Design of footing

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This document provides the design of an isolated square footing with uniform thickness to support a column bearing a vertical load of 600KN. It outlines the 8 step process to size and design the footing and reinforcement. The key details are: 1) The footing is designed as a 2.4m x 2.4m square footing with a uniform thickness of 250mm 2) It requires 18 numbers of 12mm diameter bars at 91mm center-to-center spacing as reinforcement 3) All checks for bending, shear, development length and bearing capacity are satisfied

FOOTING
• A Footing is the Substructure, which safely transfers
the load from the superstructure.
• The common types of footing are Isolated, continuous
& Combined Footing.

Isolated Footing of uniform depth For RC column
Question
Design an Isolated footing of Uniform thickness of a R.C. column
bearing a vertical load of 600KN and having a base of size 500mm*500
mm. The safe bearing capacity of soil may be taken as 120 KN/m2. Use
M20 Concrete and Fe415 steel.

STEP- 1 DESIGN CONSTANTS
Fck = 20 N/mm2
Fy = 415 N/mm2
size of column (b)=0.5m
xu,max/d = 700/(1100+0.87 Fy) =0.479
Ru = 0.36 *Xu,max/d (1- 0.42* Xu,max/d)* Fck =2.755

STEP - 2 SIZE OF FOOTING
W =600KN (given)
Let W' = 10% of W =60KN
SBC (qo) =120KN/m2
Area (A) = (w+w')/qo =6m2
Side of square footing (B) = √A =2.35m
Adopt B = 2.4m
Provide a Square Footing of 2.4m * 2.4m
Net upward pressure (p ) = W/A =104KN/m2

STEP – 3 DEPTH ON THE BASIS OF
BENDING COMPRESSION
M = po*(B/8)*(B-b)2=112812500N-mm
Mu = 1.5*M =169218750N-mm
Mu = Ru*Bd2
d=160mm
D = d+60 =220mm

STEP - 4 DEPTH ON THE
BASIS OF ONE-WAY SHEAR
d=160mm
Shear Force (V) = po*B*(((B-b)/2)-d) = 197500N
Vu = 1.5*V=296250N
Nominal shear stress (τv) =Vu / b*d
= 0.77N/mm2
For URS Pt = 0.3%. For M20 concrete
τc =0.384N/mm2
τc' = K* τc = 1.15*0.384 = 0.442N/mm2
τc' > τv REVISE

Not satisfied with 160m depth, increasing depth we get,
Assume d=250mm
V=175000N
Vu=262500N
τv =0.438N/mm2
τc =0.384N/mm2
τc'=0.442N/mm2
τc > τv OK

STEP - 5 DEPTH ON THE
BASIS OF TWO-WAY SHEAR
d=250mm
For two way shear, the section lies at d/2 from the column
face all around as shown in fig.

bo=b+d=750mm
F=po(B2-bo2)
i.e. F = po( Area of ABCD - Area of PQRS) = 541406N
Fu = 1.5F =812109N
τv= Fu/4bod=1.083N/mm2
Permissible shear stress= ks* τc
Ks= 1
τc = 0.25√ Fck = 1.118N/mm2
τc > τv OK

STEP - 6 DESIGN OF STEEL
REINFORCEMENT
Ast = 0.5(Fck/Fy)(1-√1-(4.6Mu)/(Fck*d^2))bd = 2016mm2
Provide bars of12mm
Now,
n*(π*d^2)/4 =2016mm2
n =18
Spacing (S) =91mm c/c
Provide 18 no. of 12 mm Φ bars @ 91 mm c/c

STEP - 7 CHECK FOR
DEVELOPMENT LENGTH
Ld = 0.87FyΦ/4τBd = 564mm
Length of Bar (Lb) = (B-b)/2 = 950mm
Lb > Ld OK

STEP - 8 TRANSFER OF LOAD
AT COLUMN BASE
Allowable bearing pressure ( σall ) = 1.5W/A = 3.6N/mm2
Permissible bearing stress (σper) = 0.45*√Fck = 2.012N/mm2
σall > σper OK

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Design of footing

1. DESIGN OF FOOTING Mr. NRIPESH JHA

2. FOOTING • A Footing is the Substructure, which safely transfers the load from the superstructure. • The common types of footing are Isolated, continuous & Combined Footing.

3. Isolated Footing of uniform depth For RC column Question Design an Isolated footing of Uniform thickness of a R.C. column bearing a vertical load of 600KN and having a base of size 500mm*500 mm. The safe bearing capacity of soil may be taken as 120 KN/m2. Use M20 Concrete and Fe415 steel.

4. STEP- 1 DESIGN CONSTANTS Fck = 20 N/mm2 Fy = 415 N/mm2 size of column (b)=0.5m xu,max/d = 700/(1100+0.87 Fy) =0.479 Ru = 0.36 *Xu,max/d (1- 0.42* Xu,max/d)* Fck =2.755

5. STEP - 2 SIZE OF FOOTING W =600KN (given) Let W' = 10% of W =60KN SBC (qo) =120KN/m2 Area (A) = (w+w')/qo =6m2 Side of square footing (B) = √A =2.35m Adopt B = 2.4m Provide a Square Footing of 2.4m * 2.4m Net upward pressure (p ) = W/A =104KN/m2

6. STEP – 3 DEPTH ON THE BASIS OF BENDING COMPRESSION M = po*(B/8)*(B-b)2=112812500N-mm Mu = 1.5*M =169218750N-mm Mu = Ru*Bd2 d=160mm D = d+60 =220mm

7. STEP - 4 DEPTH ON THE BASIS OF ONE-WAY SHEAR d=160mm Shear Force (V) = po*B*(((B-b)/2)-d) = 197500N Vu = 1.5*V=296250N Nominal shear stress (τv) =Vu / b*d = 0.77N/mm2 For URS Pt = 0.3%. For M20 concrete τc =0.384N/mm2 τc' = K* τc = 1.15*0.384 = 0.442N/mm2 τc' > τv REVISE

8. Not satisfied with 160m depth, increasing depth we get, Assume d=250mm V=175000N Vu=262500N τv =0.438N/mm2 τc =0.384N/mm2 τc'=0.442N/mm2 τc > τv OK

9. STEP - 5 DEPTH ON THE BASIS OF TWO-WAY SHEAR d=250mm For two way shear, the section lies at d/2 from the column face all around as shown in fig.

10.

11. bo=b+d=750mm F=po(B2-bo2) i.e. F = po( Area of ABCD - Area of PQRS) = 541406N Fu = 1.5F =812109N τv= Fu/4bod=1.083N/mm2 Permissible shear stress= ks* τc Ks= 1 τc = 0.25√ Fck = 1.118N/mm2 τc > τv OK

12. STEP - 6 DESIGN OF STEEL REINFORCEMENT Ast = 0.5(Fck/Fy)(1-√1-(4.6Mu)/(Fck*d^2))bd = 2016mm2 Provide bars of12mm Now, n*(π*d^2)/4 =2016mm2 n =18 Spacing (S) =91mm c/c Provide 18 no. of 12 mm Φ bars @ 91 mm c/c

13. STEP - 7 CHECK FOR DEVELOPMENT LENGTH Ld = 0.87FyΦ/4τBd = 564mm Length of Bar (Lb) = (B-b)/2 = 950mm Lb > Ld OK

14. STEP - 8 TRANSFER OF LOAD AT COLUMN BASE Allowable bearing pressure ( σall ) = 1.5W/A = 3.6N/mm2 Permissible bearing stress (σper) = 0.45*√Fck = 2.012N/mm2 σall > σper OK

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