**Introduction**

Slabless, sawtooth staircase (Figure 1) is one type of the stair that produces a lot of aesthetics appeal. Aesthetics and function are prime considerations in the design of buildings so if you learn how to design and build this type of stair, it would be good to actualize it. Different codes have their way of designing this type of stair but I would dwell on the method outlined in the book by Reynolds and Steedman (1988) which is in consonance with BS 8110 and Eurocode 2 (EC 2).

According to Reynolds and Steedman (2010), Cusens (1966) showed that if axial shortening is neglected and the strain energy due to bending only is considered, the midspan moment for the sawtooth stair case can be given by the general expression.

Where k = stiffness of thread/stiffness of riser and **j** is the number of threads

**If j is odd,**

**If j is even**,

I refer my reader to study principles of strain energy to understand more the concept defined here.

**Steps in the design of slabless stair case**

Step 1; Carry out the functional design of the staircase to determine length of the span based on the stairwell and using Blondel formula to determine the suitable length of thread (going) and riser height.

The Blondel formula states thus, 2R+T = 600 mm where R = Riser and T = Thread or Going (See table 1).

Table 1; Optimum dimensions of stair case (mm) – BS 5395

**Step 2**; Carry out the load analysis based on EC 2 guide to determine the design load, n_{d}. The design load is considered as concentrated load at the mid span as shown below (Figure 2)

**Step 3**; Determine k from the expression below, (See Figure 3 for explanation of expressions)

Where L_{t }= length of going/thread; L_{r} = length of riser; h_{t} = height of thread; h_{r }= height of riser

**Step 4**; Determine support moment coefficient from the chart in Figure 4 based on k and j

Using the support moment coefficient, calculate the support moment based on the expression below,

Support moment, M_{s} = coefficient x n_{d}L^{2}

Otherwise, follow **step 4a**

**Step 4a**; Determine the parameters, k_{11}, k_{12}, k_{13} and k_{14} from j, which is the number of goings/threads and determine the support moment from the expression.

**Step 5**; Determine the free bending moment from the expression

**Step 6**; Determine the maximum moment at the midspan (M_{o}) with the expression

M_{o} = M – M_{s}

Where M = free bending moment and M_{s} = support moment

**Step 7**; Draw the bending moment and shear force diagrams if required. They should be as shown

**Step 8**; Determine the reinforcement required at mid span and support based on the mid span moment and support moments and detail the slab as shown below. Due to the stair profile, concentrations of stress occur in the re-entrant corners, and the actual stresses to be resisted will be larger than those calculated from the moment. Note that re-entrant corner is any inside corner that forms an angle of 180^{0} or less. In a solid object subjected to internal or external loads, re-entrant corners create high stress concentrations. To resist such stresses, Cusens recommends providing twice the reinforcement theoretically required (at the re-entrant corners) unless suitable fillets or haunches are incorporated at these junctions. If this can be done, the actual steel provided should be about 10% more than the theoretically necessary. The possible reinforcement patterns are shown below (Figure 7)

Figure 7 is very suitable but practically, if haunches are provided otherwise the bars should be arranged for wall-to-wall corners as shown below (Figure 8).

If Figure 7 would not be possible, Figure 9 can be adopted for the reinforcement

**Example**

A proposed slabless stair case has the following geometrical details and design information

Length of going/thread, L_{t}= 300 mm

Length of Riser, L_{r} = 150 mm

Height of thread, h_{t }= 125 mm

Height of riser, h_{r} = 125 mm

Number of threads = 6

Width of stair = 1200 mm

Length of stair, L = 300 x 6 = 1800 mm

Variable action (public access) = 3.0 kN/m^{2}

Weight due to finishes as 1.5 kN/m^{2}

Unit weight of concrete = 25 kN/m^{3}

Characteristic strength of concrete, f_{ck} = 25 N/mm^{2};

