This article presents a worked example on the design of half-turn 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 Span/d = 20 d = Span/20 = 3/20 = 0.15m Since d = 150mm, make h = 175mm

Estimate of Loads

Waist slab = 0.175 x 25 = 4.38KN/m2

Convert waist self-weight to horizontal load

4.38 x √(R² + T²)/T

4.38 x √(150² + 200²)/200

4.38 x 1.25 = 5.48KN/m2  

Steps = 0.5 x 0.15 x 25 = 1.88KN/m2  

Finishes = 1.2KN/m2

Live Load = 1.5KN/m2

Total dead load on stair = 5.48 + 1.88 + 1.2 = 8.5KN/m2 

Ultimate Load on staircase

= 1.3(gk) + 1.5(qk) = 1.3 x 8.8 + 1.5 x 1.5 = 16KN/m2  

Compute the Internal Forces:

Moment (M) =   wl²/8 = 16 x 3²/8 = 18.46KNm

V = Vl/2   = 16 x 3/2   = 24KN  

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/bd²fck

(18.46 x 10^6)/ (1000 x 149² x 25) =0.03

Since K (0.03) < K’ (0.167); design as singly reinforced.

• Calculate the Lever arm (z)

$ Z\,\,=\,\,d\left( 0.5+\sqrt{\text{0.25}-\,\,\frac{K}{1.134}} \right) $

$ Z\,\,=\,\,149\left( 0.5+\sqrt{\text{0.25}-\,\,\frac{0.003}{1.134}} \right)$

= 144.5mm

It is not according to Eurocode 2 that a limiting value of 0.95d should be maintained for lever arm (z), this is however recommended in the UK Annex and is a good practice, so we maintain it here.

Limiting value for lever arm = 0.95d

= 0.95 x 149 = 141.55mm

Since the value of Z (144.5mm) is greater than the upper limit value 0.95d (141.6), use Z = 141.6mm as lever arm.

• Calculate the area of Steel

$A_{st\,\,=\,\,\frac{M_{Ed}}{0.87f_{yk}Z}} $

$A_{st\,\,=\,\,\frac{18.46 x 10^6}{0.87 x 500 x 141.6}} $

As = 299.8mm²

Provide 5T12 (562.32mm²) @200 c/c

• Check minimum reinforcement requirement by the code.

Asmin = 0.26 x fctm/fyk x b x d ≥ 0.0013bd

Fctm = 0.3 x fck^(2/3) = 2.56

Asmin = 0.26 x 2.56/500 x 1000 x 149 ≥ 0.0013 x 1000 x 149

= 198.7mm2

Since Asmin (198.7mm²) < (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

• Check whether the concrete section can resist the shear force without shear reinforcement

V_Rdc = (0.12K(100_ρLf_ck)^1/3 ) b_wd

K     = (1 +√200/d) = 1 + (200/149)^0.5 = 2.16

The limiting value of K is 2. Since the 2.16 is > 2, then k is taken as 2

ρL = Asl/b_wd = 299.8/1000 x 149 = 0.002

Now plug the value of K and ρL into the equation

V_Rdc = (0.12 x 2 (100 x 0.002 x 25) ^1/3) 1000 x 149

V_Rdc = 71.96KN

Since V_Rdc (71.96KN) is greater than V_Ed (24KN) then shear reinforcement is not required.

 

Deflection Check

1. Calculate the actual span-effective depth ratio

Span/depth ratio = 3000/149   = 20.3

2. Calculate the limiting Span-effective depth ratio

ρ = Asrequired/b x d

ρ = 299.8/1000 x 149

= 0.002

ρ^o = 10^-3√fck

ρ^o = 10^-3√25

= 0.005

K = 1 (for simply supported)

Since ρ < ρ^o  = then we will use

l/d = [11 + 1.5√f_ck ρ0/ρ + 3.2√f_ck (ρ^o/ρ – 1)3/2]

l/d = [11 + 1.5√25 0.005/0.002 + 3.2√25 (0.005/0.002 – 1)3/2]

= 58.6mm

Since actual span-effective depth ratio (20.3) is less than the limiting span-effective depth ratio (58.6), the stair passes deflection check.

