This article presents a worked example on the analysis and design of continuous one-way reinforced concrete slab. The bending moments and shear forces are obtained using tabular coefficients from BS 8110-1-1997, while the design is performed according EN 1992-1-1:2004.
The plan of the slab is provided in the figure below.
Design Data:
Variable load on slab = 2.5KN/m2
Finishes = 1.0KN/m2
Unit weight of concrete = 25KN/m3
Compressive strength of concrete (fck) = 20N/mm2
Characteristic Strength of main reinforcement (fyk) = 500N/mm2
Characteristic Strength of Shear reinforcement (fyv) = 500N/mm2
Analysis
Permanent action
Characteristic Self-weight of slab = 0.2 x 25 = 5KN/m2
Partition Load on Slab = 1.0KN/m2
Characteristic Permanent Load on Slab = 5.0 + 1.0 = 6.0KN/m2
Variable action
Variable load on slab = 2.5KN/m2
Total Design Load
Ultimate Load acting on Slab =1.35(6) + 1.5(2.5) = 11.85KN/m2
Applicability of table 3.12 of BS 8110
Before table 3.12 of BS 8110 can be used for evaluating the effects of loading, three conditions must be met:
- Condition 1: Bay area > 30m2
Check: Bay area = (6+6+6) x 4 = 72m2
72m2 > 30m2.
Result: Condition met
- Condition 2: qk/gk ≤ 1.25
Check: = 2.5/6 = 0.4
0.4 < 1.25.
Result: Condition met
- Condition 3: qk ≤ 5KN/m2
Check: qk = 2.5KN/m²
Result: Condition met
Since the three conditions are satisfied, table 3.12 can be used.
Loading Arrangement
Since table 3.12 shall be used, a maximum ultimate load arrangement (of 1.35gk + 1.5qk) shall be used over the entire span without additional redistribution.
Computation of Internal Forces:
The beam is taken to be continuous rather than simple at the end span to account its monolithic connection with the end beams
At Outer Support
Moment = 0.04FL = 0.04 x 11.85 x 6 x 6 = 17.06KNm
Shear force = 0.46 x F = 0.46 x 11.85 x 6 = 32.7KN
At First Interior Support
Moment = 0.086FL = 0.086 x 11.85 x 6 x 6 = 36.7 KNm
Shear force = 0.6 x F = 0.6 x 11.85 x 6 = 42.7KN
Near Middle of end Span
Moment = 0.075FL = 0.075 x 11.85 x 6 x 6 = 32.0KNm
Middle interior span
Moment = 0.063FL = 0.063 x 11.85 x 6 x 6 = 26.9KNm
(Note: F = wL = 11.85 x 6)
Design
Note:
For simplicity and practicality, since the end spans have higher moment than the middle span, the end span will only be designed and same reinforcement as the end spans is to be provided for the middle span.
Likewise, since the first interior supports have higher moment and shear force than the outer supports, then the first interior support will only be designed, and same reinforcement is to be provided for outer support.
Near Middle of end Span (End Span)
flexural strength design
- Calculate the cover to reinforcement
Cover
Minimum cover = 15 (XC1 exposure condition table (4.4N of EC2)
Nominal Cover = minimum cover + allowable deviation
Nominal cover = 15mm + 10mm = 25mm
2. Calculate the effective depth
Assume reinforcement diameter = 12mm
Effective depth = h-c-ᴓ/2
= 200-25-12/2
= 169mm
3. Check whether section is to be designed as singly or doubly reinforced beam
$ K\,=\,\,\frac{M}{bd^2f_{ck}} $
$ K\,=\,\,\frac{32×10^6}{1000x 169^2 x20} $
= 0.06
Since K (0.06) < K’ (0.168); design as singly reinforced.
