Design of Staircase

Design of staircase

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Introduction

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Components of stair staircase case

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Types of staircase

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Guidelines

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Distribution of Loading on Stairs

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Design Problem

Introduction Why do we need stairs ? To protect people from injury and to facilitate access during movement from one level to another in a building STAIRS CONSIST OF STEPS ARRANGED IN A SERIES FOR THE PURPOSE OF GIVING ACCESS TO DIFFERENT FLOORS OF BUILDING.

STAIR IS OFTEN THE ONLY MEANS OF COMMUNICATION BETWEEN THE VARIOUS FLOORS OF BUILDING,THE LOCATION REQUIRES GOOD AND CAREFUL CONSIDERATION.

Components of staircase (Assignment)

Types •

Geometrical classification       

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Straight Stair Dog legged Stair Quarter turn Stair Open newel Stair Three quarter turn Stair Bifurcated Stair Circular Stair

Structural Classification    

Stair slab spanning transversely or horizontally between stringer beams or walls Stair slab spanning longitudinally Slab less stair or stair either only riser and treads Spiral staircase

Geometrical classification

Straight Stair

Dog legged Stair

Geometrical classification

Quarter turn Stair

Open Newel Stair

Geometrical classification

Bifurcated Stair Three Quarter Turn Stair

Spiral or Circular Stair

Structural Classification

Stair slab spanning transversely or horizontally between stringer beams or walls

Here the main steel is provided transversely and the distribution steel is in the longitudinal direction.

Stair slab spanning longitudinally

Structural Classification

Here the main steel in provided longitudinally and the distribution steel is in the transverse direction.

(i) Supported on edges AE and DH (Fig.b) (ii) Clamped along edges AE and DH (Fig.c) (iii) Supported on edges BF and CG (Fig.d) (iv) Supported on edges AE, CG (or BF) and DH (Fig.e)

(v) Supported on edges AE, BF, CG and DH (Fig.f)

Stair slab spanning longitudinally

Structural Classification

Guide lines •

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Rise (R) is 150mm to 180mm and tread (T) is 220 mm to 250 mm for a residential building. For public building rise is kept between 120 to 150 mm and tread between 250 to 300 mm Sum of tread (T) and twice the rise (2R) should be between 500 mm to 650 mm The width of the stair is dependent on the usage and is between 0.8 m to 1 m for residential building and 1.8 m to 2 m for public building. The width of the landing is equal to the width of stairs. The number of steps in each flight should not be greater than 12 The pitch of the stair should not be more than 38 degrees. The head room measured vertically above any step or below the mid landing shall not be less than 2.1 m.

Distribution of Loading on Stairs

Cl33.2, pg63

DESIGN STEPS 1) GEOMETRICAL DESIGN:-

ASSUME SUITABLE TREAD AND RISER NO OF RISER= (F/F HEIGHT)⁄ RISE

NO OF RISERS IN ONE FLIGHT= (depend upon no. of flights provided NO OF TREAD = (NO OF RISERS ―1)

GOING DISTANCE= (NO OF TREAD) × (TREAD WIDTH) WIDTH OF LANDING ≥ WIDTH OF STAIR

2) STRUCTURAL DESIGN:-

EFFECTIVE SPAN CALCULATION:Stairs spanning horizontally EFFECTIVE SPAN CALCULATION=C/C DISTANCE BETWEEN SUPPORTS

IF NOT GIVEN WIDTH OF SUPPORT CAN BE TAKEN IN BETWEEN 200 TO 300 MM Stairs spanning Longitudinally CL33.1

TRIAL DEPTH OF WAIST SLAB:ACCORDING TO IS 456:2000 ARTICLE 23.2.1 BY CALCULATING RATIO OF SPAN TO EFFECTIVE DEPTH AND AFTER THAT RATIO IS MULTIPLYING BY THE MODIFICATION FACTOR

ALTERNATE METHOD:1) THE THICKNESS OF WAIST SLAB(t) NORMAL TO SLOPE CAN BE ASSUMED AS (L/20) FOR SIMPLY SUPPORTED SLAB (L/25) FOR CONTINUOUS SLAB WHEN WAIST SLAB SPANNING IN LONGITUDINAL DIRECTION

2) IN CASE OF TREAD-RISER STAIRS SPANNING LONGITUDINALLY THE THICKNESS OF RISER & TREAD SLAB KEPT SAME WITH VALUES OF (SPAN/25) FOR SIMPLY SUPPORTED AND (SPAN/30) FOR CONTINUOUS STAIRCASES (MINIMUM THICKNESS OF 80 MM SHOULD BE PROVIDED)

3) LOAD CALCULATION:-

Stairs spanning horizontally W1- Dead load of a step = ½ x T x R x 25 kN/m W2- Dead load of waist slab = b x t x 25 kN/m W3-Live load = LL (KN/m2) x T kN/m W4- Floor finish = assume 0.5 kN/m Stairs spanning Longitudinally CALCULATIONS SHOULD BE MADE BY CONSIDERING WIDTH OF SLAB EQUAL TO 1 METRE

W1- SELF WEIGHT OF SLAB on Plan = 25 × D × √(R²+T²)/T (KN/M² /M) W2-WT OF STEPS = 25 × 0.5 × R (KN/M² /M) W3-WT OF FLOOR FINISH = 1 × 1 (KN/M² /M) (ASSUME) W4-LIVE LOAD = 3 KN/M² (RESIDENTIAL BUILDING) = 4-5 KN/M² (PUBLIC BUILDING) W5) NET LOAD(W) =W₁ +W₂ +W₃+W₄

FACTORED LOAD = W’=1.5 ×W

4) CALCULATION OF DESIGN MOMENTS:-

FIND MAX BENDING EITHER BY DRAWING SHEAR FORCE AND BENDING 5)CHECK FOR EFFECTIVE DEPTH:-

6)CHECK FOR REINFORCEMENT:CALCULATE MAIN STEEL(Ast) Ast =0.5Fck/Fy × (1 – √(1 – (4.6M/Fckbd²))bd

FOR Fe 250 FOR Fe 415

Ast ≥ Ast minimum Ast min = 0.12 % of GROSS AREA Ast min = 0.15 % of GROSS AREA

PROVIDE SUITABLE DISTRIBUTION STEEL=Ast min

7) CHECK FOR SHEAR:-

7.1)CALCULATE MAX DESIGN SHEAR FORCE:Vud = 0.5 × W‘L 7.2)CALCULATE SHEAR RESISTED BY CONCRETE:Vuc = k × τc × b×d Vud < Vuc VALUES OF “τc” AND “k” is obtained from IS 456

8) PROVISION

OF DEVELOPMENT LENGTH:Ld req = (0.87 × Fy × Φ)/4 × τ

WHERE

Φ = DIAMETRE OF BAR PROVIDED τ = DESIGN BOND STRESS

9) CHECK FOR DEFLECTION:CALCULATE ACTUAL % OF REINFORCEMENT L/d provided < L/d max

10) SUMMARY AND DETAILING

Ex. 1 Design a dog legged stair case for a residential building hall measuring 2.2m x 4.7 m. The width of the landing is 1m. The distance between floor to floor is 3.3 m. The rise and tread may be taken as 150mm and 270mm respectively. The weight of floor finish is 1 kN/m2. The materials used are M20 grade concrete and Fe415 grade steel. Sketch the details of steel. Here flight and the landing slabs spans in the same direction i.e, Flight spans longitudinally.

Solution

Q. For the same problem, if the flight is supported at its ends by means of landing slab (i.e. at the junction of flight and landing) Design the Staircase.

Hint: If the support conditions are different, then the effective span will be different. All other design calculations remain same for the problem le = 2700 mm (distance b/n first and last risers) for both flights. Here the flight and the landing spans in opposite directions. Flight spans longitudinally and landing spans transversely). Landing is to be designed separately depending on its edge conditions.