DESIGN OF STAIRCASE. Dr. Izni Syahrizal bin Ibrahim. Faculty of Civil Engineering Universiti Teknologi Malaysia
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1 DESIGN OF STAIRCASE Dr. Izni Syahrizal bin Ibrahim Faculty of Civil Engineering Universiti Teknologi Malaysia
2 Introduction T N T G N G R h Flight Span, L Landing T = Thread R = Riser G = Going h = Waist N = Nosing = Slope
3 Introduction Public building R 180 mm & G 255 mm Private building R 200 mm & 250 mm G 400 mm For comfort: (2 R) + G = 600 mm (UBBL, BS 5395, Reynold et al. 2007)
4 Types of Staircase Straight stair spanning longitudinally Free-standing stair Helical stair
5 Types of Staircase Spiral stair Straight stair spanning horizontally Slabless stair
6 General Design Considerations Loads Permanent action: Weight of steps & finishes. Also consider increased loading on plan (inclination of the waist) Stairs with open well: Two intersecting landings at right-angles to each other, loads on areas common to both spans may be divided equally between spans Bending Moment & Shear Force Stair slab & landing to support unfavourable arrangements of design load Continuous stairs: Bending moment can be taken as FL/10 (F is the total ultimate load)
7 General Design Considerations Effective Span Stairs between beam or wall: Centreline between the supporting beam or wall Stairs between landing slab: Centreline of the supporting landing slab, or the distance between edges of supporting slab m (whichever is the smaller) Detailing Ensure that the tension bar may not break through at the kink
8 General Design Considerations Correct detailing
9 General Design Considerations Incorrect detailing
10 Design Procedure Step Task Standard 1 Determine design life, Exposure class & Fire resistance 2 Determine material strength 3 Select the waist, h and average thickness, t of staircase EN 1990 Table 2.1 EN : Table 4.1 EN : Sec. 5.6 BS : Table A.3 EN 206-1: Table F1 EN : Table 7.4N EN : Table Calculate min. cover for durability, fire and bond requirements EN : Sec Estimate actions on staircase EN Analyze structure to obtain maximum bending moments and shear forces EN : Sec. 5 7 Design flexural reinforcement EN : Sec Check shear EN : Sec Check deflection EN : Sec Check cracking EN : Sec Detailing EN : Sec.8 & 9.3
11 Example 1 STRAIGHT STAIRCASE SPANNING LONGITUDINALLY
12 Example 1: Straight Staicase G = 255 mm R = 175 mm Permanent action, g k = 1.0 kn/m 2 (excluding selfweight) Variable action, q k = = 4.0 kn/m 2 f ck = 25 N/mm 2 f yk = 500 N/mm 2 RC density = 25 kn/m 3 Cover, c = 25 mm bar = 8 mm h = 110 mm 250 mm mm = 2550 mm L = 2800 mm 250 mm
13 Example 1: Straight Staircase Determine Average Thickness of Staircase y = h G 2 +R 2 G = = 133 mm Average thickness: t = y+(y+r) 2 = 133+( ) 2 = 221 mm G y t R y
14 Example 1: Straight Staircase Action Slab selfweight = = 5.52 kn/m 2 Permanent action (excluding selfweight) = 1.00 kn/m 2 Characteristics permanent action, g k = = 6.52 kn/m 2 Characteristics variable action, q k = 4.00 kn/m 2 Design action, n d = 1.35g k + 1.5q k = kn/m 2 Consider 1 m width, w d = n d 1 m = kn/m/m width
