Staircase Design to BS 8110-1:1997

Basic Structural Element Design to BS 8110-1:1997

4.0

Staircase Design

Figure 4.1: Plan view of the stair case in the General Arrangement Drawing

Edifice Consultants Pvt.Ltd

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S

Figure 4.2: Plan view of the stair case


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Reference

Calculations

Out Put

Layout

Riser

= R = 150mm

Going

= G = 300mm

As per “ Graded Examples in Reinforced Concrete Design “ by Prof. W.P.S Dias 2R+G should kept close as possible to 600mm. =

2R + G = 2 × 150 + 300 = 600mm

Nosing is 25mm. Hence Tread

=

T = 300 + 25 = 325mm

Cover

BS 8110-1:1997

For durability requirements: Assume mild exposure conditions.

Table 3.3

Nominal Cover required is 20mm.

BS 8110-1:1997

For fire requirements: Assume 2 hour fire resistance

Table 3.4

Nominal Cover required is 25mm Hence Select a cover of 25mm.

Cover 25mm

Loads 2

Finishes on Tread

= 1kN/m

Finishes on soffit plaster

=

Imposed loads

= 3kN/m

2 0.25kN/m

2

Cl.3.10.1.2

Distribution of Loading

BS 8110-1:1997

Staircase is built monolithically at its ends in to structural members spanning right angles to its span. Hence Effective depth can be evaluated as follows.


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Reference

Calculations

Out Put

la = 3.3

l b,1

= 2.8m > 1.8m

l b,1

= 1.8m

l b,2

= 1.325m < 1.8m

l b,2

= 1.8m = l + 0.5(l

Effective span

a

b1

+l

b2

)

= 3.3 + 0.5(1.8 + 1.325) = 4.8625m

Initial Waist Thickness

Assume a trial span/ effective depth ration of 32, for continuous one way spanning stair case. d=

4862.5 32

= 151.95mm

Assume T10 bars are to be used. h = 151.95 + 25 +

10 2

= 181.9mm

Hence select the depth of the waist slab as 175mm. Loading (For 1m strip of the staircase) Factor for slope Overlap =

T G

=

2 R =

325 300

+

2 G

2 150 =

G

+

2 300

300

=

1.12

= 1.08

Load Evaluation

Waist

= 0.175 × 24 × 1.12 = 4.70kN/m

Steps

= × 0.150 × 24 = 1.80kN/m

Top finishes

= 1 × 1.08 = 1.08kN/m

Soffit plaster

= 0.25 × 1.12 = 0.28kN/m


1 2

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Reference

Calculations

Total dead load (g k )

= 7.86kN/m

Imposed load (qk )

= 3kN/m

Hence design load

=

1.4g k

+

Out Put

1.6q k

= 1.4 × 7.86 + 1.6 × 3 = 15.80kN/m Design for Bending

The stair case can be idealized as shown in Figure 10.2

About A

15.80 × 3.3 × (0.9 + RB =

3.3 2

)

= 27.34kN

4.8625

R A = 15.80 × 3.3 − 27.34 = 24.8kN M =R ×X − x A dMx dx

w × [X − 0.9] 2 2

= 0 when R = w(X − 0.9) A

R 24.8 + 0.9 = 2.47m X = A + 0.9 = w 15.8 M = 24.8 × 2.47 − x

24.8 × [2.47 − 0.9]2 2

= 30.69kNm

Figure 4.2: Stair Case Idealization for bending


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Reference

Calculations

Out Put

Maximum moment =30.69kNm/m Assume T10 is to be used as main reinforcement. Effective depth (d) d = 175 − 25 −

10

= 145mm

2

K=

M 2 f cubd

=

30.69 × 10

6

2 30 × 1000 × 145

= 0.049 < 0.156

Hence section can be designed as a singly reinforced section.



z = 0.5 + 0.25 −



 d 0.9  K



z = 0.5 + 0.25 −

0.049 



0.9

d = 0.94d < 0.95d 

Hence Area of reinforcement required , A s,req =

M

=

0.87f y Z

30.69 × 10

6

0.95 × 460 × 0.94 × 145

= 515

Table 3.25

Checks for minimum area of reinforcement

BS 8110-1:1997

100A s,min bh

A s,min =

2 mm m

= 0.13

0.13 × 1000 × 175 100

= 227.5

2 mm m

Provide T10 @ 150mm C/C

A

2 mm

= 524 s,provided m

Hence minimum steel is Ok.

T10 @ 150 B1

Minimum steel Ok.


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Reference

Calculations

Out Put

Checks for maximum spacing Clause

Maximum spacing is lesser of 3 time’s effective depth or

3.12.11.2.7

750mm.

BS 8110-1:1997

3d = 3 × 145 = 435mm Hence provided spacing of 200mm is satisfactory. Distribution Steel

Distribution steel needs to satisfy the minimum area of reinforcement requirement. A s,min =

0.13 × 1000 × 175 100

= 227.5

2 mm m

Hence provide T10 @ 200mm C/C. A s,provided = 393

2 mm m

Maximum spacing is lesser of 3 times effective depth or 750mm.

3d = 3 × 145 = 435mm Hence provided spacing of T10 @ 200mm spacing is

T10@ 200 B2

satisfactory

Checks for Deflection

For deflection check end span slab panel is critical. Span =4862.5mm Table 3.9

For a Continuous slab panel

BS 8110-1:1997 Basic

  Span  Effective Depth  = 26  

Allowable

    Span Span Basic =  Effective Depth   Effective Depth  × F1 × F2    

F1 − Modificati on Factor for Tension Reinforcem ent F2 − Modificati on Factor for Compressi o n Reinforcem ent


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Reference

Cl.3.10.2.2 BS 8110-1:1997

Calculations

Out Put

Length of the flight/ Span

3300 4862.5

× 100 = 67.87%

Flight occupies more than 60% of the span. Hence allowable deflection limit may be increased by 15%. f s =

2 3

× f y ×

A s,req

×

A s,pro

1 β b

β =1

b

f s =

2 3

× 460 ×

515 524

1

× = 301.39N/mm2 1

6 M 30.69 × 10 = = 1.46 2 2 bd 1000 × 145 F1 = 0.55 +

(477 − f s ) M 120(0.9 + ) 2 bd

= 0.55 +

(477 − 301.39) 120(0.9 + 1.46)

= 1.17 < 2

No Compression reinforcement is provided. