**Introduction**

The conventional rigid combined footing approach is a method of analysing raft foundation using simple statics without any consideration of the elastic properties of the raft and the soil. Here the raft is analysed as a large beam member independently in both directions. The row of column loads perpendicular to the length of the beam are coupled together in single column load. Then for these column loads acting on the beam, the upward soil pressure is calculated and the moments and the shears at any section is determined by simple statics. Hence, the moment per unit width of the raft is determined by dividing the moment values by corresponding width of the section

*(Gupta, 1997)*.

In order to obtain the upper bound values of the stresses the raft is divided into strips bounded on the centre line of the column bays in each direction. Each of these strips is then analysed as independent combined footing by simple statics. Using the column loads on each strip the soil pressure under each strip is determined without reference to the planar distribution determined for the raft as a whole.

The eccentricity of the load and the pressure distribution below the raft which is considered to be linearly varying are taken into account in this analysis.

**Design Example**

It is desired to design the raft slab shown below to support the column loads for a building as given below. All columns are 230 x 450mm, the grade of concrete (f_{ck}) is 30 N/mm^{2}, and the yield strength of the reinforcement (f_{yk}) is 500 N/mm^{2}. The allowable bearing capacity for the supporting soil is 60 KN/m^{2}.

Total axial load = 18296 KN

**Eccentricity along the x-direction**

This is obtained by taking moment about grid 5

*x*= [22.4(770 + 1050 + 776 + 350) + 16.8(870 + 1450 + 860 + 660) + 11.2(1211 + 1850 + 1000 + 779) + 5.6(875 + 1400 + 865 + 660)] / 18296 = 11.258m

e

_{x}=*x*- (L/2) = 11.258 - (22.4/2) = 0.058m**Eccentricity along the y-direction**

This is obtained by taking moment about grid D

*y*= [15.1(770 + 870 + 1211 + 875 + 770) + 9.1(1050 + 1450 + 1850 + 1400 + 1050) + 3.1(776 + 860 + 1000 + 865 + 700) ] / 18296 = 7.8045m

e

_{y}=*x*- (B/2) = 7.8045 - (15.1/2) = 0.2545m**Moment Due to Eccentricity**

M

*x*= P.e_{x}= (18296 × 0.058) = 1061.168 kNm
M

*y*= P.e_{y}= (18296 × 0.2545) = 4656.332 kNm**Read Also**

**Other geometrical properties**

Moment of inertia of the raft slab about the x-direction;

I

_{x}= (17.1 × 24.4^{3}) / 12 = 20700.6672 m^{4}
Moment of inertia of the raft slab about the y-direction;

I

_{y}= (24.4 × 17.1^{3}) / 12 = 10167.0957 m^{4}A = Area of raft slab = (17.1m × 24.4m) = 417.24 m

^{2}

The soil pressure at any point is given by the equation below;

P/A = (18296 KN / 417.24 m

^{2}) = 43.85 KN/m^{2}My/I

_{y}= (4656.332 kNm / 10167.0957 m

^{4}) = 0.45798 KN/m

^{2}

Mx/I

_{x}= (1061.168 kNm / 20700.6672 m

^{4}) = 0.05126 KN/m

^{2}

**σ = 43.85 ± 0.45798**

*x*± 0.05126*y*

**At corner A1;**σ

_{A1}= 43.85 - (0.45798 × 12.2) + (0.05126

*× 8.55) = 38.7 KN/m*

^{2}

**At corner A5;**σ

_{A5}= 43.85 + (0.45798 × 12.2) + (0.05126

*× 8.55) = 49.975 KN/m*

^{2}

^{}

**At corner B1;**σ

_{B1}= 43.85 - (0.45798 × 12.2) + (0.05126

*× 1.55) = 38.342 KN/m*

^{2}

^{}

**At corner B5;**σ

_{B5}= 43.85 + (0.45798 × 12.2) + (0.05126

*× 1.55) = 49.516 KN/m*

^{2}

^{}

**At corner C1;**σ

_{C1}= 43.85 - (0.45798 × 12.2) - (0.05126

*× 4.45) = 38.07 KN/m*

^{2}

^{}

**At corner C5;**σ

_{C5}= 43.85 + (0.45798 × 12.2) - (0.05126

*× 4.45) = 49.209 KN/m*

^{2}

^{}

**At corner D1;**σ

_{D1}= 43.85 - (0.45798 × 12.2) - (0.05126

*× 8.55) = 37.824 KN/m*

^{2}

^{}

**At corner D5;**σ

_{D5}= 43.85 + (0.45798 × 12.2) - (0.05126

*× 8.55) = 48.999 KN/m*

^{2}

^{}

**At corner A2;**σ

_{A2}= 43.85 - (0.45798 × 5.6) + (0.05126

*× 8.55) = 41.723 KN/m*

^{2}

^{}

**At corner A3;**σ

_{A3}= 43.85 + (0.05126

*× 8.55) = 44.288 KN/m*

^{2}

^{}

**At corner A4;**σ

_{A4}= 43.85 + (0.45798 × 5.6) + (0.05126

*× 8.55) = 46.852 KN/m*

^{2}

^{}

**At corner D2;**σ

_{D2}= 43.85 - (0.45798 × 5.6) - (0.05126

*× 8.55) = 40.847 KN/m*

^{2}

^{}

**At corner D3;**σ

_{D3}= 43.85 - (0.05126

*× 8.55) = 43.411 KN/m*

^{2}

^{}

**At corner D4;**σ

_{D4}= 43.85 - (0.45798 × 5.6) - (0.05126

*× 8.55) = 40.847 KN/m*

^{2}

^{}A little consideration will show that the bearing capacity checks are satisfactory.

