Design of Continuous Slab and Beam Strip Footing

Continuous beam and slab strip footings normally consist of inverted T-beams that are used to support a series of columns or continuous walls. They are a cheaper alternative to raft foundations and can be used where individual pad foundations will overlap or where the soil is expected to undergo differential settlement or excessive shrinkage and swelling. The design of a continuous beam and slab strip footing involves the design of the base slab and upstand beam by providing adequate concrete sections and reinforcements to resist the earth pressure reaction from the applied load.

The design presented here is an excerpt from my final year project ‘Structural Analysis and Design of 35,000 Capacity Reinforced Concrete Stadium’. After the analysis and design of the superstructure (see a section of the stadium in the picture below), it was realized that a very large magnitude of axial loads and moments are being transferred to the foundation.

Foundations must be designed to resist geotechnical and structural failure, and at the same time should be economical. The ultimate bearing capacity of the supporting soil at 2.00m depth was very good at 380 KN/m² (gravely sand), so a shallow foundation was adopted.

However, adopting a pad footing proved very uneconomical given the large area of excavation required (columns are spaced at 6.0m), and the depth of concrete needed to handle shear forces was much. Raft foundation proved to be too expensive for soil with such a good bearing capacity.

After much consideration, it was realized that chaining the columns continuously will do the trick, but at the same time, it is possible to combine the slab with upstand beams running continuously along the axis of the column. The whole aim was to reduce the great quantity of concrete that would have been required to control diagonal shear by using shear reinforcements (stirrups) in the beams. This eventually proved to be much cheaper.

In the excerpts of the design presented below, the following data were used;

Concrete grade = 30 N/mm²
Yield strength of reinforcement = 460 N/mm²
Concrete cover of reinforcement = 50mm
Bearing capacity of soil = 380 kN/m²

For this post, let us consider an axis from the structure in which the intermediate columns approximately have equal ultimate axial loads of 3081.075 kN while the end columns have an axial load of 1680.3 kN (see Figure below).

From the symmetrical arrangement of the loads, it is quite obvious that the centroid will pass through the middle column, hence, soil pressure can be assumed to be uniform under the whole length of the footing.

Dimension of all columns = (500 x 300 mm)
Total Ultimate Limit State (ULS) Load = 2(1680.3) + 5(3081.075) = 18765.975 kN
Axial load conversion factor to Serviceability Limit State (SLS) α = 1.45
N_SLS = 18765.975/(1.45 ) = 12942.05 kN

Assume 12% of service load to be the self-weight (S_W) of the footing
S_W = (12/100) × 12942.05 = 1553.04 kN
Area of footing required A_req = N_SLS+ S_W /(Bearing Capacity) = (12942.05 + 1553.04)/380 = 38.14 m²

Taking a 36.3m long base, the width B = 38.14/36.3 = 1.051 m
Therefore, provide a 1.1 m x 36.3m base (A_prov = 39.93 m²)
Let the thickness of the footing be 1100mm
Earth Pressure intensity (q) = N/A_prov = (18765.975 KN)/(39.93 m² ) = 469.97 kN/m²

Design of the footing cantilever slab portion (per meter strip)
Note that the bending moment in the slab is maximum at the face of the column (in this case at the face of the upstand beams)

Width of the upstand beam = 500mm = 0.5m

Hence, Moment arm (j_xx) = (1.1– 0.5)/(2 ) = 0.30 m
Assume depth of slab h = 300mm, concrete cover (C_c) = 50mm and assuming that (ϕ) Y20mm bars will be used;
Hence, the effective depth (d) = h – Cc – ϕ/2 = 300 – 50 – (20/2) = 240mm

M_x-x = (q × j_xx²) / 2 = (469.97 × 0.3²)/2 = 21.14 KN.m
k = M/(F_cubd² ) = (21.14 × 10⁶) / (30 × 1000 × 240² ) = 0.012
k < 0.156, no compression steel needed
Lever arm (l_a) = 0.5 + (0.25 – K/0.9)^0.5 Hence la = 0.95

The area of tenseion steel required; AS_req = M/(0.95f_y.l_a.d ) = (21.14 × 10⁶) / (0.95 × 460 × 0.95 × 240) = 212.17 mm²

AS_min = 0.13bh/100 = (0.13 × 1000 × 300)/100 = 390 mm²
Therefore, Provide Y12 @ 200 mm c/c (As_prov = 566 mm²/m)

