Design of column footing:Design 2: A square column 500mm X 500mm carries an axial load of 1500 KN . Design the column and the square footing for the column. The safe bearing capacity of the column is 225 KN/m2. Use M20 and Fe 415 steel.

Design of Column:

Load on the column W = 1500 KN

Factored load Pu= 1.5 x 1500 = 2250 KN

Over all area of the column section Ag= 500 x 500 = 250000 mm2

Area of the steel = Asc

Area of the concrete = Ac = Ag- Asc

= 250000-Asc

Ultimate load Pu=0.4 fck Ac + 0.67 fy Asc

2250000 = 0.4 X 20 X ( 250000 Asc) + 0.67 X 415 x Asc

Asc= 925.75 mm2

Assume 29 mm dia bars

Provide 4 bars of 20 mm dia

Lateral ties :

(i) diameter of the longitudinal bar

(ii) 5mm

From the above two take the greater one so provide the diameter of 6 mm dia bar.Pitch of lateral ties:

(i)Least lateral dimension of the column = 500mm

(ii)16 times the diameter of the longitudinal bars =16 X 20 = 320 mm

(iii)48 times the diameter of the ties = 48 X6

Provide 6 mm dia ties at 250 mm c/c.

Design of the foundation:

Load on the column = 1500 KN

Approximate weight of the footing at 10 % of the column load = 150 KN

Total load = 1650 KN

Safe bearing capacity of the soil = 225 KN/m2

Area of foundation = 1650/225 = 7.333 m2

BX B = 7.333

Breadth of foundation B = (7.333) = 2.71 say 2.75 mSo the area of the foundation is 2.75 X 2.75 m

Net upward pressure = load on the column / area of the footing

= 1500000/(2.75 X 2.75) = 198347.11 N/m2.

Depth of the foundation =

Minimum depth of the foundation = (p/r){(1-sin0)/(1+sin0)}2

= (225/18) {(1-sin30)/(1+sin30)}2= 1.4 mDetermination of the depth of the concrete slab below the footing.

Critical section for bending moment is = (2750-500)/2 = 1125 mm = 1.125 m

Maximum bending moment = M = 198347.11 x 2.75 X 1.125 x( 1.125/2) = 345170 Nm

Factored moment Mu = 1.5 X M

= 1.5 X 345170 = 517755 Nm.To find the depth of the slab in the foundation.

Mu= 0.138 fck bd2517755000 = 0.138 X 20 X 500 (width of column) X d2d= 613 mm

D= 613 + 12/2 +60 = 691 mm(12- dia of bar , 60 clear cover for footing)

The depth of slab of the foundation is increased by 30 %

D= 691 X ( 0.3 X 691) = 900 mm

d=900 (12/2)- 60 =822 mm.

Determination of quantity of steel required:

Ast = Pt X b x d= (0.47/100) x 500 x 822 = 1932 mm2Assume 12 mm dia bars so provide 18 bars of 12 mm diameter.

Here the column is square so provide the reinforcement on both the directions.

If it is rectangular find the bending in moment for opposite direction also.2. A rectangular column footing 600 mm X 400 mm carries an axial load of 800 KN . Design a rectangular footing to support the column . The safe bearing capacity of the soil is 200 KN/m2 . Use M20 concrete and Fe415 steel.Load on the column = 800000N

Approximate weight of the foundation take 10 % of the weight of the column = 80000N

Total load = 880000N

Safe bearing capacity of the soil is given as 200 KN/m2 = 200000N/m2

Area of the foundation = Total load / Safe bearing capacity of the soil.

= 880000/200000 = 4.40 m2

To find the length and breadth of the foundation

in case of square footing its easy because by taking square root we get all the values Area = 4.4

BL = 4.4

B= 4.4 /L

Equating the projections on both sides beyond the footing

( B-0.4) = (L-0.6)

Sub B Value

( {4.4/L}-0.4) = (L-0.6)

Solving the above equation we get L = 2.2 m

Sub this is B value we get

B= 4.4/L = 4.4/ 2.2 = 2 m

Now find the projections on both the axis

= 0.8 mNet upward pressure = column load / Area of the footing

= 800000/4.4

=181820 N/m2

Determination of reinforcement in section xx axis and YY axis.

Bending moment Mxx = 181820 X 2.2X 0.8 X (0.8/2).

= 128000 Nm

Factored Moment Mux = 128000 X 1.5

= 192000 Nm

Determine the depth

Mux = 0.138 Fck bd2192000 = 0.138 x 20 x 600 x d2d= 341 mm.

Bending moment Mxx = 181820 X 2 X 0.8 X (0.8/2).

= 116364.8 Nm

Factored Moment Mux = 116364.8 X 1.5

= 174547.2Nm

Determine the depth

Mux = 0.138 Fck bd2174547.2 = 0.138 x 20 x 400 x d2d= 398 mm.

From the above two depth take the greater one

d= 398 mm

Providing 10 mm dia bars at a clear cover of 70 mm

D = 398 + (10/2) + 70 = 473 mm

The overall depth may increased by 30%

= 473 + (0.3 X473)

= 614.9 mm

D=620 mm.

d= 620 (10/2) 70

= 545 mm.

Take Muy and find Longer direction steel. (similar to above problem)

For long b = 600 mm , d = 545 mm.

Ast = Pt b dSize of bars and number of bars

Take Mux and find the Shorter direction steel .(similar to above problem)

For shorter b = 400 mm , d = 545 mm

Size of bars and number of bars