15 Example 1: Straight Staircase Analysis L = 2.8 m M = FL/10 = 11.6 knm M = FL/10 = 11.6 knm M = FL/10 = 11.6 knm Note: F = w d L = m = kn
16 Example 1: Straight Staircase Main Reinforcement Effective depth, d = /2 = 81 mm K = M = = K f ck bd bal = Compression reinforcement is NOT required z = d 0.25 K = 0.93d 0.95d A s = M 0.87f yk z = = 353 mm2 /m
17 Example 1: Straight Staircase Minimum & Maximum Area of Reinforcement A s,min = 0.26 f ctm bd = bd bd f yk 500 A s,min = bd = = 108 mm 2 /m A s,max = 0.04A c = 0.04bh = = 4400 mm 2 /m Secondary Reinforcement A s = 20% of the main reinforcement = = 71 mm 2 /m Main bar H8-125 (A s = 402 mm 2 /m) Secondary bar H8-350 (A s = 144 mm 2 /m)
18 Example 1: Straight Staircase Shear 11.6 knm/m kn/m 11.6 knm/m V A 2.8 m V B B = V A ( ) = 0 V A = 20.7 kn/m V B = 20.7 kn/m
19 Example 1: Straight Staircase Shear Maximum design shear force, V Ed = 20.7 kn/m k = = = Use k = 2.0 d 81 ρ l = A sl bd = 402 = V Rd,c = 0.12k 100ρ l f ck 1/3 bd = / = N = 45.0 kn/m V min = 0.035k 3/2 f ck bd = / = N = 40.1 kn/m V Ed (20.7 kn/m) V Rd,c (45.0 kn/m) OK
20 Example 1: Straight Staircase Deflection Percentage of required tension reinforcement: ρ = A s,req bd = = Reference reinforcement ratio: ρ o = f ck 10 3 = = Since o Use Eq. (7.16a) in EC 2 Cl
21 Example 1: Straight Staircase Factor or structural system, K = 1.5 l d = K f ck ρ o ρ f ck ρ o ρ 1 3/2 (l/d) basic = 1.5 ( ) = 30.8 Modification factor for span less than 7 m = 1.00 Modification for steel area provided = A s,prov = 402 = A s,req 353 (l/d) allow = = 35.0 (l/d) actual = 2800/81 = 34.6 (l/d) allow Deflection OK
22 Example 1: Straight Staircase Cracking h = 110 mm 200 mm Main bar: S max, slab = 3h (330 mm) 400 mm Max bar spacing = 125 mm S max, slab Secondary bar: S max, slab = 3.5h (385 mm) 450 mm Max bar spacing = 350 mm S max, slab 330 mm OK 385 mm OK Max bar spacing Cracking OK
23 Example 1: Straight Staircase = 1750 mm Detailing 840 mm 840 mm H8-350 H8-125 H8-125 H mm mm = 2550 mm 250 mm
24 Example 2 STAIRCASE WITH LANDING & CONTINUOUS AT ONE END
25 Example 2: Staircase with Landing & Continuous at One End G = 260 mm h = 160 mm R = 170 mm Permanent action, g k = 1.2 kn/m 2 (excluding selfweight) Variable action, q k = = 3.0 kn/m 2 f ck = 25 N/mm 2 f yk = 500 N/mm 2 RC density = 25 kn/m 3 Cover, c = 25 mm bar = 10 mm 200 mm mm = 2600 mm 1500 mm 200 mm L 1 = 2700 mm L 2 = 1600 mm L = 4300 mm
26 Example 2: Staircase with Landing & Continuous at One End Determine Average Thickness of Staircase y = h G 2 +R 2 G = = 191 mm Average thickness: t = y+(y+r) 2 = 191+( ) 2 = 276 mm G y t R y
27 Example 2: Staircase with Landing & Continuous at One End Action & Analysis Landing Slab selfweight = = 4.00 kn/m 2 Permanent action (excluding selfweight) = 1.20 kn/m 2 Characteristics permanent action, g k = = 5.20 kn/m 2 Characteristics variable action, q k = 3.00 kn/m 2 Design action, n d = 1.35g k + 1.5q k = kn/m 2 Consider 1 m width, w d, landing = n d 1 m = kn/m/m width
28 Example 2: Staircase with Landing & Continuous at One End Action & Analysis Flight Slab selfweight = = 6.90 kn/m 2 Permanent action (excluding selfweight) = 1.20 kn/m 2 Characteristics permanent action, g k = = 8.10 kn/m 2 Characteristics variable action, q k = 3.00 kn/m 2 Design action, n d = 1.35g k + 1.5q k = kn/m 2 Consider 1 m width, w d, flight = n d 1 m = kn/m/m width
