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Lecture Notes 15 July 2010 Dr. B. Rajeevan 1 Dr. B. Rajeevan Senior Lecturer Department of Civil Engineering Govt. College of Engineering Kannur E-mail: [email protected] Mob: 9495 333 088 Contact Time: 4 pm 5 pm DESIGN OF STAIRS STRUCTURAL DESIGN 15 July 2010 Dr. B. Rajeevan 2 Geometrical Classification 15 July 2010 Dr. B. Rajeevan 3 Geometrical Classification 15 July 2010 Dr. B. Rajeevan 4 15 July 2010 Dr. B. Rajeevan 5 Structural Classification-Stair spanning longitudinally 15 July 2010 Dr. B. Rajeevan 6

Stairs rajesh sir

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Page 1: Stairs rajesh sir

Lecture Notes 15 July 2010

Dr. B. Rajeevan 1

Dr. B. RajeevanSenior Lecturer

Department of Civil EngineeringGovt. College of Engineering Kannur

E-mail: [email protected]: 9495 333 088

Contact Time: 4 pm – 5 pm

DESIGN OF STAIRSSTRUCTURAL DESIGN

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Geometrical Classification

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Geometrical Classification

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Structural Classification-Stair spanning longitudinally

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Lecture Notes 15 July 2010

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Structural Classification-Stair

spanning transversely

Parts of a stair

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Public BuildingsRise = 150 mmTread = 270 mm to 300 mm

Residential BuildingsRise = 160 mmTread = 250 mm

Number of rise in one flight < 12

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Empirical relation

550 < (2R + G) < 750 (in mm)

(2R + G) = 600 (approx)

Principles of Design

• Cl. 33 of IS 456-2000

– Stair spanning horizontally (transversely)

– Stair spanning longitudinally

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Stair spanning horizontally

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Lecture Notes 15 July 2010

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Stair spanning horizontally

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Stair spanning horizontally

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Stair spanning longitudinally

Slab between beams

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Stair spanning longitudinally

Slab between landings

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Stair spanning longitudinally

Slab between walls

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Dead Loads

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Lecture Notes 15 July 2010

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Live Loads

• Live Load (IS 875)

– Residential building without overcrowding – 2 kN/m2

– Public building with overcrowding – 5 kN/m2

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Depth of Section

• One way reinforced slab

– Longitudinally spanning – min thickness = W

– Transverse spanning – average depth of step; D/2

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Structural Detailing

• Maximum BM – simply supported = wl2/8– One half of tension steel taken to supports

• Maximum BM – slab built-in to beams at supports = (0.9) wl2/8; – 2/3 of tension steel to support as negative steel

• Maximum BM – slab continuous over beams at supports = wl2/10; – Full tension steel to support as negative steel

• Distributors– 0.0012bD or 0.0015bD in transverse direction; – Full tension steel to support as negative steel

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Calculation of Dead Loads and Effective Depth

• Transverse Stair

– Effective span = width of stair

– Individual steps of width = G and span = width of stair

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Lecture Notes 15 July 2010

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Calculation of Dead Loads and Effective Depth

• Longitudinally spanning Stair

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Design Example 1

A staircase has the following dimensions: waist W = 75 mm; nosing N = 25 mm; rise R =175 mm; going G = 225 mm. The live load expected is 2 kN/m2. Calculate the factored loads and effective depths for design if the staircase is spanning (a) transversely , and (b) longitudinally

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Ref Step Calculations Output

1

wu=2.3kN/m

2

2 2 2 2250 175 305

tan

125 1.08 /

1 2

0.225 2 0.45 /

, 1.5( ) 2.295 /u

B G R mm

Slab spanning seversely

RTDL WB kN m

LL kN m

Design Load w DL LL kN m

covercos 2 2

161.5 ( 12 )

W Rd

mm Use mmdia161.5d mm

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Ref Step Calculations Output

1

wu=10.22 kN/m

2

2 2 2 2250 175 305

tan

125 4.81 /

2

2 /

, 1.5( ) 10.22 /u

B G R mm

Slab spanning seversely

RTDL WB kN m

G

LL kN m

Design Load w DL LL kN m

cover2

54 ( 12 )

d W

mm Use mmdia54d mm

Design Example 2

A longitudinal type of stair spans a distance of 3.75 m c/c of beams. R = 175 mm, T =250 mm. The treads have 15 mm granite finish and consists of 15 steps. Use M25 grade concrete and Fe 415 steel. Take Live load = 5 kN/m2. Width of stair = 1.5 m.

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Ref Step Calculations Output

IS 456Cl. 26.4

1 Thickness of waist slab, WAssume l/d =30, d = 3750/30 = 125W = 125 + cover + /2 = 125 + 15 + 6 = 146 mm = 150 mm (say)

d = 150 -15 -6 =129 mm W = 150 mmd = 129 mm

2 Loads – Dead Load

DL =

Design Load = 1.5(DL + LL) = 18.48 kN/m2

SP 16 Table D

3 Bending Moment, Mu, Check for Mu,lim

SP16Table 3

4 Area of main steel

2 2 2 2250 175 305B G R mm

2125 23.5 7.32 /

2

RT FTWB kN m

G G

218.48 /uw kN m

2

26 /10

wkN m

,limu uM M

2

2

1.56, 0.472

606100

u

st

Mp

bd

pbdA mm

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Ref Step Calculations Output

SP 16 Table 96

5 Compute Spacing Provide 12 mm dia bars at 180 mm c/cMax spacing 3d or 300 mm

12 mm dia bars at 180 mm c/c

IS 456Cl. 25.5.2.1

SP 16 Table 96

6 Distributors

Provide 8 mm dia @ 270 mm c/c

Max spacing 5d or 450 mm

8 mm dia @ 270 mm c/c

IS456Table 19

7 Shear Check

No shear reinforcement necessary

IS 456Fig 4

8 Check for Deflection(l/d)allowed = 1.2 x 26 =31.2(l/d)assumed = 30 < 31.2

(l/d) ratio OK

2

,min 0.12 /100 180stA bD mm

34.65 ; / 0.272

0.36 ; Noshear reinforcement

v

c v c

wV kN V bd MPa

MPa

Design Example 3

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Design a waist slab type stair shown in Figure. Landing slab A is supported on beams along JK and PQ, while the waist slab and landing slab B are spanning longitudinally as shown in Figure. The finish load and Live load are 1 kN/m2 and 5 kN/m2 , respectively. Use R = 160 mm and T = 270 mm, concrete grade = M20 and steel grade = Fe 415

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Homework

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LOADS AT SECTION 1-1

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