Design of Shallow Foundation (Isolated Footings)

CHALLENGE - 11

ANSWERS

1. AIM & INTRODUCTION:

• To know the general design procedure of spread footing with manual calculations.
• To practice how the inputs can be collaborated and arrived as data for calculations with assumptions.

Given data are:

• Column size 400x400mm
• Compression axial load (1.50DL+1.5 LL) = 2000 KN
• Safe bearing capacity of soil = 150 KN/M2 at 2m depth
• Grade of concrete = M25

Steps to be involved:

• Calculations of gross soil pressure for footing area.
• Calculations of net soil pressure for flexure and shear design
• Initial depth of footing calculation
• Flexural reinforcement calculation
• One way shear check
• Two-way shear check.

1. Gross soil pressure & footing area:

= Unfactored load (i/c surcharge + self wt of foundation/safe bearing capacity

Add 10% for surcharge and self weight

Unfactored load = 2000/1.50

S.B.C = 150

= (2000/1.5)x1.10 /150

= 9.78 Sqm

Assume square footing

√9.78 = 3.12

Say 3.20 m

2. Net soil pressure & initial footing depth :

= Factored load/actual area of footing

= 2000/(3.2x3.2)

= 195.30 KN/M²

Assume initial overall depth of footing as 500mm and checking for bearing stress as per IS 456-200 clause 34.4

• Permissible bearing stress at Jn = 0.45x Fck = 0.45x25 = 11.25 N/M²

Permissible bearing stress x area factor = 11.25x2 = 22.50 N/M²

Actual bearing stress at base of column

= (2000x1000)/(400x400)

= 12.50 N/M²

Bearing stress at base, 12.50 < 22.50 N/M²

Hence Ok with this depth 500mm.

Taking effective cover as 50mm

Ie effective depth = 450mm

3. Flexural reinforcement calculation :

• Conversion of net pressure into udl

= 195.30 x 3.20 = 625 KN/M

• BM due to udl over cantilever section

= wl2/2 = 625x1.6x1.6/2 (1 = 3.20/2

= 800 KN-m

• Reinforcement calculation as per IS :456 :200 , Annex – G P-96

M = 800 KN-m

Fy = 500 N/mm2

b = 3.20 m

d = 450 mm

fck = 25 N/mm2

Ast = ?

800 = 0.87 x 500 xAstx450 ((1- Astx500/(3200x450x25))

= 195750 Ast – 2.72 Ast

Ast = 4350 mm2

Taking 16mm bar = 4350/201 = 22

16mm bar @ 130mmc/c

Ie 24 x 201 = 4824 mm2, Hence OK

4.One way shear check:

Shear force in the one way shaded area

= 0.95 x 3.20 x 195.30 = 593.70 KN

Design shear stress at one way cut face,

= 593.70x1000 / (3200x450) = 0.412 N/mm2

Shear stress capacity as per IS 456 Table 19

For 100Ast/bd = 0.335

From interpolation = 0.425 N/mm²

Design shear < Shear stress capacity, OK

5.Two way shear check:

• Shear force in the two way shaded area

= (3.2x3.2) – (0.85x0.85)x 195.30 = 1858.80 KN

Design shear stress at one way cut face,

= 1858.80x1000 / (3400x450) = 1.215 N/mm2

Shear stress capacity as per IS 456 clause 31.6.3.1

= 0.25 x 5 = 1.25 N/mm2

Design shear (1.215)< Shear stress capacity (1.25),

Hence OK