4) Calculate the lever arm (Z)
$ Z\,\,=\,\,d\left( 0.5+\sqrt{\text{0.25}-\,\,\frac{K}{1.134}} \right) $
$ Z\,\,=\,\,169\left( 0.5+\sqrt{\text{0.25}-\,\,\frac{0.06}{1.134}} \right) $
Since 160.2 < 0.95d (160.6): use Z = 160.2
5. Calculate the area of steel
$ A_{st\,\,=\,\,\frac{M_{Ed}}{0.87f_{yk}Z}} $
$ A_{st\,\,=\,\,\frac{32×10^6}{0.87x500x160.2}} $
Ast = 459.2mm2
Provide Y12@200mm (562mm2)
6. Traverse reinforcement
Let’s calculate the transverse reinforcement using 0.0013bh ≥ 20%
Ast 0.0013 x 1000 x 200 = 260mm2
Provide T10@200mm2 (390mm2)
Deflection Check
- Calculate the actual span-effective depth ratio
Span/depth ratio = 6000/169 = 35.5
2. Calculate the limiting Span-effective depth ratio
l/d = K[11 + 1.5√fck ρ0/ρ + 3.2√fck (ρo/ρ – 1)3/2] if ρ ≤ ρo
l/d = K[11 + 1.5√fck ρo/ρ + 3.2√fck √ρo/ρ ] if ρ > ρo
ρ = Asprovided/b x d
ρ = 459.2/1000 x 169
= 0.003
ρo = 10-3√fck
ρo = 10-3√20
= 0.004
K = 1.3 (for end spans)
Since ρ ≤ ρo = then we will use
l/d = K[11 + 1.5√fck ρ0/ρ + 3.2√fck (ρo/ρ – 1)3/2]
l/d = 1.5[11 + 1.5√20 0.004/0.003 + 3.2√20 (0.004/0.003 – 1)3/2]
= 38.3mm
Since actual span-effective depth ratio is less than the limiting span-effective depth ratio, the end spans pass deflection check.
First Interior Supports
flexural strength design
- Calculate the cover to reinforcement
Cover
Minimum cover = 15 (XC1 exposure condition table (4.4N of EC2)
Nominal Cover = minimum cover + allowable deviation
Nominal cover = 15mm + 10mm = 25mm
2. Calculate the effective depth
Assume main reinforcement diameter = 12mm
Effective depth = h-c-ᴓ/2 = 200-25-12/2 = 169mm
3. Check whether section is to be designed as singly or doubly reinforced beam
$ K\,=\,\,\frac{M}{bd^2f_{ck}} $
$ K\,=\,\,\frac{32×10^6}{1000x 169^2 x20} $
= 0.06
Since K (0.06) < K’ (0.168); design as singly reinforced.
4) Calculate the lever arm (Z)
$ Z\,\,=\,\,d\left( 0.5+\sqrt{\text{0.25}-\,\,\frac{K}{1.134}} \right) $
$ Z\,\,=\,\,169\left( 0.5+\sqrt{\text{0.25}-\,\,\frac{0.06}{1.134}} \right) $
Since 160.2 < 0.95d (160.6): use Z = 160.2
5. Calculate the area of steel
$ A_{st\,\,=\,\,\frac{M_{Ed}}{0.87f_{yk}Z}} $
$ A_{st\,\,=\,\,\frac{36.7 x 10^6}{0.87x500x160.2}} $
Ast = 459.2mm2
Provide Y12@200mm (562mm2)
Traverse reinforcement/Distributed bar
Let’s provide as we have provided for the end span, hence:
Provide T10@200mm2 (390mm2)
Shear Strength Design
- Check whether the concrete section can resist the shear force without shear reinforcement
VRdc = (0.12K(100ρLfck)1/3 + K1σcp) bwd
K = (1 +√200/d) = 1 + (200/169)0.5 = 2.09
Since 2.09 > 2, hence K is taken as 2
ρL = Asl/bwd = 459.2/1000 x 169 = 0.003
VRdc = (0.12 x 2 (100 x 0.003 x 20) 1/3) 1000 x 169
VRdc = 88.6KN
Since VRdc (88.6KN) is greater than VEd (42.7KN) then shear reinforcement is not required
Detailing
The detailing of the top and bottom reinforcement is provided separately to avoid congestion
