raft foundation design

Example of Raft Foundation design:

The higher structural loads and low bearing capacity of soil demands for raft foundation design. The example of raft foundation design is discussed here in details.

Example: Design a raft foundation of a raft foundation shown below. Some of the design constrains required for raft foundation design are as,

Design Constants:

Unit weight of soil (Ξ³) = 18 KN/m3

Service Load (P) = 188190.67 kN

Service load includes the total axial forces applied from column, lift wall weight , and load from basement walls.

Grade of Concrete = M30

Grade of steel = Fe 500D

Bearing Capacity (q) = 150 KN/m2

Angle of Repose of soil (Ø) = 300

raft foundation design

The raft foundation design with step wise procedure are as given:

As per IS 1893: 1 Cl 6.3.5.2 allowable bearing pressure in soil can be increased depending upon type of foundation thus bearing capacity of soil is increased by 50% assuming it will be Raft foundation

Then, q = 1.5Γ—150 = 225 kN/m2 and applying factor of safety of 1.2, q = 187.5 kN/m2

Generally, the depth of raft foundation from ground level shall be not less than 1 m (IS 2950 Part 1, Cl. 4.3)

Df = π‘žπ‘’/𝛾𝑠 Γ— (1βˆ’π‘ π‘–π‘›Γ˜)2/(1+π‘ π‘–π‘›Γ˜)2 = 187.5/18Γ—(1βˆ’π‘ π‘–π‘›30)2/(1+𝑠𝑖𝑛30)2 = 1.157 m > 1 m OK.

If the value was less than 1 m, the lower face of the designed footing shall have been placed at a level of 1 m below which soil is free from seasonal volumetric change.

The area of foundation required for safe transmission of load, assuming above reaction from superstructure is taken as non-eccentric surcharge to the soil, is given by:

Area of foundation required = Σ𝑃+10% π‘œπ‘“ Ξ£π‘ƒπ‘ž = 1.1Γ—188190.67 * 187.5 π‘š2 = 1104 m2

Plinth area of the building = 1356.78 m2

Hence, % Area of required footing = (1104/1356.78)Γ—100 = 81.37 % > 50%

Consider 500 mm projection for the raft having same shape of the superstructures along the building periphery for critical shear section consideration.

∴ Area of Foundation provided = 1430.97 m2

Location of geometric C.G.: X = 15.24 m, Y = 21.265m

Calculation of Eccentricity:

ColumnService loadXYF*XF*Y
IDKNmm
A13057.2980000
A23433.3906707.16024583.08
A33546.08333010.32036595.58
A44078.51267017.48071292.4
A54265.566027.50117303.1
A64845.374036.230175547.9
A72491.35333042.530105957.3
B14537.913337.62034578.90
B25397.606677.627.1641129.7638646.86
B35567.017337.6210.3242420.6757451.62
B46836.089337.6217.4852091119494.8
B56945.637337.6227.552925.76191005
B67042.533337.6236.2353664.1255151
B74210.067337.6242.5332080.71179054.2
C14673.9093315.24071230.380
C25609.9633315.247.1685495.8440167.34
C35733.0833315.2410.3287372.1959165.42
C46737.1686715.2417.48102674.5117765.7
C57416.84615.2427.5113032.7203963.3
C67247.2113315.2436.23110447.5262566.5
C74579.0626715.2442.5369784.92194747.5
D14547.1633322.860103948.20
D25485.9806722.867.16125409.539279.62
D35611.9453322.8610.32128289.157915.28
D47946.60822.8617.48181659.5138906.7
D58692.1253322.8627.5198702239033.4
D67090.1093322.8636.23162079.9256874.7
D74454.2746722.8642.53101824.7189440.3
E13118.1346730.48095040.740
E23641.1786730.487.16110983.126070.84
E33718.39230.4810.32113336.638373.81
E45271.2646730.4817.48160668.192141.71
E55760.8573330.4827.5175590.9158423.6
E64714.1626730.4836.23143687.7170794.1
E73115.5193330.4842.5394961.03132503
A82641.38267022.49059404.7
Total184060.78628451103849620

The total service load from the table and initially taken is different because the load of staircase and shear wall should also be taken for the area calculation.

XΜ… = (Σ𝐹𝑖×𝑋𝑖)/Σ𝑃 = 15.457 m

Θ²= (Ξ£πΉπ‘–Γ—π‘Œπ‘–)/Σ𝑃 = 20.915 m

Load Centroid (X, Y) = (15.457m, 20.915m)

ex = 15.457 – 15.24 = 0.217 m

ey = 21.265 – 20.915 = 0.35 m

Mx = P * ey = 188190.67 * 0.35 = 65866.74 KN-m

My = P * ex  = 188190.67 * 0.207 = 40837.37 KN-m

Ixx = BD3/12 = 32.28 * 44.333 /12 = 234340 m4

Iyy = DB3/12 = 32.283 * 44.33 / 12 = 124256 m4

Soil Pressure at different Points (corner and edge) are as follows:

Column IDP/AIxxIyyXYMxx*Y /IxxMyy * X/IyySoil pressure
Corner A1131.5234340124256-16.14-22.165-6.230-5.304119.966
 A2131.5234340124256-16.14-14.105-3.965-5.304122.231
 A3131.5234340124256-16.14-10.945-3.076-5.304123.119
A4131.5234340124256-16.14-3.785-1.064-5.304125.132
A5131.5234340124256-16.146.2351.752-5.304127.948
 A6131.5234340124256-16.1414.9654.206-5.304130.402
 Corner A7131.5234340124256-16.1422.1656.230-5.304132.425
Corner E1131.523434012425616.14-22.165-6.2305.304130.575
E2131.523434012425616.14-14.105-3.9655.304132.840
E3131.523434012425616.14-10.945-3.0765.304133.728
E4131.523434012425616.14-3.785-1.0645.304135.741
E5131.523434012425616.146.2351.7525.304138.557
E6131.523434012425616.1414.9654.2065.304141.011
Corner E7131.523434012425616.1422.1656.2305.304143.034
 B1131.5234340124256-7.62-22.165-6.230-2.504122.766
C1131.52343401242560-22.165-6.2300.000125.270
 D1131.52343401242567.62-22.165-6.2302.504127.774
 B7131.5234340124256-7.6222.1656.230-2.504135.226
 C7131.5234340124256022.1656.2300.000137.730
D7131.52343401242567.6222.1656.2302.504140.234

Hence maximum downward stress (140.234 kN/m2) is less than safe bearing capacity (187.5 kN/m2) so OK.

In X-direction raft is divided in seven strips that is into 5 equivalent beam, the beam with the respective soil pressure and moment are as follows.

Bending moment is obtained by coefficient (1/12) and β€˜L’ as center to center distance, from IS 456 Cl. 22.5.1

+M = -M = wl2/12

BeamsStrip widthLength LCoeff.Soil pressureEquivalent soil pressureSoil pressure per strip KN/m2
A-A4.7110.021/12132.425132.4251107.960
B-B7.6210.021/12135.225133.8251119.674
C-C7.6210.021/12137.73136.47751141.866
D-D7.6210.021/12140.23138.981162.804
E-E4.7110.021/12143.03141.631184.976
     Maximum1184.970

In the Y-direction the raft is divided into 7 strips i.e. into seven equivalent beams.

BeamsStrip widthLength LCoeff.Soil pressureEquivalent soil pressureSoil pressure per strip KN/m2
14.487.621/12130.57130.57631.79
25.167.621/12132.84131.71637.28
35.167.621/12133.73133.28644.92
48.597.621/12135.74134.73651.94
59.3757.621/12138.56137.15663.62
67.527.621/12141.01139.78676.37
74.057.621/12143.03142.02687.20
     maximum 687 .20

Therefore, maximum moment is 1184.97 kNm/m per strip.

Calculation of Depth of Foundation:

i. Calculation of Depth from Moment Criterion (IS 456 : 2000, ANNEX G 1.1):

Mu = 0.133 fck b d2 [for fe500]

1184.97 * 106 = 0.133 * 30 * 1000 * d2

d = 545 mm

ii. Calculation of Depth from Two Way Shear:

Depth of raft will govern by two-way shear at one of the exterior column. In case, location of critical shear is not obvious it may be necessary to check all locations. When shear reinforcement is not provided, the calculated shear stress at critical section shall not exceed KsΓ—Ο„c. i.e. Ο„v ≀ KsΓ—Ο„c. (IS 456 : 2000, Cl. 31.6.3.1)

Where,

Ks = (0.5 + Ξ²c) but not greater than 1, Ξ²c being the ration of short side to long side of the column/capital; and

Ο„c = 0.25 βˆšπ‘“π‘π‘˜ in limit state method of design.

Here, Ξ²c = 1

Ks = 1+0.5 = 1.5 > 1

Hence, Ks = 1

Shear strength of concrete (Ο„c) = 0.25Γ—βˆš30 = 1.37 N/mm2

For column D5

Factored Column Load = 13038.19 kN

Perimeter (po) = 4 (d + 800)

The nominal shear stress in flat slabs shall be taken as V/(poΓ—d) where V is the shear force due to design, P0 is the periphery of the critical section and d is the effective depth of the slab. (IS 456 : 2000, Cl. 31.6.2.1)

πœπ‘£=𝑉𝑒/(π‘π‘œΓ—π‘‘)

1.37 = 13038.19Γ—103/[4(𝑑+540)×𝑑]

d = 1193.87 mm

For corner column E7

Factored Column Load = 4677.2 kN

Perimeter po = 2Γ—(0.5d + 800 + 500 + 0.5d + 800+500) = 2Γ—(d + 2600)

πœπ‘£=𝑉𝑒/(π‘π‘œΓ—π‘‘)=4677.2*103/ 2(𝑑+2600)×𝑑

π‘‚π‘Ÿ,1.37=4677.2Γ—103/ [2(𝑑+2600)×𝑑]

Hence d = 543.33 mm

Hence, the thickness of raft foundation design is governed by two-way shear.

