This article presents a worked example on the design of half-turn staircase to BS 8110-1:1997 (Click here to study the design of the same staircase to Eurocode 2). The staircase to be designed connects the ground floor and the first floor of a building. The flights of the staircase span a horizontal distance of 1.8m while the first landing supported by end beam span a distance of 1.115m. The final landing connecting to the first floor spans a distance of 1.138m. The layout of the staircase is shown below. 
Worked Example
First Flight and Landing
Riser = 150mm
Tread = 200mm
Landing Span = 1115mm
Breadth = 1000mm
Calculate the Effective Span
Effective span = Flight horizontal Span + Landing Span + Support breadth/2
= 9 x 200 + 1115 + 225/2 = 3m
Choose the slab waist thickness (h)
Span/d = 20
d = Span/20 = 3/20 = 0.15m
Since d = 150mm, make h = 175mm
Estimate Loads on First Flight
Waist slab = 0.175 x 24 = 4.2KN/m2
Convert waist self-weight to horizontal load
4.2 x √(R² + T²)/T
4.2 x √(150² + 200²)/200
4.2 x 1.25 = 5.25KN/m2
Steps = 0.5 x 0.15 x 24 = 1.8KN/m2
Finishes = 1.2KN/m2
Live Load = 1.5KN/m2
Total dead load on stair = 5.25 + 1.8 + 1.2 = 8.25KN/m2
Ultimate Load on staircase
= 1.4(gk) + 1.6(qk) = 1.4 x 8.25 + 1.6 x 1.5 = 16.84KN/m2
Compute the Internal Forces:
Moment (M) = wl²/8 = 16.84 x 3²/8 = 19.29KNm
Shear force (V) = Vl/2 = 16.84 x 3/2 = 25.26KN
Design flexural strength design
- Calculate the effective depth
Assumptions
Cover = 20mm
Main reinforcement diameter = 12mm
Effective depth = h-c-ᴓ/2 = 175-20-12/2 = 149
- Check whether section is to be designed as singly or doubly reinforced beam
K = M/bd²fcu
=(19.29 x 10^6)/ (1000 x 149² x 25) =0.04
Since K (0.04) < K’ (0.156); design as singly reinforced.
- Calculate the Lever arm (z)
$ Z\,\,=\,\,d\left( 0.5+\sqrt{\text{0.25}-\,\,\frac{K}{0.9}} \right) $
$ Z\,\,=\,\,149\left( 0.5+\sqrt{\text{0.25}-\,\,\frac{0.004}{0.9}} \right)$
= 143mm
Limiting value for lever arm = 0.95d
= 0.95 x 149 = 141.55mm
Since the value of Z (143mm) is greater than the upper limit value 0.95d (141), use Z = 141.6mm as lever arm.
- Calculate the area of Steel
$A_{st\,\,=\,\,\frac{M}{0.87f_{y}Z}} $
$A_{st\,\,=\,\,\frac{19.29 x 10^6}{0.87 x 460 x 141.6}} $
As = 340.54mm²
Provide 5T12 (562.32mm²) @200 c/c
-
Check minimum reinforcement requirement by the code.
Asmin = 0.0013bd Asmin
= 0.0013 x 1000 x 149 = 193.7mm2
Since Asmin (193.7mm²) < (562.32mm²), minimum reinforcement requirement is satisfied.
-
Check maximum reinforcement requirement by the code.
Asmax = 0.04Ac
= 0.04 x b x h = 0.04 x 1000 x 175 = 7000mm2
Since As < 7000mm², maximum area of reinforcement criteria is satisfied.
Shear Strength Design
- Calculate the shear stress
v = V/ bd
$\frac{25.26×10^3}{1000×149}\,\,\,\,$
= 0.17N/mm2
- Check whether the concrete section can resist the shear force without shear reinforcement.
$ v_{c\,\,=}\,\,\text{0.79}\left( \frac{100As}{b_vd} \right) ^{\text{1/}3}\frac{\left( \frac{400}{407} \right) ^{\text{1/}4}}{\gamma _m} $
$ v_{c\,\,=}\,\,\text{0.79}\left( \frac{100 x 562.32}{1000 x 149} \right) ^{\text{1/}3}\frac{\left( \frac{400}{149} \right) ^{\text{1/}4}}{1.5} $
vc = 0.58N/mm2
Since Vc (0.58N/mm2) is greater than V (0.17N/mm2) then there is no need for shear reinforcement.
Deflection Check
- Calculate the actual span-effective depth ratio
Span/depth ratio = 3 x 1000/149 = 20.13
2. Calculate the limiting Span-effective depth ratio
From table 3.9 of BS 8110, the basic Span/depth ratio for a simply supported beam is 20.
Tension reinforcement modification factor
$ \text{0.55}+\frac{477-\,\,f_s}{120\left( \text{0.9}+\frac{M}{bd^2} \right)} $
where:
$ f_s\,\,=\,\,\frac{2f_yA_{sreq}}{3A_{sprov}} $
$ f_s\,\,=\,\,\frac{2 x 460 x 340.54}{3 x 562.3} $
fs = 185.72
Substitute fs and values of other parameters into the tension modification formular
$ \text{0.55}+\frac{477-\,\,185.72}{120\left( \text{0.9}+\frac{19.29×10^6}{1000 x 149^2} \right)} $
= 2.11
Limiting Value of tension modification factor is 2. Since tension modification factor (2.11) is greater than the limiting value, the limiting value is adopted. Hence m.d = 2
Modified basic span-effective depth ratio = basic span-effective depth ratio x M.F tension rebars
Modified basic span-effective depth ratio = 20 x 2
= 40
Since Actual L/d (20.13) < modified Basic L/d (40); the staircase passes deflection check.
