Design of Shallow Foundation (Isolated Footings)

Question

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 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

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.

Please do go through the solved examples in the document provided with the challenge.

Answer

Aim:To 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.

Introduction:

Footings are structural members used to support columns and walls and transmit their loads to the underlying soils. Reinforced concrete is a material admirably suited for footings and is used as such for both reinforced concrete and structural srteel buildings, bridges, towers, and other structures.The permissible pressure on a soil beneath a footing is normally a few tons per square foot. The compressive stresses in the walls and columns of an ordinary structure may run as high as a few hundred tons per square foot. It is therefore necessary to spread these loads over sufficient soil areas to permit the soil to support the loads safely.Not only is it desired to transfer the superstructure loads to the soil beneath in a manner that will prevent excessive or uneven settlements and rotations, but it is also necessary to provide sufficient resistance to sliding and overturning.

Procedure:

Given data:

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

Assuming the self weight of footing and soil above it as 10% load.

Grade of concrete = 25kN/m

Assuming fy = 500N/mm

Here a sloped footing is designed and the steps invoived in the design are as tollows

Step 1) Size of footing

a)Load from column = 2000

10% of column axial load = 200

2000+200 = 2200

b) Area of footing = Load on footing / SBC

= 2200/150

= 14.66 m2

To get the dimension = root of 14.66 = 3.82m

Assuming the size of = 4m X 4m = 16m² > 14.66

Hence ok.

Step 2) Net upward pressure

σ = Forece /Area = (1.5 × Total load on footing)/ 16

σ = (1.5 × 2200)/16

= 206.25 kN/m²

Step 3) Bending Moment

Critical section of footing as that force of the column

SL = [(400/2) - (400/2)]

= 1800 mm

Load on critical section = σ × bf

= 206.25 × 4 = 825 kN/m

Bending moment = wl²/2 = (825 × 1.8²)/2 = 1336.5 kNm

Step 4) Depth of footing

M_u = 0.138 f_ckbd² (Moment of restraint)

1336.5 × 10⁶ = 0.138 × 25 × 4000 × d²

d = 311.2 mm

We take effective depth 2 to 2.5 times greater

Take,d = 650 mm

Toatal depth D_f = 650 + 50 = 700 mm

Step 5) Reinforcement of footing

Bending moment = 1336.5 × 10⁶ Nmm

P_t = 50 f_ck / f_y [1-√1-(4.6M_u/f_ckbd²)] = 50 × 25/500[1-√1-(4.6 × 1336.5 × 10⁶/25 × 4000 × 650²)]

= 0.189

A_st = (P_t/100) × b × d = (0.189 × 4000 × 650) /100 = 4914 mm²

Hence A_st(req) = 4914 mm²

Take 16 mm dia bars,

Checking dia of bars of 16 mm

No of bars =4914/(π/4) × 16² = 24.45 ≅ 25 Nos

We provide 25 Nos of 16 mm dia bars.

Check sapacing for allowable as per IS 456:2000

= (4000 - (50 +50) - (8+8)) / 25 = 155.36 mm < 180 mm

Hence its okay.

A_st (prov) = 25 × (π/4) × 16² = 5026.54 mm² > 5902 mm²

Hence its okay.

Step 6) Check for one way slab

V_u = Upward pressure × Area of height(b_cs)

b_cs = 4000/2 - 400/2 - 650 = 1150 mm = 1.15 m

V_u = 206.25 × (4 × 1.15) = 948.75 kN

Now, τ_v = V_u /bd = 948.75 × 10³ /(4000 × 650) = 0.36 N/mm²

Pt = 0.19

Here τc > τ_v .Hence its Safe.

Check for development length

As per IS456:2000 Ld = Φσ_s / 4τ_bd, τ_bd value from table 2.6.2.1.1

σ_s = 0.55 f_y as per table 22

Ld = (20 × 0.55 × 500) / 4 × 1.4

From calculation, Φ ia used for column = 20 mm

Ld = (20 × 0.55 × 500) / 4 × 1.4 = 982 ≅ 980 mm

Step 7) Check for two way shear

Area of heighted portion = 4² -1.05² = 14.89 m²

bcs = 400+650 =1050 mm = 1.05 m

Perimeter of section = 4 × 1050

Now,V_u = 206.25 × 14.89 = 3071.06 kN

τ_v = V_u /bd = 3071.06 × 10³ / (4 × 1050) × 650 = 1.12 N/mm²

τ_c' = k_s.τ_c from IS456:2000 Pg.No:58,59

k_s = 1 for square column

τ_c = 0.25√f_ck = 1.25 N/mm²

τ_v = 1.25 × 1 = 1.25 N/mm²

Since τ_c = τ_v ,Hence Safe in shear.

Result:

The calculation of square fooring by using given details has been designed successfully.