Analysis and Design of Continuous Continuous Slab and Beam Footing… Footing…
Ubani Obinna U. (2016)
EXAMPLE ON THE ANALYSIS AND DESIGN OF CONTINUOUS SLAB AND BEAM FOOTING PER BS 8110-1:1997 Ubani Obinna Uzodimma Department of Civil Engineering, Nnamdi Azikiwe University PMB 5025, Awka, Anambra State, Nigeria E-mail:
[email protected]
1.1 Introduction This design is just an excerpt from my final year project ‘Structural Analysis and Design of 35,000 Capacity Reinforced Concrete Stadium’. After analysis and design of the superstructure (see figure 1.1), I realized I had a very large magnitude of axial loads and moments at 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 at Naze, Owerri) so shallow foundation was adopted.
Figure 1.1: A section through the stadium
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 a soil with such good bearing capacity. After much consideration, I realized that chaining the columns continuously will do the trick, but at the same time, I could combine the slab with upstand beams running continuously along the axis of the column. My whole aim was to reduce the great quantity of concrete that would have been required to control diagonal shear by using shear r einforcements einforcements (stirrups) in the beam b eamss which pr oved to be much cheaper cheaper (see Figure 1.1).
Figure 2.2: Schematic form of continuous beam and slab foundation
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Analysis and Design of Continuous Continuous Slab and Beam Footing… Footing…
Ubani Obinna U. (2016)
For this paper, I selected an axis from the structure in which the intermediate columns had approximately an ultimate axial load of 3081.075 KN each while the end columns had an axial load of 1680.3 KN (see Figure 1.3).
Figure 3.3: Loading configuration of the foundation.
From the symmetrical arrangement 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.
1.2 Design Data Concrete cover = 50mm; Fy = 460 N/mm 2; Fyv = 460 N/mm2; Fcu = 30 N/mm2 Dimension of all columns = (500 × 300 mm); Bearing Capacity = 380 KN/m2 Total Ultimate Limit State (ULS) Load = 2(1680.3) + 5(3081.075) = 18765.975 KN Axial load conversion factor to Serviceability Limit State (SLS) = 1.45
. = 12942.05 KN NSLS = .
Assume 12% of service load to be the self weight of the footing
× 12942.05 = 1553.04 KN = . . Area of footing required Areq = . Taking a 36.3m long base, the width B = . = 1.051 m SW =
= 38.14 m2
Provide a 1.1 m x 36.3m x 1.1m (trial depth) base (Aprov = 39.93 m2) Earth Pressure intensity (q) =
=
. = 469.97 KN/m2 .
1.3 Design of the slab (per meter strip) Bending moment on the slab is maximum at the face of the column (in this case at the face of the upstand beams) (clause 3.11.2.2 BS 8110-1:1997 ) ) Width of the beam = 500mm = 0.5m Hence, Moment arm (jxx) =
. . = 0.30 m
Assume depth of slab h = 300mm; concrete cover = 50mm and assuming that Y12mm bars will be used at the slab; Hence, the effective depth (d) = 300 – 50 (12/2) = 244mm
k=
=
−
. × . = 21.14 KN.m . × = 0.01 2; k < 0.156, no compression steel needed (clause 3.4.4.4) × × 0.012
The design moment Mx-x =
×
=
Lever arm (la) = 0.5 + (0.25 – k/0.9)0.5; Hence la = 0.95
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Analysis and Design of Continuous Continuous Slab and Beam Footing… Footing…
Area of tension steel required ASreq = Minimum area of steel ASmin =
Ubani Obinna U. (2016)
. × 212. 2.17 17 mm = ... . × × . × = 21
. = . × × = 390 mm2 (Table 3.25 BS 8110-1:1997)
Provide Y12 @ 200 mm c/c (Asprov = 566 mm2/m) Distribution bars on slab ASmin =
. = . × × = 390 mm2 (Table 3.25 BS 8110-1:1997)
Provide Y12 @ 200 mm c/c (Asprov = 566 mm2/m) in longitudinal direction as distribution bars Check for shear
(Table 3.8 BS 8110-1:1997) The concrete resistance shear stress (Vc) = 0.632 × × Vc = 0.632 × × ×
Vc = 0.632 × 0.6144 × 1.1315 = 0.439 N/mm2 1 2
For Fcu = 30N/mm , Vc = 0.439 ×
30 3 25
= 0.4665 N/mm2
Critical diagonal shear force at d from face of support = V = q(jxx – d) V = 467.97 × (0.30 – 0.244) = 28.078 KN/m The shear stress v =
=
. × = 0.1151 N/mm2 (clause 3.4.5.2 BS 8110-1:1997) ×
0.1151 N/mm2 < 0.4665 N/mm2 Diagonal Shear is ok Punching shear is also ok (perimeter falls outside the footing dimensions). Hence design is ok.