29 Example 2: Staircase with Landing & Continuous at One End Analysis kn/m L 1 = 2.7 m L 2 = 1.6 m M = FL/10 = 25.9 knm M = FL/10 = 25.9 knm Note: F = w d L = ( m) + ( m) = 60.1 kn
30 Example 2: Staircase with Landing & Continuous at One End Main Reinforcement Effective depth, d = /2 = 130 mm K = M = = K f ck bd bal = Compression reinforcement is NOT required z = d 0.25 K = 0.94d 0.95d A s = M 0.87f yk z = = 485 mm2 /m
31 Example 2: Staircase with Landing & Continuous at One End Minimum & Maximum Area of Reinforcement A s,min = 0.26 f ctm bd = bd bd f yk 500 A s,min = bd = = 173 mm 2 /m A s,max = 0.04A c = 0.04bh = = 6400 mm 2 /m Secondary Reinforcement A s = 20% of the main reinforcement = = 97 mm 2 /m Main bar H (A s = 524 mm 2 /m) Secondary bar H (A s = 196 mm 2 /m)
32 Shear Example 2: Staircase with Landing & Continuous at One End 25.9 knm/m kn/m kn/m V A 2.7 m 1.6 m V B B = V A 25.9 ( ) ( ) = 0 V A = 38.0 kn/m V B = 22.1 kn/m
33 Shear Example 2: Staircase with Landing & Continuous at One End Maximum design shear force, V Ed = 38.0 kn/m k = = = Use k = 2.0 d 130 ρ l = A sl bd = 524 = V Rd,c = 0.12k 100ρ l f ck 1/3 bd = / = N = 67.4 kn/m V min = 0.035k 3/2 f ck bd = / = N = 64.3 kn/m V Ed (38.0 kn/m) V Rd,c (67.4 kn/m) OK
34 Example 2: Staircase with Landing & Continuous at One End Deflection Percentage of required tension reinforcement: ρ = A s,req bd = = Reference reinforcement ratio: ρ o = f ck 10 3 = = Since o Use Eq. (7.16a) in EC 2 Cl
35 Example 2: Staircase with Landing & Continuous at One End Factor or structural system, K = 1.3 l d = K f ck ρ o ρ f ck ρ o ρ 1 3/2 (l/d) basic = 1.5 ( ) = 31.5 Modification factor for span less than 7 m = 1.00 Modification for steel area provided = A s,prov = 524 = A s,req 485 (l/d) allow = = 34.0 (l/d) actual = 4300/130 = 33.1 (l/d) allow Deflection OK
36 Example 2: Staircase with Landing & Continuous at One End Cracking h = 160 mm 200 mm Main bar: S max, slab = 3h (480 mm) 400 mm Max bar spacing = 150 mm S max, slab Secondary bar: S max, slab = 3.5h (560 mm) 450 mm Max bar spacing = 400 mm S max, slab 400 mm OK 450 mm OK Max bar spacing Cracking OK
37 Example 2: Staircase with Landing & Continuous at One End = 170 mm Detailing 0.3L = 1290 mm H H H H H mm mm = 2600 mm 1500 mm 200 mm
38 Example 3 STAIRCASE SUPPORTED BY LANDING
39 Example 3: Staircase Supported by Landing
40 Example 3: Staircase Supported by Landing Plan View =
41 Example 3: Staircase Supported by Landing Permanent action, g k = 1.2 kn/m 2 (excluding selfweight) Variable action, q k = = 3.0 kn/m 2 f ck = 25 N/mm 2 f yk = 500 N/mm 2 RC density = 25 kn/m 3 Cover, c = 25 mm bar = 10 mm G = 260 h = 150 R = 170 Section h = 150
42 Example 3: Staircase Supported by Landing Determine Average Thickness of Staircase y = h G 2 +R 2 G = = 179 mm Average thickness: t = y+(y+r) 2 = 179+( ) 2 = 264 mm G y t R y
43 Example 3: Staircase Supported by Landing For this type of staircase, design for LANDING and FLIGHT should be done SEPARATELY!!!
