Design of Shallow Foundation (Isolated Footings)

Design a square footing for a column size of 400x400. The compression axial load for the load combination of (1.5 DL + 1.5 LL) is 2000 KN. The safe soil bearing capacity is 150 KN/m2 at a depth of 2 meters below E.G.L. Participants are free to go for either a tapered or stepped footing. Besides the total axial load, also account for the self-weight of the footing and soil above it. Assume M25 concrete grade. Do the following checks:

• If the gross bearing pressure under the footing is within the 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 the column-footing joint is within 0.45 times the maximum concrete compressive strength

Provide a simple sketch showing the size of the footing, maximum and minimum depth of the section, flexural reinforcement in both directions, and anchorage of column steel.

ANSWER:

COMPLETE STEPWISE PROCEDURE TO DESIGN AN ISOLATED SQUARE FOOTING BASED ON THE LOADING CONDITIONS OF A BUILDING

STEP ONE: ASSEMBLE THE BUILDING DATA

So, the very first step in the process is to collect all the relevant information like column load and soil bearing capacity regarding the building site. In our case, we have been provided with the following information about the structure:

1. Size of column = 400mm x 400mm
2. Factored load due to the column = 2000kn
3. SBC at the site = 150kn/m^2
4. Concrete grade = M25
5. Steel grade = Fe415

STEP TWO: OBTAIN THE SIZE OF THE FOOTING

As we all know that the formula for pressure is:

P = F/A

here, p-> soil pressure

F-> unfactored load of column + self-weight of footing and surcharge pressure

A-> Area of footing

So, we can easily obtain the area of footing from the above formula.

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

Unfactored load due to column = 2000/1.5 = 1333.33 kn

Similarly, assuming the self-weight of the footing as 10% of the total axial load

Therefore, footing self. wt. = 0.1 x 20000 = 200

Now, the total unfactored load on the footing slab = 1333.33+ 200= 1533.33 kn

substituting the above values in the pressure formula, we get

A = F/P = 1533.33/150 = 10.22 m^2

breadth of footing = √A = 3.19 = 3.20m

And since we are adopting a square footing, the footing width will also be the same.

B = L = 3.20 m

STEP THREE: CALCULATE THE NET UPWARD PRESSURE (Po)

The expression for the net upward pressure is as follows. Also this time, the total factored axial load plus self-weight of the footing and surcharge pressure shall be considered.

Po = W/B^2 = 2200 / (3.2)^2 = 214.84 kn/m^2

STEP FOUR: DETERMINE THE EFFECTIVE DEPTH OF FOOTING BASED ON THE MAXIMUM BENDING MOMENT

The maximum bending moment for footings is given by the following expression,

M = Po x B/8 x (B-b)^2

Here, b -> column width. Substituting the values we get

M = 673.74 KNm

Getting the ultimate moment,

Mu = 1.5 X M = 1.5 X 673.74 = 1010.61 KNm

Mu = 0.138fckBd^2

solving for d we get,

d = 302 mm

However, this depth is insufficient as the code states to carry the column confinement into the footing which roughly accounts for 300mm. Therefore we need to go for a higher effective depth value.

So, based on practice, we adopt 550mm as the effective depth of footing.

d = 550mm

Assuming 75mm cover, total depth of the footing, D = 550 + 75 = 625 mm

STEP FIVE: OBTAIN THE REQUIRED AREA OF STEEL BASED ON THE ULTIMATE MOMENT CALCULATED

The expression for Ast required goes as follows,

Solving the above expression, we get

Ast = 5363 mm^2

Now we need to provide the flexural reinfrocement based on the Ast claculated which is done as follows:

So, adopting the 16mm dia bars (201mm^2)

No. of bars required = Ast/area of 16mm dia bars = 5363/201 = 26.68 = 28 bars.

28 x 201 = 5628>5363. Hence OK.

Spacing = B/no. of bars = 3200/28 = 114.28 = 115mm c/c.

Now, as we know the minimum Ast for footing as per the codebook is equal to 0.12% of the gross footing area.

Therefore, pt required= (Ast/BD) X 100 = (5628/3200x625) x 100 = 0.32%>0.12%. Hence OK.

STEP SIX: CHECK FOR ONE WAY SHEAR

As we know, the critical section for the one-way shear lies at a distance d from the face of the column.

So, the net shear force at the critical section for one-way shear is given by

solving for V, we get

Vu = 584.36 kn

Therefore, the nominal shear stress, (τv)

τv = Vu/Bd = 584.36 X 10^3/3200x550 = 0.33 N/mm^2

• Now we need to calculate the tau c value to see if our concrete section is capable of taking this nominal shear stress or not.

So, τc' = τc x k

here, τc-> shear strength of concrete section as per IS456-2000...(TABLE 19)

k-> it is taken as the unit value for section depth more than 300mm

So, from table 19 by interpolation, as we have provided pt as 0.32%, we get

τc = 0.39 N/mm^2

Therefore, τc > τv. Hence the section is safe for one-way shear.

STEP SEVEN: CHECK FOR TWO-WAY SHEAR

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

So, the expression to calculate net shear force Vu for that particular section is given by:

Solving this, we get

Vu = 2006 KN

Therefore, nominal shear stress due to punching shear (τv),

τv = Vu/Bd = 2006 x 10^3 / 3200 x 550 = 1.14 N/mm^2

• Now, obtain the design shear strength based on two-way shear (Ks x τc)

here, Ks -> mathematical constant based on the L/B ratio of footing

Ks = 0.5 + βc .. (βc->L/B ratio)

since we have adopted a square footing, βc = 1

Therefore, Ks = 0.5 + 1 = 1.5. However, the maximum Ks value shall not exceed 1 as per the codebook.

Therefore take Ks = 1

Now, τc = 0.25√fck = 0.25√25 = 1.25

So, τc > τv. Hence our section is OK for the two-way shear as well.

STEP EIGHT: DRAWING REINFORCEMENT DETAILS OF THE FOOTING USING AUTOCAD

• Reinforcement Bars Placement Plan

• Section of the Footing