Distribution bars on slab
AS_min = 0.13bh/100 = (0.13 × 1000 × 300)/100 = 390 mm²
Therefore, Provide Y12 @ 250 mm c/c (As_prov = 452 mm²/m)

Shear design of slab
The concrete resistance shear stress (Vc) = 0.632 × (100As/bd)^1/3 × (400/d)^1/4
Vc = 0.632 × [(100 × 566)/(1000 × 240)]^1/3 × (400/240)^1/4
Vc = 0.632 × 0.617 × 1.136 = 0.443 N/mm²
For F_cu = 30N/mm²
Vc = 0.443 × (30/25)1/3 = 0.470 N/mm²

Critical diagonal shear force at ‘d’ from the face of support (Q)
Q = q (j_xx – d)
Therefore the shear force Q= 467.97 × (0.30 – 0.24) = 28.078 KN/m
The resultant shear stress V = (28.078 × 10³)/(1000 × 240) = 0.1169 N/mm²
We can see that V < Vc, ie (0.1169 N/mm² < 0.470 N/mm²)
Hence, diagonal Shear is ok
A little consideration will show that punching shear at 1.5d from the face of column is also ok (falls outside the footing dimensions). Hence design is ok.

Design of ground longitudinal upstand beam

Width of beam = width of column = 500mm = 0.5m
Uniformly distributed soil reaction on beam = Earth pressure intensity × Width of footing = 467.97 x 1.1 = 514.77 KN/m

Bottom Reinforcement Design (Column Support Points)
Total depth of beam = 1100mm
Effective depth (d) = 1100 – 50 – 16 – 10 = 1024mm (assuming the use of Y32mm bars and Y10mm links)

Design Moment (M) = 1960.09 kN.m
k = M/(F_cubd² ) = (1960.09 × 10⁶)/(30 × 500 × 1024²) = 0.124

Since k < 0.156, no compression steel needed
Hence the lever arm, la = 0.5 + (0.25 – K/0.9)^0.5 Hence la = 0.834

The area of tension steel required; AS_req = M/(0.95F_y.la.d ) = (1960.09 × 10₆)/(0.95 × 460 × 0.834 × 1024) = 5252 mm²
AS_min = 0.26bh/100 = (0.26 × 500 × 1100)/100 = 1430 mm²
Provide 6Y32mm + 2Y25mm BOT (Asprov = 5806 mm²) around the penultimate supports. (See detailed drawings)

Central Column Support Bottom Reinforcement Design
M = 1603.71 kN.m
k = M/(F_cubd²) = (1603.71 × 10⁶) / (30 × 500 × 1024²) = 0.1019, k < 0.156, no compression steel needed.
la = 0.5 + (0.25 – K/0.9)^0.5 Hence la = 0.869

The area of tension steel required; AS_req = M/(0.95F_y.la.d ) = (1603.71 × 10⁶)/(0.95 × 460 × 0.869 × 1024) = 4124 mm²
AS_min = 0.26bh/100 = (0.26 × 500 × 1100)/100 = 1430 mm²
Provide 4Y32mm + 3Y25mm (As_prov = 4689 mm²) at the central column support. (See detailed drawings)

Top Reinforcement Design (Beam Span Design)
M = 1440.08 kN.m
k = M/(F_cubd²) = (1440.08 × 10⁶/(30 × 500 × 1024²) = 0.0916
l_a = 0.5 + (0.25 – K/0.9)^0.5 Hence la = 0.885
The area of tension steel required; AS_req = M/(0.95F_y.la.d ) = (1440.08 × 10⁶)/(0.95 × 460 × 0.885 × 1024) = 3387.5 mm²
AS_min = 0.18bh/100 = (0.18 × 500 × 1100)/100 = 990 mm²
Provide 4Y32mm + 2Y16mm (Asprov = 3618 mm²)

Mid-span Sections Design
M = 802.71 kN.m
k = M/(F_cubd²)) = (802.71 × 10⁶)/(30 × 500 × 1024²) = 0.051; k < 0.156, no compression steel needed
l_a = 0.5 + (0.25 – k/0.9)^0.5 Hence la = 0.939

The area of tension steel required; AS_req = M/(0.95F_y.la.d ) = (802.71 × 10⁶)/(0.95 × 460 × 0.939 × 1024) = 1910 mm²
AS_min = 0.18bh/100 = (0.18 × 500 × 1100)/100 = 990 mm²
Provide 4Y25mm + 2Y16mm (As_prov = 2366 mm²)

Longitudinal Face Side Bars
Provide Y16mm @ 200mm c/c both faces

Shear design of the longitudinal ground beams

Punching Shear at column face
Taking the maximum shear force for punching shear verification at column face
Punching shear at column face Vf = (1870 × 3.4846)/3.6346 = 1792.8 kN
Punching shear stress = (1792.8 × 1000)/(500 × 1024) = 3.5016 N/mm²
We can see that 3.5016 N/mm² < 0.8√f_cu < 5 N/mm²
Therefore, the depth of the section is adequate for shear.