44 Example 3: Staircase Supported by Landing Action Landing Slab selfweight = = 3.75 kn/m 2 Permanent action (excluding selfweight) = 1.20 kn/m 2 Characteristics permanent action, g k = = 4.95 kn/m 2 Characteristics variable action, q k = 3.00 kn/m 2 Design action, n d = 1.35g k + 1.5q k = kn/m 2 Consider 1 m width, w d, landing = n d 1 m = kn/m/m width
45 Example 3: Staircase Supported by Landing Action Flight Slab selfweight = = 6.61 kn/m 2 Permanent action (excluding selfweight) = 1.20 kn/m 2 Characteristics permanent action, g k = = 7.81 kn/m 2 Characteristics variable action, q k = 3.00 kn/m 2 Design action, n d = 1.35g k + 1.5q k = kn/m 2 Consider 1 m width, w d, flight = n d 1 m = kn/m/m width
46 Example 3: Staircase Supported by Landing Analysis for Staircase Effective span, L e = L a (L b1 + L b2 ) L a = Clear distance between supports = 2600 mm L b1 = The lesser of width support 1 or 1.8 m = 200 mm L b2 = The lesser of width support 2 or 1.8 m = 1500 mm L e = ( ) = 3450 mm
47 Example 3: Staircase Supported by Landing Analysis for Staircase Support 1 M = FL/10 = 14.0 knm L 1 = 2.7 m L e = Effective span L 2 = 0.75 m Support 2 M = FL/10 = 14.0 knm Note: F = w d L = ( m) = 40.6 kn
48 Example 3: Staircase Supported by Landing Self Study Moment Design Shear Check Deflection Check Cracking Check Detailing
49 Example 3: Staircase Supported by Landing Analysis for Landing w kn/m L = 3.4 m w = w landing + Load from staircase = ( ) (reaction from support 2) = 28.6 kn/m V max = wl 2 M max = wl kn = knm
50 Example 3: Staircase Supported by Landing Main Reinforcement Effective depth, d = /2 = 120 mm K = M = = K f ck bd bal = Compression reinforcement is NOT required z = d 0.25 K = 0.93d 0.95d A s = M 0.87f yk z = = 854 mm2 /m
51 Example 3: Staircase Supported by Landing Minimum & Maximum Area of Reinforcement A s,min = 0.26 f ctm bd = bd bd f yk 500 A s,min = bd = = 240 mm 2 /m A s,max = 0.04A c = 0.04bh = = 9000 mm 2 /m Main bar 17H10 (A s = 1335 mm 2 /m)
52 Example 3: Staircase Supported by Landing Shear Maximum design shear force, V Ed = 48.6 kn/m k = = = Use k = 2.0 d 120 ρ l = A sl bd = 1335 = V Rd,c = 0.12k 100ρ l f ck 1/3 bd = / = N = kn/m V min = 0.035k 3/2 f ck bd = / = N = 89.1 kn/m V Ed (48.6 kn/m) V Rd,c (114.4 kn/m) OK
53 Example 3: Staircase Supported by Landing Deflection Percentage of required tension reinforcement: ρ = A s,req bd = = Reference reinforcement ratio: ρ o = f ck 10 3 = = Since o Use Eq. (7.16a) in EC 2 Cl
54 Example 3: Staircase Supported by Landing Factor or structural system, K = 1.0 l d = K f ck ρ o ρ f ck ρ o ρ 1 3/2 (l/d) basic = 1.0 ( ) = 19.1 Modification factor for span less than 7 m = 1.00 Modification for steel area provided = A s,prov = 1335 = A s,req 854 (l/d) allow = = 28.7 (l/d) actual = 3400/120 = 28.3 (l/d) allow Deflection OK