1.4 Design of the longitudinal beam Uniformly distributed soil reaction on beam = 467.97 KN/m2 × 1.1m = 514.77 KN/m
This is just like a continuous beam that is turned upside down, and analysing it by Clapeyron’s three moment equation gives the bending moment and shear force diagram shown in figure 1.4.
Figure 4.4: Internal stresses diagram
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Analysis and Design of Continuous Continuous Slab and Beam Footing… Footing…
Ubani Obinna U. (2016)
Bottom reinforcement design Width of beam (b) = 500mm; Total depth h = 1100mm, Concrete Cover Cc = 50mm; Assume Y32mm bars
∅
Effective depth d = h – Cc - / 2 -
∅
links =
1100
− 50 – 16 − 10 = 1024mm
Design of points B and F M = 1960.09 KN.m
k=
= . × 0.124 4 (clause 3.4.4.4 BS 8110-1:1997) × × = 0.12
k < 0.156, no compression steel needed
Lever arm (la) = 0.5 + (0.25 – k/0.9)0.5; Hence la = 0.834 ASreq =
. × = 52 5252 52m mm = . × × . × ...
ASmin =
. = . × × = 1430mm2 (Table 3.25 BS 8110-1:1997)
Provide 7Y32mm BOT (Asprov = 5628 mm2) Design of point D (Bottom) to cover all bottom midspan areas M = 1603.71 KN.m
k=
= . × = 0.10 19 (clause 3.4.4.4 BS 8110-1:1997) × × 0.1019
k < 0.156, no compression steel needed la = 0.5 + (0.25 – K/0.9)0.5 Hence la = 0.869 ASreq =
...
ASmin =
=
. × 124 4 mm . × × . × = 412
. = . × × = 1430 mm2 (Table 3.25 BS 8110-1:1997)
Provide 4Y32mm + 3Y25mm BOT (Asprov = 4689mm2) mainly around the mid span supports. (See detailed drawings) Top reinforcement design Width of flange (bf) = 1100mm Span A-B and F-G M = 1440.08 KN.m
k=
= . × = 0.04 16 (clause 3.4.4.4 BS 8110-1:1997) × × 0.0416
k < 0.156, no compression steel needed la = 0.5 + (0.25 – K/0.9)0.5 Hence la = 0.95 ASreq =
. × 3387. 87.5 5 mm = ... . × × . × = 33
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Analysis and Design of Continuous Continuous Slab and Beam Footing… Footing…
ASmin =
Ubani Obinna U. (2016)
. = . × × = 990 mm2 (Table 3.25 BS 8110-1:1997)
Provide 4Y32mm + 2Y16mm TOP (Asprov = 3618 mm 2)
Mid span areas (between spans B and F) M = 802.71 KN.m
k=
= . × = 0.02 3 (clause 3.4.4.4 BS 8110-1:1997) × × 0.023
k < 0.156, no compression steel needed la = 0.5 + (0.25 – K/0.9)0.5; Hence la = 0.95 ASreq =
...
ASmin =
. × 1888.224 24 mm = . × × .× = 1888.2
. = . × × = 990mm2 (Table 3.25 BS 8110-1:1997)
Provide 4Y25mm + 2Y16mm TOP (Asprov = 2366 mm 2) Longitudinal side bars Provide Y16mm @ 200mm c/c both faces. (Clause 3.12.11.2.6 BS 8110-1:1997) Shear design
Maximum shear force on the whole foundation = 1870.99 KN Hence maximum shear stress v =
√
=
3.654 N/mm2 < 0.8 30 < 5 N/mm2
. × = 3.654 N/mm2 (clause 3.4.5.2 BS 8110-1:1997) ×
Hence, section is ok for shear Shear design of support of column A
Shear force at support of column A = 1217.63 kN At d from the face of column, Vd =
. × . = 613.29 KN (clause 3.4.5.10 BS 8110-1:1997) .
. × = 1.197 N/mm2 (clause 3.4.5.2 BS 8110-1:1997) Shear stress v = =
×
(Table 3.8 BS 8110-1:1997) Vc = 0.632
× × Vc = 0.632 × × Vc = 0.632 × 0.891× 1 = 0.5629 N/mm2
For Fcu = 30N/mm , Vc = 0.5629 × = 0.5978 N/mm2 2
Vc + 0.4 < v
Hence shear reinforcement is required
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Analysis and Design of Continuous Continuous Slab and Beam Footing… Footing…
Ubani Obinna U. (2016)
Trying 4 legs of Y10mm (Asv = 314mm2), we have; Spacing Sv =
. × × = . × × = 458 mm (Table 3.7 BS 8110-1:1997) () (..)
Maximum spacing = 0.75d = 0.75 × 1024 = 768 mm (clause 3.4.5.5 BS 8110-1:1997) Provide 4Y10mm @ 300mm c/c links as shear reinforcement (adopted as nominal reinforcement)
At column support B (Span A-B)
Shear force on footing = 1870.99 KN
. × . = 1266.64 KN (clause 3.4.5.10 BS 8110-1:1997) . . × = 2.474 N/mm2 Shear stress at d from the face of column v = = × At d from the face of column, V d =
Vc = 0.632
(Table 3.8 BS 8110-1:1997)
× × Vc = 0.632 × ×
Vc = 0.632 × 1.032 × 1 = 0.652 N/mm2
For Fcu = 30N/mm , Vc = 0.652 × = 0.692 N/mm2 2
Vc + 0.4 < v
Hence shear reinforcement is required
Trying 4 legs of Y10mm, we have Spacing Sv =
. × × = . × × = 154 mm (Table 3.7 BS 8110-1:1997) () (..)
Maximum spacing = 0.75d = 0.75 × 1024 = 768 mm (clause 3.4.5.5 BS 8110-1:1997) Provide 4Y10mm @ 150mm c/c links as shear reinforcement Extent of shear links Total shear force resistance of the reinforced concrete Vn =
Fyv + bw × Vcd × 0.95 × Fyv
Vc = 0.692 N/mm2 Vn =
× 0.95 × 460 + 500 × 0.692 × 1024
Vn = 706.326 KN
+ d . – . + 1.024 = 2.112 m Extent of shear links Sn = . Extent of shear links Sn =
Stop shear links at 2.2 m from face of column and provide nominal reinforcement At column support B (Span B-C)
Shear force = 1633.4 KN At d from the face of column, Vd =
. × . = 1029 KN (clause 3.4.5.10 BS 8110-1:1997) .
× = 2.00 N/mm2 (clause 3.4.5.2 BS 8110-1:1997) v= =
×
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Analysis and Design of Continuous Continuous Slab and Beam Footing… Footing…
Ubani Obinna U. (2016)
Vc = 0.632
(Table 3.8 BS 8110-1:1997)
× × Vc = 0.632 × ×
Vc = 0.632 × 1.032 × 1 = 0.652 N/mm2
For Fcu = 30N/mm , Vc = 0.652 × = 0.692 N/mm2 2
Vc + 0.4 < v
Hence shear reinforcement is required
Trying 4 legs of Y10mm, we have Spacing Sv =
. × × = . × × = 209.9 mm () (..)
(Table 3.7 BS 8110-1:1997)
Maximum spacing = 0.75d = 0.75 × 1024 = 768 mm (clause 3.4.5.5 BS 8110-1:1997) Provide 4Y10mm @ 200mm c/c links as shear reinforcement
At column support D (middle column) col umn)
Shear force at middle column = 1574.01 kN At d from the face of column, Vd =
. × . = 969.668 KN (clause 3.4.5.10 BS 8110-1:1997) .
. × = 1.893 N/mm2 (clause 3.4.5.2 BS 8110-1:1997) Shear stress v = =
×
Vc = 0.632
(Table 3.8 BS 8110-1:1997)
× × Vc = 0.632 × × Vc = 0.632 × 0.9711 × 1 = 0.613 N/mm2
For Fcu = 30N/mm , Vc = 0.613 × = 0.6517 N/mm2 2
Vc + 0.4 < v
Hence shear reinforcement is required
Trying 4 legs of Y10mm, we have Spacing Sv =
. × × = . × × = 221.087 mm (Table 3.7 BS 8110-1:1997) () (..)
Maximum spacing = 0.75d = 0.75 × 1024 = 768 mm (clause 3.4.5.5 BS 8110-1:1997) Provide 4Y10mm @ 200mm c/c links as shear reinforcement Extent of shear links Total shear resistance of the concrete Vn =
× 0.95 × Fy + bw × Vc d
Vc = 0.6517 N/mm2 Vn =
× 0.95 × 460 + 500 × 0.6517 × 1024
Vn = 685.69 KN
+ d . – . + 1.024 = 1.049 m Extent of shear links Sn = . Extent of shear links Sn =
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Analysis and Design of Continuous Continuous Slab and Beam Footing… Footing…
Ubani Obinna U. (2016)
Stop shear links at 1.1 m from face of column and provide nominal reinforcement = 3Y10 @ 300mm
Detailing Typical sections cut through points Band C is shown in Figure 1.5. Kindly download full detailed drawing from www.structville.blogspot.com
Figure 5.5: Sections through points B and C
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