**Limit State Calculations**

A factor of 1.37 has been used to convert the load from service load to ultimate load.

**Check for punching shear at column perimeter**

^{We can determine the thickness of the raft slab by considering the punching shear at the column perimeter. At the column perimeter, the maximum punching shear stress should not be exceeded.}

^{}

^{VEd < VRd,max}

^{}

^{Where;}

V

_{Ed}= βV

_{Ed}/u

_{0}d

V

_{Rd,max}= 0.5

*v*f

_{cd}

Considering Column B3 bearing the maximum axial load, V

_{Ed}= 1.37 × 1850 kN = 2534.5 kN

β = 1.15 (an approximate value from clause 6.4.3(6))

u

_{0 }= column perimeter = 2(230) + 2(450) = 1360mm

*v*= 0.6[1 - fck/250] (strength reduction factor for concrete cracked in shear)

*v*= 0.6[1 - 30/250] = 0.528

f

_{cd}= α

_{cc}f

_{ck}/γ

_{c }f

_{cd}= (1.0 × 30)/1.5 = 20 N/mm

^{2}

_{}V

_{Rd,max}= 0.5 × 0.528 × 20 = 5.28 N/mm

^{2}

_{}

V

_{Ed}= (1.15 × 2534.5 × 1000) / (1360mm × d)

Therefore;

2914675/1360d = 5.28

On solving; dmin = 405.89 mm

_{ The basic control perimeter for punching shear check is normally taken at 2.0d, but when the concentrated force is resisted by high pressure as can be found in foundations, the punching control perimeter is taken at less than 2d.}

_{}

_{Let us consider a trial footing depth of 700mm.}

_{}

_{Effective depth (d) = 700 - 70 - (20/2) = 620mm (concrete cover is taken as 70mm and assumed diameter of bar is 20mm) }

_{ Analysis of the raft strips}

*For strip A - A bearing the most critical load;*

Width of strip = 4m

Taking the average load = (38.7 + 49.975) / 2 = 44.3375 KN/m

^{2}

Multiplying by the width of the strip = 44.3375 KN/m

^{2 }× 4m = 177.35 KN/m

On factoring at ultimate limit state = 1.37 × 177.35 = 242.9695 KN/m

Effective depth = 620mm (as previously calculated).

**Design of bottom reinforcement**

M

_{Ed}= 781.67 KNm

A

_{s1}= M

_{Ed}/(0.87f

_{yk}z)

A

_{s1 }= (781.67 × 10

^{6})/(0.87 × 500 × 0.95 × 620) = 3050.836 mm

^{2}

Minimum area of reinforcement AS

_{min }= 0.0013bd = 0.0013 × 1000 × 620 = 806 mm

^{2}/m

Provide 18H16 @ 225mm c/c BOT (A

_{Sprov}= 893 mm

^{2}/m or 3216 mm

^{2}) along the strip.

**Design of top reinforcement**

M

_{Ed}= 529.47 KNm

A

_{s1}= M

_{Ed}/(0.87f

_{yk}z)

A

_{s1 }= (529.47 × 10

^{6})/(0.87 × 500 × 0.95 × 620) = 2066.506 mm

^{2}

Minimum area of reinforcement = 0.0013bd = 0.0013 × 1000 × 620 = 806 mm

^{2}/m (clause 9.2.1.1(1))

Also provide 18H16 @ 225mm c/c TOP (A

_{Sprov}= 893 mm

^{2}/m or 3216 mm

^{2}) along the strip.

**Check for punching shear**

Let us check for shear at d from the face of the most heavily loaded column.

Using the maximum shear force V

_{Ed}= 1.37 × 1211= 1659.07 kN

V

_{Ed,red}= V

_{Ed}- ∆V

_{Ed}

Where ∆V

_{Ed}is the net upward force within the control perimeter considered i.e. upward pressure from soil minus self weight of base.

Control area = 1.69m × 1.47m = 2.4842 m

^{2}

Soil pressure = 242.97 KN/m

^{2}

Unit weight of concrete = 25 kN/m

^{3}

Hence, ∆V

_{Ed}= (242.97 × 2.4842) - (25 × 0.7 × 2.4842) = 560.1125 kN

V

_{Ed,red}= 1659.07 - 560.1125 = 1098.9575 kN

v

_{Ed,red}= V

_{Ed,red}/ ud

where u is the control perimeter = 2(1690) + 2(1470) = 6320 mm

v

_{Ed,red}= (1659.07 × 1000) / (6320 × 620) = 0.423 N/mm

^{2}

**Shear Resistance of section**

V

_{Rd,c}= [C

_{Rd,c}.k.(100ρ

_{1}f

_{ck})

^{(1/3)}] × 2d/a ≥ (V

_{min}× 2d/a)

C

_{Rd,c}= 0.18/γ

_{c}= 0.18/1.5 = 0.12

k = 1 + √(200/d) = 1 + √(200/620) = 1.568 > 2.0, therefore, k = 1.568

V

_{min}= 0.035k

^{(3/2)}f

_{ck}

^{0.5}

V

_{min}= 0.035 × (1.568)

^{1.5}× 30

^{0.5}= 0.391 N/mm

^{2}

^{}ρ

_{1}(in the transverse direction) = As/bd = 804/(1000 × 620) = 0.00129

V

_{Rd,c}= [0.12 × 1.568(100 × 0.00129 × 30 )

^{(1/3)}] × 2 = 0.5908 N/mm

^{2}

0.423 N/mm

^{2}< 0.5908N/mm

^{2}

This shows that shear is ok.

Using this approach, the entire reinforcement for the mat foundation can be obtained in the longitudinal and transverse directions.

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