sq.mm.

Now we need to provide the reinforcement based on the Ast that we already calculated:

So, assume 16 mm diameter bars. π/4× (16^2) = 201mm^2

Number of bars required = Ast/Area of 16mm diameter bars

where Ast = 5280 sq.mm

Area of 16mm diameter bars = 201 sq.mm.

Number of bars required = 5280/201

Number of bars required = 26.26 = 28 bars.

So here the Required number of bars = 28 bars.

Now we have to check if the required number of bars is sufficient or we need to redesign, so for doing this the multiplication of number of bars and the area of the 16mm bars must be greater that the Ast.

Therefore, Amount of bars/Area of 16mm bars = 28×201

= 5628.

which is greater than the Ast, 5628>5280Hence OK.

Spacing of the bars = B/No. of bars

= 3200/28

Spacing of bars = 114.28 mm c/c = 115 mm c/c

Spacing of bars = 115 mm c/c

As we know that the minimum Ast for footing is equal to 0.12% of the gross footing area according to the IS code.

Therefore, Pt required = [Ast/(B×D)] ×100

where Ast = 5280 sq.mm

B = 3200 sq.mm

D = 625 sq.mm.

Now substitute all these values

Pt required = [Ast/(B×D)] ×100

Pt required = [5280/ (3200×625)] ×100

Pt required = 0.27%

0.27% which is greater than 0.12%.

0.27%>0.12% Hence OK.

5. Check for One-Way Slab:

Now we know that the critical section for the one-way shear lies at the distance d from the face of the column.

Therefore, the net shear force at the critical section for the one-way shear is given by the below expression:

V=Po×B[1/2[B−d] −d]

where Po = 214.84 KN/sq.m

B = 3.20 m

d =0.55 m

Now put all these values in the given expression, then we will the values for V:

V=Po×B[1/2[B−d] −d]

V=214.84×3.20[1/2[3.20−0.55] −0.55]

V=687.5[1/2[2.65] −0.55]

V=687.5[0.775]

V=532.81KN

To find the Nominal stress, below is the expression which can be used to find out the nominal stress:

Tu=V/Bd

where V = 532.81×103N532.81×103N

B =3200 mm

d = 550 mm

Now put all these values

Tu=V/Bd

Tu =532.81/ (3200×550)

Tu =0.310N/sq.mm

And after finding the Tu now we have to find the Tc value in order to find out that the concrete section is capable of taking this nominal shear stress or not

So T’c = Tc xk

T’c =shear strength of concrete

and k = Unit value for section whose depth is more than 300 mm

So, from table 19 of IS 456:2000 by the interpolation, we provided pt as 0.32%, then we will get:

T’c =0.39 N/sq.mm.

Therefore Tc > Tu

Hence the section is safe for the one-way shear.

6. Check of two-way slab:

As we know that the critical section for two-way shear is spread across the footing at a distance of d/2 from the face of the column.

Below is the given expression to calculate the net shear force Vu:

Vu= Po×[B^2−(b+d) ^2]

where Po = 214.84 KN/sq.m

B = 3.20 m

d =0.55 m

b =0.4 m

Vu= 214.84× [3.20^2−(0.4+0.55) ^2]

Vu= 2005 KN

Nominal stress shear due to the punching shear (Tu) expression is given below:

Tu =Vu/Bd

Here 2005 X 10^3 N

B =3200 mm

d = 550 mm.

And now put all these values

Tu =Vu/Bd

Tu = (2005×103^3)/ (200×550)

Tu = 1.14 N/sq.mm.

Now design shear strength based on two-way shear:

Ks X Tc

where Ks = mathematical constant based on L/B ratio of footing.

Ks = 0.5+βc

where βc is the L/B ratio of the footing.

Because we choose square footing βc is taken as 1.

Therefore Ks = 0.5+1

Ks = 1.5.

But according to the IS code we cannot take the value of Ks more than 1,so here we will choose Ks as 1.

So, Ks =1.

Now, Tc =0.24√fck

Tc =0.24√25

Tc = 1.25

Therefore Tc > Tv

Hence the section is safe for the two-way shear also.

REINFORCEMENT DRAWINGS:

Reinforcement bars placement plan:

Section of the footing: