design of staircase - people.utm.my · • stairs with open well: two intersecting landings at...

DESIGN OF STAIRCASE

Dr. Izni Syahrizal bin Ibrahim Faculty of Civil Engineering

Universiti Teknologi Malaysia

Email: [email protected]

mailto:[email protected]

T

G N

Introduction

T

G

R

N

h

Span, L

Flight Landing

T = Thread R = Riser G = Going h = Waist N = Nosing = Slope

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)

Types of Staircase

Straight stair spanning longitudinally

Free-standing stair

Helical stair

Types of Staircase

Slabless stair

Straight stair spanning horizontally Spiral stair

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)


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 + 1.8 m (whichever is the smaller)

Detailing • Ensure that the tension bar may not break through at the kink


Correct detailing


Incorrect detailing

Design Procedure

Step Task Standard

1 Determine design life, Exposure class & Fire resistance

EN 1990 Table 2.1 EN 1992-1-1: Table 4.1 EN 1992-1-2: Sec. 5.6

2 Determine material strength BS 8500-1: Table A.3 EN 206-1: Table F1

3 Select the waist, h and average thickness, t of staircase EN 1992-1-1: Table 7.4N EN 1992-1-2: Table 5.8

4 Calculate min. cover for durability, fire and bond requirements EN 1992-1-1: Sec. 4.4.1

5 Estimate actions on staircase EN 1991-1-1

6 Analyze structure to obtain maximum bending moments and shear forces

EN 1992-1-1: Sec. 5

7 Design flexural reinforcement EN 1992-1-1: Sec. 6.1

8 Check shear EN 1992-1-1: Sec. 6.2

9 Check deflection EN 1992-1-1: Sec. 7.4

10 Check cracking EN 1992-1-1: Sec. 9.3

11 Detailing EN 1992-1-1: Sec.8 & 9.3

Example 1

STRAIGHT STAIRCASE SPANNING LONGITUDINALLY

Example 1: Straight Staicase

G = 255 mm

R = 175 mm

h = 110 mm

L = 2800 mm

10 255 mm = 2550 mm 250 mm 250 mm

• Permanent action, gk = 1.0 kN/m2 (excluding selfweight)

• Variable action, qk = = 4.0 kN/m2 • fck = 25 N/mm2

• fyk = 500 N/mm2

• RC density = 25 kN/m3

• Cover, c = 25 mm • bar = 8 mm

Example 1: Straight Staircase

Determine Average Thickness of Staircase

𝑦 = ℎ𝐺2+𝑅2

𝐺= 110

2552+1752

255= 133 mm

Average thickness:

𝑡 =𝑦+(𝑦+𝑅)

2=

133+(133+175)

2= 𝟐𝟐𝟏 mm G

R y t

y


Action Slab selfweight = 25.0 0.221 = 5.52 kN/m2

Permanent action (excluding selfweight) = 1.00 kN/m2

Characteristics permanent action, gk = 5.52 + 1.00 = 6.52 kN/m2

Characteristics variable action, qk = 4.00 kN/m2

Design action, nd = 1.35gk + 1.5qk = 14.81 kN/m2

Consider 1 m width, wd = nd 1 m = 14.81 kN/m/m width


Note: F = wd L = 14.81 2.8 m = 41.47 kN

M = FL/10 = 11.6 kNm

M = FL/10 = 11.6 kNm M = FL/10 = 11.6 kNm

L = 2.8 m

Analysis


Main Reinforcement Effective depth, d = 110 – 25 – 8/2 = 81 mm

𝐾 =𝑀

𝑓𝑐𝑘𝒃𝑑2 =11.6×106

25×𝟏𝟎𝟎𝟎×812 = 0.071 Kbal = 0.167

Compression reinforcement is NOT required

𝑧 = 𝑑 0.25 −𝐾

1.134= 0.93𝑑 0.95d

𝐴𝑠 =𝑀

0.87𝑓𝑦𝑘𝑧=

11.6×106

0.87×500×0.93×81= 𝟑𝟓𝟑 mm2/m


Minimum & Maximum Area of Reinforcement

𝐴𝑠,𝑚𝑖𝑛 = 0.26𝑓𝑐𝑡𝑚

𝑓𝑦𝑘𝑏𝑑 = 0.26

2.56

5000.0013𝑏𝑑 ≥ 0.0013𝑏𝑑

As,min = 0.0013bd = 0.0013 1000 81 = 108 mm2/m As,max = 0.04Ac = 0.04bh = 0.04 1000 110 = 4400 mm2/m