Adopt effective Depth (d) = 1200 mm

Adopt effective cover of 65 mm (IS456: 2000, Cl. 26.4.2.2)

Overall Depth = 1200 + 65= 1265 mm

Reinforcement calculation:

In shorter Direction

We have from (IS 456 : 2000, Annex G 1.1)

BM = 0.87Γ—fyΓ—AstΓ—(d – 𝑓𝑦×𝐴𝑠𝑑/π‘“π‘π‘˜Γ—π‘)

Or, 687Γ—106 = 0.87Γ—500Γ—AstΓ—(1200 – 500×𝐴𝑠𝑑/ 30Γ—1000)

Solving we get,

Ast = 1341.07 mm2

The mild steel reinforcement in either direction in slab shall not be less than 0.15% of the total cross-sectional area. However this value can be reduced to 0.12% when high strength deformed bar or welded wire fabric are used. (IS 456 : 2000, Cl. 26.5.2.1)

Minimum reinforcement in slab = 0.12%Γ—1000Γ—1265 = 1518 mm2

i.e. Ast, required < Ast, Min.

So, Ast, required = 1518 mm2

Using 16 mm Ø bars,

Spacing of Bar (Sv) = 𝐴𝑏/𝐴𝑠𝑑×1000=201.062/1518Γ—1000=132.45 π‘šπ‘š

Hence, provide 16 mm Ø Bars @ 120 mm C/C in Shorter direction.

Therefore, Ast, Provided = (201.062/120)Γ—1000 = 1675.52 mm2

In Longer Direction:

Adopt effective depth = 1200 – 16 = 1184 mm

Reinforcement in longer direction is given by:  

BM = 0.87Γ—fyΓ—AstΓ—(d – 𝑓𝑦×𝐴𝑠𝑑/π‘“π‘π‘˜Γ—π‘)

Or, 1184.97Γ—106 = 0.87Γ—500Γ—AstΓ—(1184 – 500×𝐴𝑠𝑑/30Γ—1000)

Solving we get Ast = 2380.50 mm2

Minimum reinforcement in slab = 0.12%Γ—1265Γ—1000 = 1518 mm2

i.e. Ast required > AstMin. OK.

Using 16 mm Ø bars,

Spacing of Bar (Sv) = 𝐴𝑏/𝐴𝑠𝑑×1000=201.062/2380.50Γ—1000= 84.46 π‘šπ‘š

Hence, Provide 16 mm Ø @ 80 mm C/C in Longer direction.

Therefore, Ast provided = (201.062/80)Γ—1000 = 2513.275 mm2

Check for Development Length:

Bond Stress (πœπ‘π‘‘)=1.4 𝑁/π‘šπ‘š2, For M30 Concrete. This value can be increased by 60% for Tor Steel. (IS 456 : 2000, Cl. 26.2.1.1)

The development length (Ld) is given by (IS 456 : 2000, Cl. 26.2.1)

𝐿𝑑=βˆ…Γ—πœŽπ‘ /4Γ—πœπ‘π‘‘= βˆ…Γ—0.87Γ—500/(4Γ—1.6Γ—1.6)=42.48βˆ…

Ld = 42.48Γ—16 = 679.68 mm

𝐿𝑑≀ 1.3×𝑀1/𝑉+π‘™π‘œ (IS 456 : 2000, Cl. 26.2.1)

lo = Effective depth or 12βˆ…, Whichever is greater.

𝐿𝑑≀ 1.3Γ—1184.97Γ—106 /4677.2Γ—1000+1200=1529.08 π‘šπ‘š OK.

Load Transfer from Column to footing:

Nominal bearing stress in column concrete (Οƒbr) = Pu/Ac

= (13038.19Γ—1000)/(800Γ—800) = 20.37 N/mm2

Allowable bearing stress = 0.45Γ—fck = 0.45Γ—30 = 13.5 N/mm2 (IS 456: 2000, Cl. 34.4)

If the permissible bearing stress on the concrete within the supporting or supported members exceeded, reinforcement shall be provided for developed surplus force by dowels. (IS 456: 2000, Cl. 34.4.1)

Dowel of at least 0.5% of the cross-sectional area of the supported column and a minimum of four bars shall be provided. Diameter of the Dowels shall not exceed the diameter of column bar by more than 3 mm. (IS 456 : 2000, Cl. 34.4.1)

Area of Dowels Bar = 0.5%Γ—800Γ—800 = 3200 mm2

Provide 28 mm βˆ… as Dowel bar.

Development length for dowel bar = βˆ…Γ—πœŽπ‘ /4Γ—πœπ‘π‘‘= 28Γ—0.87Γ—500/4Γ—1.6Γ—1.6=1189.45 π‘šπ‘š

Use 6 – 28 mm βˆ… Bars as Dowel Bar, then As, provided = 6Γ—PIΓ—282/4 = 3694.5 mm2 > 3200 mm2

Also read:

Design of isolated footing

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