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 metres 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.

AIM

To design a square footing for the column size of 400x400 mm². And to check whether the gross bearing pressure of the footing is within the safe bearing capacity. And to provide sufficient depth and the longitudinal steel for the resisting bending and shear and to also check the sufficient anchorage for the column steel into the footing.

PROCEDURE

Given Data:

Size of column = 400mm x 400 mm²

Factored load due to the column = 2000kN

SBC at the site = 150kN/m²

Concrete grade = M25

Steel grade = Fe415

Determining the size of footing

To find the size of the footing, we have the expression

P = F/A

Where,

P: soil pressure

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

A: Area of footing

Unfactored load due to column = Factored load/1.5

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

Assuming the self-weight of the footing as 10% of the total axial load

Therefore, self-weight of footing = 0.1 x 2000 = 200 kN

Hence the total unfactored load on the footing slab = 1333.33+ 200= 1533.33 kN

Substituting the above values in the formula, we get

A = F/P = 1533.33/150 = 10.22 m²

Width of footing = √A = 3.19 = 3.20m

And since we are adopting a square footing, the footing length would also be the same.

B = L = 3.20 m

Therefore the size of the footing = 3.20 m x 3.20 m

CALCULATING THE NET UPWARD PRESSURE (P_O)

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.

P_O = W/B² = 2200 / (3.2)² = 214.84 kN/m²

DETERMINING THE EFFECTIVE DEPTH OF FOOTING BASED ON THE MAXIMUM BENDING MOMENT

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

M = P_o x B/8 x (B-b)²

Here,

b is the column width

Substituting the values we get

M = 673.74 kNm

The ultimate moment

M_u = 1.5 x M = 1.5 x 673.74 = 1010.61 kNm

Also, M_u = 0.138 f_ck B d²

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

OBTAIN THE REQUIRED AREA OF STEEL BASED ON THE ULTIMATE MOMENT CALCULATED

The expression for A_st is as follows,

`A_(st)= frac{0.5f_(ck)}{f_y}(1-sqrt (1-frac{4.6M_u}{f_(ck)Bd^2) ]) Bd`

Where,

f_ck= 25

f_y = 415

M_u = 1010.61x10⁶ Nmm.

B = 3200 mm

d = 550 mm

Solving the above expression, we get

A_st = 5364.8 mm²

Now we need to provide the flexural reinforcement based on the A_st calculated which is done as follows:

So, adopting the 16mm dia bars (201mm²)

No. of bars required = A_st/area of 16mm dia bars = 5365/201 = 26.6 = 28 bars

28 x 201 = 5628>5365. Hence OK.

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

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

Therefore, Pt required= (A_st/BD) x 100 = (5628/3200x625) x 100 = 0.28% > 0.12%. Hence OK.

CHECK FOR ONE-WAY SHEAR

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

`V= P_oB[1/2[B-d]-d]`

Where,

P_o = 214.84 kN/m²

B = 3.20 m

d = 0.55 m

Solving for V, we get

V_u = 532.8 kN

Therefore, the nominal shear stress, (τv)

τv = V_u/Bd = 532.8 X 10³/3200x550 = 0.30 N/mm²

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

So, τ_c' = τ_c x k

Here,

τ_c shear strength of concrete section as per IS 456:2000 (TABLE 19)

k 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

τ_c = 0.39 N/mm²

Therefore, τ_c > τ_v. Hence the section is safe for one-way shear.

CHECK FOR TWO-WAY SHEAR

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 V_u for that particular section is given by:

`V= P_o[B^2-(b+d)^2]`

where

P_o = 214.84 kN/m²

B = 3.20 m

d = 0.55 m

b = 0.4 m

Solving this, we get

V_u = 2006 kN

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

τ_v = V_u/Bd = 2006 x 10³ / 3200 x 550 = 1.14 N/mm²

Now, obtain the design shear strength based on two-way shear (K_s x τ_c)

Here,

K_s is a 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, K_s = 0.5 + 1 = 1.5

However, the maximum K_s value shall not exceed 1 as per IS code

Therefore taking K_s = 1

Now, τ_c = 0.25√f_ck = 0.25√25 = 1.25

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

Reinforcement Bars Placement Plan

Section of the Footing