Secondary Reinforcement As = 20% of the main reinforcement = 0.20 353 = 71 mm2/m

Main bar H8-125 (As = 402 mm2/m) Secondary bar H8-350 (As = 144 mm2/m)


Shear

14.81 kN/m 11.6 kNm/m 11.6 kNm/m

VA VB 2.8 m

M @ B = 0 2.80VA – 11.6 + 11.6 – (14.81 2.80 1.4) = 0 VA = 20.7 kN/m VB = 20.7 kN/m


Shear Maximum design shear force, VEd = 20.7 kN/m

𝑘 = 1 +200

𝑑= 1 +

200

81= 2.57 2.0 Use k = 2.0

𝜌𝑙 =𝐴𝑠𝑙

𝒃𝑑=

402

𝟏𝟎𝟎𝟎 × 81= 0.0050 ≤ 0.02

𝑉𝑅𝑑,𝑐 = 0.12𝑘 100𝜌𝑙𝑓𝑐𝑘1/3 𝒃𝑑

= 0.12 × 2.0 100 × 0.0050 × 25 1/3 𝟏𝟎𝟎𝟎 × 81 = 45011 N = 45.0 kN/m

𝑉𝑚𝑖𝑛 = 0.035𝑘3/2 𝑓𝑐𝑘 𝒃𝑑

= 0.035 × 2.03/2 25 𝟏𝟎𝟎𝟎 × 81 = 40093 𝑁 = 40.1 kN/m

VEd (20.7 kN/m) VRd,c (45.0 kN/m) OK


Deflection Percentage of required tension reinforcement:

𝜌 =𝐴𝑠,𝑟𝑒𝑞

𝒃𝑑=

353

𝟏𝟎𝟎𝟎 × 81= 0.0044

Reference reinforcement ratio:

𝜌𝑜 = 𝑓𝑐𝑘 × 10−3 = 25 × 10−3 = 0.0050

Since o Use Eq. (7.16a) in EC 2 Cl. 7.4.2


Factor or structural system, K = 1.5

𝑙

𝑑= 𝐾 11 + 1.5 𝑓𝑐𝑘

𝜌𝑜

𝜌+ 3.2 𝑓𝑐𝑘

𝜌𝑜

𝜌− 1

3/2

(l/d)basic = 1.5 (11 + 8.6 + 0.90) = 30.8 Modification factor for span less than 7 m = 1.00

Modification for steel area provided = 𝐴𝑠,𝑝𝑟𝑜𝑣

𝐴𝑠,𝑟𝑒𝑞=

402

353= 1.14 1.50

(l/d)allow = 30.8 1.00 1.14 = 35.0 (l/d)actual = 2800/81 = 34.6 (l/d)allow

Deflection OK


Cracking h = 110 mm 200 mm Main bar: Smax, slab = 3h (330 mm) 400 mm 330 mm Max bar spacing = 125 mm Smax, slab OK Secondary bar: Smax, slab = 3.5h (385 mm) 450 mm 385 mm Max bar spacing = 350 mm Smax, slab OK

Cracking OK

Max bar spacing


Detailing

10

1

75

= 1

75

0 m

m

10 255 mm = 2550 mm 250 mm 250 mm

840 mm 840 mm

H8-125

H8-350

H8-125

H8-350

Example 2

STAIRCASE WITH LANDING & CONTINUOUS AT ONE END

Example 2: Staircase with Landing & Continuous at One End



• fyk = 500 N/mm2



G = 260 mm

R = 170 mm

h = 160 mm

L1 = 2700 mm

10 260 mm = 2600 mm 1500 mm 200 mm

L2 = 1600 mm

200 mm

L = 4300 mm




𝐺= 160

2602+1702

260= 191 mm

Average thickness:


2=

191+(191+170)

