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.

ANS:

AIM:

• TO design a square footing for a column size of 400x400 using other given details.
• To 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
• To provide a simple sketch showing required details.

INTRODUCTION:

• Footings which are provided under each column independently are called as Isolated footings.
  
  
• They are usually square, rectangular or circular in section.
  
  
• Footing is laid on PCC. Before laying PCC, termite control liquid is sprayed on top face of PCC to restrict the termites to damage the footing.
  
  
• Isolated footings are provided where the soil bearing capacity is generally high and it comprises of a thick slab which may be flat or stepped or sloped.
  
  
• This type of foothigs are most economical when compared with the
  other kind of footings.

PROCEDURE:

Design of square footing

Given:

• Soil-bearing capacity = 150 KN/m²
• Compression Axial load = 2000 KN
• Compressive strength f_ck = M25
• Assume the thickness of concrete footing to be 0.8m
• depth of 2m below E.G.L
• Soil depth = 20KN/cu.m
• RCC density = 25 KN/cu.m

SOLUTION:

STEP: 1

SIZE OF THE FOOTING:

Assume self weight of the footing equal to 10% of superimposed load.

Wd = 10% x 2000

= 0.1 x 2000

= 200 KN

Now the total load = Wl + Wd

= 2000 + 200

W = 2200 KN

Now,

Area of footing (A) = W/SBC

= 2200/150 = 14.66m²

hence, Area of footing (A) = 14.66m²

so, A = B²

:. B = `root a`

= `root 14.66`

= 3.82 ~ 4m

hence, we provide the size of footing as 4m x 4m

then Area of footing A = 16 m²

STEP: 2

NET UPWARD PRESSURE

`Po = ( W+1.5)/(size of footing)`

`= (2200 * 1.5) / (4 * 4)`

Po = 206.25 KN/m²

STEP: 3

DEPTH OF FOOTING ON BASIS OF B.M.

The maximum BM at the face of column and is given by,

M = Po (B/8) (B -b)²

M = 206.25 . (4/8) . (4 - 0.4)²

M = 1336.5 KN/m

Now, ultimate moment [Mu] = 1.5 (M)

= 1.5 x 1336.5

Mu = 2004.75 KN/m

To get the ultimate moment, we need to calculate that

Mu = 1.5 ( factor of safety) x moment.

Now, depth of footing (D) :

Mu = 0.138 f_ckBd²

2004.75 = 0.138 x 25 x 4000 x d²

d² = 145271.73

d = 381.14 mm

By the thumb rule, we take 2.5 times higher

Value of d for shear check

d = 381.14 x 2.5

d = 952.85 mm

D = d + 50

= 952.85 + 50

D = 1003mm

STEP: 4

AREA OF REINFORCEMENT:

Ast= 5985.3 mm²

Now

we have to find the % of reinforcement.

P = Ast/ Bd x 100
=( 5985.3/ (4000x 953)) X100

P = 0.157%

No -of Bars required for reinforcement

assume 20mm dia of bar.

Ast = 5985.3mm2

Pie/4 x20² = 314.5
⇒ No⋅ot bars = 5985.3/ 314.5
= 20

:. 20 nos of bars

Spacing between the bars = B/no. of Bars
= 4000/20
=200 mm - C/C

STEP: 5

CHECH FOR ONE-WAY SHEAR

For One way Shear, Critical Section is located at a distance 'd' from the face of the Columns.

Now

Shear force (V)= Po.B { ½ (B-b)-d}

= 206-25 × 4 [ 2 (4-0-4) - 0.952}

= 699.6 KN

Ultimate shear force (V) = 1.5(U)

=1.5x699.6

=1049.4 KN.

Nominal Shar smers (Tu) =Vu/Bd (As per IS 456-2000, pg:42)

= (1049.4 x10³)/ (4000 x 951)

= 0.27 N/mm²

Now,

Design Shear strength of concrete (T)

Tc'= Tc K. (As per Is 456-2000, Pg: 72)

K=1, Because depth > mm

Tc for 0.157% of P is 0.28 N/mm²

|Tc = 0-28 N/mm²]

Then Tv < Tc.
Hence OK.

Now

Tv < Tc'
Vu/ Bd = K. Tc
d = Vu/B.K.Tc
d = (1049.4×10³)/(4000 x 1 x 0.28)

d = 937mm

d provided = 953 mm > 937 mm

:. OK for 1 way Sheer.

Now, from [(Page: 57, Clause 31.6.2) I5-456-2000]

Tv = Fu/4bod

= 673.65x10^3/(4x1353x 953)

= 0.23 N/mm²

Now,

permialible Shear Stress (Ks Tc) → [IS-456-2000, Pg - 58}
Hence Ks = 1 (max)

:. Tc = 0.25 √fck
= 0.25√25
= 1.25

:. Ks.Tc = 1×1.25
= 1.25

:. Tv < Tc

→ Safe in the two way Shear,

RESULT:

• By using given details the square footing has been designed succesfully.
• All the required checks has also been done
• The sketch is also provided as required.

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