Characteristic strength of steel, f_{yk} = 460 N/mm^{2}

Determine the support moments and midspan moment and design the stair

**Step 1**; The functional design parameters are outlined above

**Step 2;** Loading

Self-weight of the staircase = {[(L_{t} x h_{t}) + (L_{r }x h_{r})]/L_{t}} x f_{ck }= {[(0.3 x 0.125) + (0.15 x 0.125)]/0.3 x 25 =

0.1875 x 25 = 4.69 kN/m^{2}

Weight of finishes = 1 kN/m^{2}

Total permanent action = 4.69 + 1 = 5.69 kN/m^{2}

Variable action = 3 kN/m^{2}

Design action, n_{d} = 1.35g_{k} + 1.5q_{k} = 1.35 (5.69) + 1.5 (3) = 12.18 kN/mper m width, say 13 kN/mper m width

**Step 3;** Determination of k

K = (h_{t}^{3}l_{r})/ (h_{r}^{3}l_{t}) = (125^{3} x 150) (125^{3} x 300) = 150/300 = 0.5

**Step 4**; Determine support moment coefficient from the chart in Figure 4 based on k and j

Since k = 0.5 and j = 6 the support moment coefficient from chart = -0.875

Support moment, Ms = -0.0875 x 13 x 1.8^{2 }= -3.69 kNm

Using formula to estimate this value, since j is even,

= ((1/48) x 6 x (6-1) x (6-2) = 2.5

= (1/48) x (6-1) x (6-2) x (6-3) = 1.25

= (1/2) x (6-1) = 2.5

= (1/2) x (6-2) = 2

M_{s} = [13 x 1.8^{2} (2.5 + (0.5 x 1.25))]/6^{2}[2.5 + (0.5 x 2)] = 131.625/126 = 1.044 kNm (the value is lower than one obtained from chart). We would proceed with the value from chart.

**Step 5**; Determine the free bending moment. Since j is even, free bending moment, M = (1/8) x 13 x 1.8^{2 }= 5.265 kNm

**Step 6**; Determine the maximum moment at the midspan (Mo) with the expression

M_{o} = 5.265 – 3.69 = 1.575 kNm

**Step 7**; Draw the bending moment and shear force diagrams if required. They should be as shown

**Step 8:** Estimation of area of reinforcement: In slabless stair case, the links are designed while the main bars are provided. It is usually 6 number of bars per link and of the same size as the link.

M_{Ed} = 1.575 kNm

b = 1000; f_{ck} = 25 N/mm^{2}

Assuming bar size of 12 mm and concrete cover of 25 mm, d = 150 – 25 – (12/2) = 119 mm

k = 1.575 / (25 x 1000 x 119^{2}) = 0.00445 < 0.167 Ok

z = d (0.5 + √0.25 – 0.882k) = 0.99d ˃ 0.95d, use 0.95d

z = 0.95 x 119 = 133 mm

A_{s1} = 1.575/ (0.87 x 460 x 113) = 34.83 mm^{2}/m provide H12@ 300 mm c/c (377 mm^{2}/m)

**Check for shear**

Maximum design shear force, M_{Ed} = 11.7 kN/m

k = 1 + √ (200/d) = 1 + √ (200/119) = 2.30 ˃2.0, use 2.0

ρ_{l }= A_{s1}/bd = 34.83/(1000 x 119) = 0.000293 ≤ 0.02

V_{Rd,c} = [0.12k(100ρ_{l}f_{ck})^{1/3}] bd = [0.12 x 2 (100 x 0.000293 x 25)^{1/3} x 1000 x 119 = 25.75 kN/m

Since V_{Ed} (11.7 kN/m) ˂ V_{Rd,c} (25.75 kN/m), shear is satisfied

**Reinforcement details**

**References**

Cusens, A. R. (1966): Analysis of slabless stairs. Concrete and Constructional Engineering 61(10), pp. 359—64.

Reynolds, C.E. and Steedman, J.C. (1988): Reinforced Concrete Designer’s Handbook, 10^{th} edition. E & FN Spon, Taylor & Francis Group 11 New Fetter Lane, London.

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