Design Calculation for a Staircase to BS 8110-1:1997Full description
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Design Calculation for a Staircase to BS 8110-1:1997
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CHAPTER SIX STAIRCASE DESIGN
INTRODUCTION
COMPONENTS OF A STAIRCASE WORKED EXAMPLE
6.0 INTRODUCTION Staircases are basically used to access or exit buildings. There exist different types of staircases, namely: 1. Simple staircases – one flight or two flights staircases. In the two flights staircase a landing is required to break the flight into two. It is normally recommended to have a maximum of 17 steps in one flight. 2. Free standing staircase 3. Slab less or saw tooth or dog leg staircase 4. Helical staircase 5. Spiral staircases – may be in reinforced concrete (in-situ or precast) or steel 6. Pre-cast stairs
For residential building, the minimum waist thickness should be taken as 125mm. For public or commercial buildings, the minimum waist thickness should be taken as 150mm. Step 3: Determine the supports and the effective span of the staircase. Step 4: Calculate the total dead load of the staircase per meter, which will include selfweight of waist, self-weight of steps and characteristic superimposed loads due to tiles and screed. Step 5: Calculate the imposed load per meter. Step 6: Determine the ultimate load, F Step 7: Determine the Ultimate moment M
6.1 COMPONENTS OF A STAIRCASE The main features of a staircase are: a). Riser, b). Thread or going and c). Waist
Threa d
Step 8: Calculate the area of main reinforcement as well as the distribution steel. Step 9: Check for deflection criteria NB: If the stair flight occupies at least 60% of the span, the Modification factor may be multiplied by 15%. Step 10: Detail the staircase reinforcement.
Riser
waist
6.3
WORKED EXAMPLE
8x250 Section through staircase The optimum dimensions of staircase as laid down in BS5395 are as tabulated below. All dimensions are in mm. Usage Public Semipublic Private
Going
Riser
300 275
150 165
Min. Width 1000 1000
250
175
800
6.2 METHODOLOGY Step 1: The Riser and thread should be determined depending on usage. Step 2: The waist of the staircase need to be specified.
Step 4: Dead loads Consider unit width of staircase, i.e. b = 1000mm Slope length of staircase, l = sqrt (horizontal distance2 + vertical distance2) l = [(32) + (1.5752)]1/2 = 3.39m CSP Characteristic superimposed load = 1.0kPa Dead load due to CSP = 1.0*3.39*(unit width of staircase = 1m) = 3.39kN Waist Self-weight of waist = 0.125*24 = 3.0 kPa Dead load due to waist= 3.0*l *(unit width of staircase = 1m) = 3.0*3.39*1.0 = 10.17 kN Steps Self-weight of steps = (Thread/1000)*vertical distance*0.5*24*(Unit width of staircase=1m) = .25*1.575*0.5*24*1.0 = 4.725kN Total dead load = load due to CSP + Load due to Waist + Load due to steps = 3.39 + 10.17 + 4.725 = 18.29 kN Step 5: Imposed load The imposed load will act vertically on the stair; hence instead of using the slope length, the projected length should be used. Projected length of staircase = 3.0m Imposed load = 3.0kPa Imposed load on staircase = 3.0*3.0*(unit width of staircase =1m) = 9kN Step 6: Ultimate load F ‘@ULS load F = 1.4*dead + 1.6*Live = 1.4*(18.29)+1.6*(9) = 40kN Step 7: Moment at ULS ‘@ULS moment M
Step 8: Area of reinforcement ‘b Cover
= Fl /8 = 40*3/8 = 15kNm
= 1000mm = 20mm
Diameter of main bar = 10mm Effective depth d = 125-20-(10/2) = 100mm Moment coefficient, k = M/f cu bd2 = 15x1000000/(30*1000*100*100) = 0.05 Lever arm Z= d[0.5 + sqrt(0.25- (k/0.9))] = 100*[0.5+sqrt(0.25 – (0.05/0.9))] = 94.09mm Area of steel reinf. = M/(0.95f yZ) = 15000000/(0.95*460*94.09) = 364.8 mm2/m Minimum area of steel= 0.13%bh = 0.13*1000*125/100 = 162.5mm2/m Area of steel provided = T10 – 200 equivalent 396mm2/m Secondary reinforcement = Area of minimum steel = 0.13%bh = 162.5mm2/m = T08-200 Step 9: Deflection check Service stress, f s = 2*f y*As req/( As prov * 3) = 2*460*364.8/(3*396) = 282.5 N/mm2 Modification factor = 0.55 + [(477-f s)/(120*(0.9+(M/bd2))] = 0.55 + [(477-282.5)/(120*(0.9+(0.05*30)))] = 1.23 Since flight occupies more than 60% of the span, the modification factor may be increased by 15%. Modification factor = 1.23*1.15 = 1.41 Actual l/d ratio = 3000/100 = 30 Basic l/d ratio = 20 Permissible l/d ratio = 20*1.41 = 28.2 Since permissible l/d ratio is less than actual l/d, the deflection criterion is not satisfied. Alternative 1: Increase the reinforcement provided from T10-200 to T10-175 Therefore area of steel provided = 449mm2/m Service stress, f s = 2*f y*As req/( As prov * 3) = 2*460*364.8/(3*449) = 249 N/mm2 Modification factor = 0.55 + [(477-f s)/(120*(0.9+(M/bd2))] = 0.55 + [(477-249)/(120*(0.9+(0.05*30)))] = 1.34
Since flight occupies more than 60% of the span, the modification factor may be increased by 15%. Modification factor = 1.34*1.15 = 1.54 Actual l/d ratio = 3000/100 = 30 Basic l/d ratio = 20 Permissible l/d ratio = 20*1.54 = 30.8 Since permissible l/d ratio exceeds the actual l/d, the deflection criterion is satisfied.