Second Flight and Landing
Riser = 150mm
Tread = 200mm
Landing Span = 1115
Second Landing = 1138
Breadth = 1000
Calculate the Effective Span
Effective span = 1st support breadth/2 + 1st Landing Span + Flight horizontal Span + 2nd Landing Span + 2nd Support breadth/2
225/2 + 1115 + 9 x 200 + 1138 + 225/2 = 4.2m
Choose the slab waist thickness (h)
Span/d = 20
d = Span/20 = 4.2/20 = 0.21m
Since d = 0.21, make h = 200mm
Estimate Loads on Second Flight
Waist slab = 0.2 x 24 = 4.8KN/m2
Convert waist self-weight to horizontal load
4.8 x √(R² + T²)/T
4.8 x √(150² + 200²)/200
4.8 x 1.25 = 6KN/m2
Steps = 0.5 x 0.15 x 24 = 1.8KN/m2
Finishes = 1.2KN/m2
Live Load = 1.5KN/m2
Total dead load on stair = 6 + 1.8 + 1.2 = 9KN/m2
Ultimate Load on staircase
= 1.4(gk) + 1.6(qk) = 1.4 x 9 + 1.6 x 1.5 = 18.15KN/m2
Compute the Internal Forces:
Moment (M) = wl²/8 = 18.15 x 4.2²/8 = 41.5KNm
Shear force (V) = Vl/2 = 18.15 x 4.2/2 = 38.12KN
Design flexural strength design
- Calculate the effective depth
Assumptions
Cover = 20mm
Main reinforcement diameter = 12mm
Effective depth = h-c-ᴓ/2 = 200-20-12/2 = 174
- Check whether section is to be designed as singly or doubly reinforced beam
K = M/bd²fcu
(41.5 x 10^6)/ (1000 x 174² x 25) =0.05
Since K (0.05) < K’ (0.156); design as singly reinforced.
- Calculate the Lever arm (z)
$ Z\,\,=\,\,d\left( 0.5+\sqrt{\text{0.25}-\,\,\frac{K}{0.9}} \right) $
$ Z\,\,=\,\,174\left( 0.5+\sqrt{\text{0.25}-\,\,\frac{0.005}{0.9}} \right)$
= 162.7mm
Limiting value for lever arm = 0.95d
= 0.95 x 174 = 165.3mm
Since the value of Z (162.7mm) is less than the upper limit value 0.95d (165.3mm), use Z = 162.7mm as lever arm.
- Calculate the area of Steel
$A_{st\,\,=\,\,\frac{M}{0.87f_{y}Z}} $
$A_{st\,\,=\,\,\frac{41.5 x 10^6}{0.87 x 460 x 162.7}} $
As = 637.9mm²
Provide 7T12 (787.2mm²) @150 c/c
-
Check minimum reinforcement requirement by the code.
Asmin = 0.0013bd Asmin
= 0.0013 x 1000 x 174 = 226.2mm2
Since Asmin (226.2mm²) < (787.2mm²), minimum reinforcement requirement is satisfied.
-
Check maximum reinforcement requirement by the code.
Asmax = 0.04Ac
= 0.04 x b x h = 0.04 x 1000 x 200 = 8000mm2
Since As < 8000mm², maximum area of reinforcement criteria is satisfied.
Shear Strength Design
- Calculate the shear stress
v = V/ bd
$\frac{38.12×10^3}{1000 x 174}\,\,\,\,$
= 0.22N/mm2
- Check whether the concrete section can resist the shear force without shear reinforcement.
$ v_{c\,\,=}\,\,\text{0.79}\left( \frac{100As}{b_vd} \right) ^{\text{1/}3}\frac{\left( \frac{400}{d} \right) ^{\text{1/}4}}{\gamma _m} $
$ v_{c\,\,=}\,\,\text{0.79}\left( \frac{100 x 787.2}{1000 x 174} \right) ^{\text{1/}3}\frac{\left( \frac{400}{174} \right) ^{\text{1/}4}}{1.5} $
vc = 0.6N/mm2
Since Vc (0.6N/mm2) is greater than V (0.22N/mm2) then there is no need for shear reinforcement.
Deflection Check
- Calculate the actual span-effective depth ratio
Span/depth ratio = 4 x 1000/174 = 24.59
2. Calculate the limiting Span-effective depth ratio
From table 3.9 of BS 8110, the basic Span/depth ratio for a simply supported beam is 20.
Tension reinforcement modification factor
$ \text{0.55}+\frac{477-\,\,f_s}{120\left( \text{0.9}+\frac{M}{bd^2} \right)} $
where:
$ f_s\,\,=\,\,\frac{2f_yA_{sreq}}{3A_{sprov}} $
$ f_s\,\,=\,\,\frac{2 x 460 x 637.9}{3 x 787.2} $
fs = 248.5
Substitute fs and values of other parameters into the tension modification formular
$ \text{0.55}+\frac{477-\,\,248.5}{120\left( \text{0.9}+\frac{41.5×10^6}{1000 x 174^2} \right)} $
= 1.63
Limiting Value of tension modification factor is 2. Since tension modification factor (1.63) is less than the limiting value, then m.d = 1.63
Modified basic span-effective depth ratio = basic span-effective depth ratio x M.F tension rebars
Modified basic span-effective depth ratio = 20 x 1.63
= 32.68
Since Actual L/d (24.59) < modified Basic L/d (32.68); the staircase passes deflection check.
Detailing