Maximum shear force on the footing = 1870.99 kN
The shear force at ‘d’ from the face of the column, V_d = (1870.99 × 2.4606)/3.6346 = 1266.64 kN (calculation was done looking at the shear force diagram)
The shear stress v = V_d/bd = (1266.64 × 1000)/(500 × 1024) = 2.474 N/mm²

The concrete resistance shear stress (V_c) = 0.632 × (100As/bd)^1/3 × (400/d)^1/4
Hence, V_c = 0.632 × [(100 × 5806)/(500 × 1024)]^1/3 × (400/1024)^1/4
V_c = 0.632 × 1.032 × 1 = 0.652 N/mm²
For F_cu = 30 N/mm², V_c = 0.652 × (30/25)^1/3 = 0.692 N/mm²

Hence shear reinforcement is required
Trying 4 legs of Y10mm, we have;
Spacing Sv = (0.95 × F_yv × Asv)/(b_v (v – Vc)) = (0.95 × 460 × 314)/(500 (2.474 – 0.692)) = 154 mm
Maximum spacing = 0.75d = 0.75 × 1024 mm = 768 mm
Provide 4Y10mm @ 150mm c/c links as shear reinforcement

The extent of shear links
Shear stress of nominal links V_n = [(A_sv/S_v) × 0.95f_y + (b_w × V_c)] d
Concrete Resistance shear stress Vc = 0.692 N/mm²
Vn = [(236/300) × 0.95 × 460 + 500 × 0.692) × 1024 × 10^-3 = 706.326 KN (Taking a nominal shear reinforcement of 3Y10 @ 300mm)
Extent of shear links S_n = (V_d – V_n)/q + d Extent of shear links Sn = (1266.64 – 706.326)/(514.77 + 1.024) = 2.112 m
Therefore, stop main shear links(4Y10mm @ 150mm) i.e. at 2.2 m from face of column and provide nominal reinforcement of 3Y10 @ 300mm till the point of zero shear force.

Shear force at the central column support point = 1574.01 kN
The shear force at d from the face of column, V_d = (1574.01 × 1.88369)/3.05769 = 969.668 kN
The shear stress v = V_d/bd = (969.668 × 1000)/(500 × 1024) = 1.893 N/mm²
The concrete resistance shear stress (V_c) = 0.632 × (100As/bd)^1/3 × (400/d)^1/4
V_c = 0.632 × ((100 × 4689)/(500 × 1024))^1/3 × (400/1024)^1/4
V_c = 0.632 × 0.9711 × 1 = 0.613 N/mm²
For F_cu = 30 N/mm², V_c = 0.613 x (30/25)^1/3 = 0.6517 N/mm²

Hence shear reinforcement is required
Trying 4 legs of Y10mm, we have;
Spacing of links Sv = (0.95 × F_yv × A_sv)/(b_v (v – Vc)) = (0.95 × 460 × 314)/(500(1.893 – 0.6517)) = 221.087 mm
Maximum spacing = 0.75d = 0.75 x 1024 =768 mm
Provide 4Y10mm @ 200mm c/c links as shear reinforcement

The extent of shear links
Shear stress of nominal links V_n = [(A_sv/S_v) × 0.95f_y + (b_w × V_c)] d
V_c = 0.6517 N/mm²
V_n = [( 236/300) × 0.95 × 460 + 500 × 0.6517)] × 1024
Hence, V_n = 685.69 kN (Taking a nominal shear reinforcement of 3Y10 @ 300mm)
Extent of shear links S_n = (V_d – V_n)/q + d
Extent of shear links S_n = (698.665 – 685.69)/514.77 + 1.024 = 1.049 m
Stop shear links at 1.1 m from the face of the column and provide nominal reinforcement = 3Y10 @ 300mm
The picture below shows the section through the first span of the continuous footing;

The picture below shows the section through the second support of the continuous footing;