55 Example 3: Staircase Supported by Landing Cracking h = 150 mm 200 mm Main bar: S max, slab = 3h (450 mm) 400 mm Max bar spacing = [ ] mm = 90 mm S max, slab OK Cracking OK
56 Example 3: Staircase Supported by Landing Detailing LETS DO IT
57 Example 4 TWO SPANS OF STAIRCASE INTERSECT AT RIGHT ANGLES
58 Example 4: Two Spans of Staircase Intersect at Right Angles = A A 255 = 2550 Plan View 200
59 Example 4: Two Spans of Staircase Intersect at Right Angles Permanent action, g k = 1.0 kn/m 2 (excluding selfweight) Variable action, q k = = 3.0 kn/m 2 f ck = 25 N/mm 2 f yk = 500 N/mm 2 RC density = 25 kn/m 3 Cover, c = 25 mm bar = 10 mm R = 170 G = 255 h = 150 Section A-A h = 150
60 Example 4: Two Spans of Staircase Intersect at Right Angles Determine Average Thickness of Staircase y = h G 2 +R 2 G = = 180 mm Average thickness: t = y+(y+r) 2 = 180+( ) 2 = 265 mm G y t R y
61 Example 4: Two Spans of Staircase Intersect at Right Angles Action & Analysis Landing Slab selfweight = = 3.75 kn/m 2 Permanent action (excluding selfweight) = 1.00 kn/m 2 Characteristics permanent action, g k = = 4.75 kn/m 2 Characteristics variable action, q k = 3.00 kn/m 2 Design action, n d = 1.35g k + 1.5q k = kn/m 2 Consider 1 m width, w d, landing = n d 1 m = kn/m/m width
62 Example 4: Two Spans of Staircase Intersect at Right Angles Action & Analysis Flight Slab selfweight = = 6.63 kn/m 2 Permanent action (excluding selfweight) = 1.00 kn/m 2 Characteristics permanent action, g k = = 7.63 kn/m 2 Characteristics variable action, q k = 3.00 kn/m 2 Design action, n d = 1.35g k + 1.5q k = kn/m 2 Consider 1 m width, w d, flight = n d 1 m = kn/m/m width
63 Example 4: Two Spans of Staircase Intersect at Right Angles Analysis Load on landing 2. WHY? 5.46 kn/m L 1 = 1.6 m L 2 = 2.65 m M = FL/10 = 20.4 knm M = FL/10 = 20.4 knm Note: F = w d L = ( m) + ( m) = 48.0 kn
64 Example 4: Two Spans of Staircase Intersect at Right Angles Main Reinforcement Effective depth, d = /2 = 120 mm K = M = = K f ck bd bal = Compression reinforcement is NOT required z = d 0.25 K = 0.95d 0.95d A s = M 0.87f yk z = = 412 mm2 /m
65 Example 4: Two Spans of Staircase Intersect at Right Angles Minimum & Maximum Area of Reinforcement A s,min = 0.26 f ctm bd = bd bd f yk 500 A s,min = bd = = 160 mm 2 /m A s,max = 0.04A c = 0.04bh = = 6000 mm 2 /m Secondary Reinforcement A s = 20% of the main reinforcement = = 82.4 mm 2 /m Main bar H (A s = 449 mm 2 /m) Secondary bar H (A s = 196 mm 2 /m)
66 Shear Check Example 4: Two Spans of Staircase Intersect at Right Angles Deflection Check Cracking Check Detailing Self Study
67 Other Types
68 Other Types
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