 

Click here to study a worked example on the design of this same staircase to BS 8110

 

Second Flight and Landing

Riser = 150mm

Tread = 200mm

Landing Span = 1115

Second Landing = 1138

Breadth = 1000

Calculate the Effective Span

Effective span = 1^st support breadth/2 + 1^st Landing Span + Flight horizontal Span + 2^nd Landing Span + 2^nd Support breadth/2

225/2 + 1115 + 9 x 200 + 1138 + 225/2 = 4.3m

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 the Loads on the stair

Waist slab = 0.2 x 25 = 5KN/m2

Convert waist self-weight to horizontal load

5 x √(R² + T²)/T

5 x √(150² + 200²)/200

5 x 1.25 = 6.25KN/m2  

Steps = 0.5 x 0.15 x 25 = 1.88KN/m2  

Finishes = 1.2KN/m2

Live Load = 1.5KN/m2

Total dead load on stair = 6.25 + 1.88 + 1.2 = 9.33KN/m2

Compute the Internal Forces:

Moment (M) =   wl²/8 = 17.4 x 4.3²/8 = 39.8KNm

V = Vl/2   = 16 x 4.3/2   = 36.54KN  

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 = 174

• Check whether section is to be designed as singly or doubly reinforced beam

K = M/bd²fck

(39.8 x 10^6)/ (1000 x 174² x 25) =0.05

Since K (0.05) < K’ (0.167); design as singly reinforced.

• Calculate the Lever arm (z)

$ Z\,\,=\,\,d\left( 0.5+\sqrt{\text{0.25}-\,\,\frac{K}{1.134}} \right) $

$ Z\,\,=\,\,174\left( 0.5+\sqrt{\text{0.25}-\,\,\frac{0.05}{1.134}} \right)$

= 165.5mm

Limiting value for lever arm = 0.95d

= 0.95 x 174 = 165.3mm

Since the value of Z (165.5mm) is greater than the upper limit value 0.95d (165.3), use Z = 165.3mm as lever arm.

• Calculate the area of Steel

$A_{s\,\,=\,\,\frac{M_{Ed}}{0.87f_{ck}Z}} $

$A_{s\,\,=\,\,\frac{39.8x 10^6}{0.87 x 25 x 165.3}} $

As = 553.6mm²

Provide 6Y12 (674.8mm²) @150 c/c

• Check minimum reinforcement requirement by the code.

Asmin = 0.26 x fctm/fyk x b x d ≥ 0.0013bd

Fctm = 0.3 x fck^(2/3) = 2.56

Asmin = 0.26 x 2.56/500 x 1000 x 174 ≥ 0.0013 x 1000 x 174

= 232mm2

Since Asmin (232mm²) < (553.6mm²), 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

• Check whether the concrete section can resist the shear force without shear reinforcement

V_Rdc = (0.12K(100_ρLf_ck)^1/3 ) b_wd

K     = (1 +√200/d) = 1 + (200/174)^0.5 = 2.07

The limiting value of K is 2. Since the 2.07 is > 2, then k is taken as 2

ρL = Asl/b_wd = 533.6/1000 x 174 = 0.003

Now plug the value of K and ρL into the equation

V_Rdc = (0.12 x 2 (100 x 0.003 x 25) ^1/3) 1000 x 174

V_Rdc = 95.82KN

Since V_Rdc (95.82KN) is greater than V_Ed (36.54KN) then shear reinforcement is not required.

Deflection Check

1. Calculate the actual span-effective depth ratio

Span/depth ratio = 4200/174   = 24.13

2. Calculate the limiting Span-effective depth ratio

ρ = Asrequired/b x d

ρ = 533.6/1000 x 174

= 0.003

ρ^o = 10^-3√fck

ρ^o = 10^-3√25

= 0.005

K = 1 (for simply supported)

Since ρ < ρ^o  = then we will use

l/d = [11 + 1.5√f_ck ρ0/ρ + 3.2√f_ck (ρ^o/ρ – 1)3/2]

l/d = [11 + 1.5√25 0.005/0.003 + 3.2√25 (0.005/0.003 – 1)3/2]

= 29.7mm

Since actual span-effective depth ratio (24.13) is less than the limiting span-effective depth ratio (29.7), the stair passes deflection check.

Detailing