2= 𝟐𝟕𝟔 mm G

R y t

y


Action & Analysis Landing Slab selfweight = 25.0 0.160 = 4.00 kN/m2





Consider 1 m width, wd, landing = nd 1 m = 11.52 kN/m/m width


Action & Analysis Flight Slab selfweight = 25.0 0.276 = 6.90 kN/m2





Consider 1 m width, wd, flight = nd 1 m = 15.44 kN/m/m width


Note: F = wd L = (15.44 2.7 m) + (11.52 1.6 m) = 60.1 kN

M = FL/10 = 25.9 kNm

M = FL/10 = 25.9 kNm

L1 = 2.7 m

Analysis

L2 = 1.6 m

11.52 kN/m



𝐾 =𝑀

𝑓𝑐𝑘𝒃𝑑2 =25.9×106

25×𝟏𝟎𝟎𝟎×1302 = 0.061 Kbal = 0.167


𝑧 = 𝑑 0.25 −𝐾

1.134= 0.94𝑑 0.95d

𝐴𝑠 =𝑀


25.9×106

0.87×500×0.94×130= 𝟒𝟖𝟓 mm2/m





2.56

5000.0013𝑏𝑑 ≥ 0.0013𝑏𝑑


Secondary Reinforcement As = 20% of the main reinforcement = 0.20 485 = 97 mm2/m



Shear

15.44 kN/m 25.9 kNm/m

VA VB 2.7 m

M @ B = 0 4.30VA – 25.9 – (15.44 2.70 2.95) – (11.52 1.6 0.80) = 0 VA = 38.0 kN/m VB = 22.1 kN/m

11.52 kN/m

1.6 m



𝑘 = 1 +200

𝑑= 1 +

200

130= 2.24 2.0 Use k = 2.0


𝒃𝑑=

524

𝟏𝟎𝟎𝟎 × 130= 0.0040 ≤ 0.02


= 0.12 × 2.0 100 × 0.0040 × 25 1/3 𝟏𝟎𝟎𝟎 × 130 = 67376 N = 67.4 kN/m


= 0.035 × 2.03/2 25 𝟏𝟎𝟎𝟎 × 130 = 64347 𝑁 = 64.3 kN/m





𝒃𝑑=

485

𝟏𝟎𝟎𝟎 × 130= 0.0037


𝜌𝑜 = 𝑓𝑐𝑘 × 10−3 = 25 × 10−3 = 0.0050




𝑙

𝑑= 𝐾 11 + 1.5 𝑓𝑐𝑘

𝜌𝑜


𝜌𝑜

𝜌− 1

3/2




524

485= 1.08 1.50


Deflection OK


Cracking h = 160 mm 200 mm Main bar: Smax, slab = 3h (480 mm) 400 mm 400 mm Max bar spacing = 150 mm Smax, slab OK Secondary bar: Smax, slab = 3.5h (560 mm) 450 mm 450 mm Max bar spacing = 400 mm Smax, slab OK

Cracking OK

Max bar spacing


Detailing

10 260 mm = 2600 mm 1500 mm 200 mm

10

1

70

= 1

70

mm

H10-150

H10-400

H10-150

H10-400

H10-400

200 mm

0.3L = 1290 mm

Example 3

STAIRCASE SUPPORTED BY LANDING

Example 3: Staircase Supported by Landing


Plan View

200

1500

1500

100

50

50 200

200 1500

200 10 @ 260 = 2600


G = 260

R = 170

h = 150

h = 150



• fyk = 500 N/mm2



Section




𝐺= 150

2602+1702

260= 179 mm

Average thickness:


2=

179+(179+170)

2= 𝟐𝟔𝟒 mm G

R y t

y


For this type of staircase, design for LANDING and FLIGHT should be done

SEPARATELY !!!


Action Landing Slab selfweight = 25.0 0.150 = 3.75 kN/m2







Action Flight Slab selfweight = 25.0 0.264 = 6.61 kN/m2







Analysis for Staircase

Effective span, Le = La + 0.5 (Lb1 + Lb2) La = Clear distance between supports = 2600 mm Lb1 = The lesser of width support 1 or 1.8 m = 200 mm Lb2 = The lesser of width support 2 or 1.8 m = 1500 mm

Le = 2600 + 0.5 (200 + 1500) = 3450 mm


Note: F = wd L = (15.04 2.7 m) = 40.6 kN

M = FL/10 = 14.0 kNm

M = FL/10 = 14.0 kNm

L1 = 2.7 m L2 = 0.75 m

Le = Effective span

Support 1

Support 2

Analysis for Staircase


Moment Design

Shear Check

Deflection Check

Cracking Check

Detailing

Self Study


Analysis for Landing

w kN/m

L = 3.4 m

w = wlanding + Load from staircase = (11.18 1.5) + 11.8 (reaction from support 2) = 28.6 kN/m

