Design of open-well stair

Design of open-well stair:

Open-well stair is that type of stair in which two or more than two straight flights are arranged around a well or space. It may be rectangular or of any geometrical shape and constructed where there is large space. The space between two flights may varies from 0.15 m to 1 m. It is extensively used in public buildings. Here we discuss about the design of open-well stair with an example.

Numerical design of open-well stair:

Design an open-well stair for an institution building of floor height 12’. Assume necessary data.

design of open-well stair

Step1: Design of space required for stair

Normally, in institution buildings, the tread ranges from 250 – 300 mm whereas the riser ranges from 100 – 150 mm. The minimum width of main-stair in institutional building is 2.0 m.

Floor height = 12’ = 3.6585 m

Take, Riser = 6” = 0.1524 m [To get number of steps in whole number, slighter greater than 0.15 m is assumed]

Tread = 280 mm

Width of stair = 2.0 m

Width of landing = width of stairs = 2.0 m

Since requirement of stair type is open-well.

Total number of riser = 12’/6” = 24

Arrange the numbers of riser such that lower and upper flight contain 10 and 10. And 4 riser in intermediate flight.

 Lower & upper flightIntermediate flight
No. of Riser104
No. of Tread93
Going9 * 0.28 = 2.523 * 0.28 = 0.84

Space required for staircase,

Length = entry space + going of lower or upper flight + landing = 2.0 + 2.52 + 2.0 = 6.52 m

[Entry space is the space required before the beginning of step and it is normally taken equal to landing]

Width = entry space + going of intermediate flight + landing = 2.0 + 0.84 + 2.0 = 4.84 m

Step 2: Effective span of stair:

The flight with the greatest span is taken for analysis and design of all flights. Since lower and upper have same span so take any one.

Let’s take lower flight for designed and analysis.

The effective span of stair is calculated as explained under the topic Design of staircase [IS 456:2000, clause 33.1]

Since the stair is supported by beam (width = 250mm) spanning parallel to riser at the start and end of steps in stair but the landing is supported in the masonry wall of thickness 200mm.

Effective span = Going + landing + half of bearing on both side.

                              = 2.52 + 2.0 + (0.2/2 + 0.25/2)

                              = 4.745 m

Step 3: Thickness of waist slab

The thickness of waist slab is taken slightly greater than the normal slab. Since the thickness of slab used in the floor is 130mm so take the thickness of waist slab equal to 150 mm.

Step 4: Load Calculation:

  1. Load on going

Self-weight of steps = 25 * (1/2 * R*T) * width of stair * Number of steps

                        = 25 * ½ * 0.1524 * 0.28 * 2.0 * 9

                        = 9.6012 KN

Self-weight of waist slab = 25 * √ (R2 + T2) * overall depth * width of stairs * Numbers of steps

                                    = 25 * √ (0.15242 + 0.282) * 0.15 * 2.0 * 9

                                    = 18.65 KN

Floor finish in going = 1.5 KN/m2 * width of stairs * going

= 1.5 KN/m2 * 2.0 * 2.52 = 7.56 KN

Total DL on going = Self-weight of steps + waist slab + Floor finish

                                    = 9.612 + 18.65 + 7.56

                                    = 35.822 KN

So, UDL due to DL on going = total load / length of going = 35.822 / 2.52 = 14.21 KN/m

Live load LL = 4.0 KN/m2

So, UDL due to LL on going = 4.0 * width of stairs = 8.0 KN/m

Total UDL on going = 14.21 + 8 = 22.21 KN/m

2. Load on landing

UDL due to DL = (25 * overall depth + floor finish)*width of stairs

                        = (25 * 0.15 + 1.5) * 2.0

                        =10.5 KN/m

Total UDL on landing = UDL due to DL + UDL due to LL

                        = 10.5 + 4 * 2

                        = 18.5 KN/m

Average UDL on the flight = (18.5 * 2 + 22.21 * 2.52) / (2+2.52) = 20.56 KN/m

Average factored UDL on the flight wu = 1.5 * 20.56 = 30.85 KN/m

 Step 5: Calculate bending moment and shear force

Assuming the flight to be simply supported on both ends and is loaded with a UDL, maximum moment is found at mid span as:

Mmax = Wul2/8

            = 30.85 * 4.7452 / 8

            = 86.82 KN-m

Vu,max = Wu l / 2

            = 30.85 * 4.745 /2 = 73.2 KN

Step 6: Check for depth

The thickness of slab is derived for calculated value of factored moment.

Mu = Mu,lim = 0.36 fck b xu [d – 0.42 xu]

For Fe 500, xu/d = 0.46 [from IS 456:2000 clause 38.1]

 Therefore, Mu = Mu,lim = 0.1336 fck b d²

86.82 * 106 = 0.1336 * 30 * 2000* d²

                              d = 104.07mm < 130 mm

Step 7: Calculate reinforcement

The area can be calculated from given formula;

            Mu = 0.87 fy As [d – fy As/ fck b]

86.82 * 106 = 0.87 * 500 * As [130 – 500 * As / 30*2000]

            As = 1726.31 mm²

Since diameter of bar = 10 mm

Spacing of bar = area of a bar * width of strip / area of reinforcement required

               = 78.54 * 2000 / 1726.31 = 91 mm

The spacing of main reinforcement bar should not be more than 3 times the effective depth of solid slabs or 300 mm whichever is smaller.

3d = 3* 130 = 390mm

So, provide spacing of 80 mm

Area provided = 78.54 * 2000 / 80 = 1963.5 mm²

And the minimum area of main steel reinforcement is equal to 0.12% of bD.

Min. area required = 0.12* 2000 * 150 / 100 = 360 mm² < 1963.5 mm² (area provided), Ok

Provide 10mm dia. bars @ 80mm c/c throughout the width of waist slab

Calculation of Distribution bars:

Provide minimum reinforcement as distribution bars:

Ast = Ast,min = 360mm2

Adopting 10mm bars for distribution reinforcement,

Number of bars = 360/(π*102/4) = 4.58

Spacing (S) = b/4.58 = 2000/4.58 = 436.33 mm; Provide 300mm < Min(5d,450mm) = 450mm Ok.

Provide 10mm dia. bars @ 300mm c/c throughout the span of waist slab

Step 8: Check for shear:

Shear stress in waist slab τuv = Vu,max/bd = 73.2*1000/(130*2000) = 0.2815N/mm2

Allowable shear stress = k τuc , where k=1.3 for D=150mm

For τuc , percentage reinforcement %pt is required

%pt = 100As/bd = 100*1963.5 /(2000*130) = 0.755%

For M25 grade concrete, from Table 19 of IS456:2000,

τuc = 0.59 N/mm2

Since τuv = 0.2815N/mm2 < k τuc = 1.3*0.59 = 0.767N/mm2, safe in shear

After completion of design of open-well stair, the reinforcement arrangement should be shown as mentioned on codes.

ALSO READ

STAIRCASE BASICS

DESIGN OF DOG-LEGGED STAIR

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