𝑽𝒎𝒂𝒙 =𝒘𝑳

𝟐 𝟒𝟖. 𝟔 𝒌𝑵

𝑴𝒎𝒂𝒙 =𝒘𝑳𝟐

𝟖= 𝟒𝟏. 𝟑 𝒌𝑵𝒎



𝐾 =𝑀

𝑓𝑐𝑘𝒃𝑑2 =41.3×106

25×𝟏𝟓𝟎𝟎×1202 = 0.077 Kbal = 0.167


𝑧 = 𝑑 0.25 −𝐾

1.134= 0.93𝑑 0.95d

𝐴𝑠 =𝑀


41.3×106

0.87×500×0.93×120= 𝟖𝟓𝟒 mm2/m





2.56

5000.0013𝑏𝑑 ≥ 0.0013𝑏𝑑


Main bar 17H10 (As = 1335 mm2/m)



𝑘 = 1 +200

𝑑= 1 +

200

120= 2.29 2.0 Use k = 2.0


𝒃𝑑=

1335

𝟏𝟓𝟎𝟎 × 130= 0.0074 ≤ 0.02


= 0.12 × 2.0 100 × 0.0074 × 25 1/3 𝟏𝟓𝟎𝟎 × 120 = 114350 N = 114.4 kN/m


= 0.035 × 2.03/2 25 𝟏𝟓𝟎𝟎 × 120 = 89095 𝑁 = 89.1 kN/m





𝒃𝑑=

854

𝟏𝟓𝟎𝟎 × 120= 0.0047


𝜌𝑜 = 𝑓𝑐𝑘 × 10−3 = 25 × 10−3 = 0.0050




𝑙

𝑑= 𝐾 11 + 1.5 𝑓𝑐𝑘

𝜌𝑜


𝜌𝑜

𝜌− 1

3/2




1335

854= 1.56 1.50


Deflection OK


Cracking h = 150 mm 200 mm Main bar: Smax, slab = 3h (450 mm) 400 mm 400 mm

Max bar spacing =[1500−2 25 −10]

16= 90 𝑚𝑚 Smax, slab OK

Cracking OK


Detailing

LETS DO IT

Example 4

TWO SPANS OF STAIRCASE INTERSECT AT RIGHT ANGLES

Example 4: Two Spans of Staircase Intersect at Right Angles

200 1500 10 @ 255 = 2550 200

200

1500

10 @ 255 = 2550

200

A A

Plan View


Section A-A



• fyk = 500 N/mm2



G = 255

R = 170

h = 150

h = 150




𝐺= 150

2552+1702

255= 180 mm

Average thickness:


2=

180+(180+170)

2= 𝟐𝟔𝟓 mm G

R y t

y

Action & Analysis Landing Slab selfweight = 25.0 0.150 = 3.75 kN/m2







Action & Analysis Flight Slab selfweight = 25.0 0.265 = 6.63 kN/m2








Note: F = wd L = (5.46 1.6 m) + (14.80 2.65 m) = 48.0 kN

Analysis

M = FL/10 = 20.4 kNm

M = FL/10 = 20.4 kNm

L2 = 2.65 m L1 = 1.6 m

5.46 kN/m

Load on landing 2. WHY?



𝐾 =𝑀

𝑓𝑐𝑘𝒃𝑑2 =20.4×106

25×𝟏𝟎𝟎𝟎×1202 = 0.057 Kbal = 0.167


𝑧 = 𝑑 0.25 −𝐾

1.134= 0.95𝑑 0.95d

𝐴𝑠 =𝑀


20.4×106

0.87×500×0.95×120= 𝟒𝟏𝟐 mm2/m





2.56

5000.0013𝑏𝑑 ≥ 0.0013𝑏𝑑


Secondary Reinforcement As = 20% of the main reinforcement = 0.20 412 = 82.4 mm2/m



Shear Check

Deflection Check

Cracking Check

Detailing

Self Study

Other Types

design of staircase - people.utm.my · • stairs with open well: two intersecting landings at...

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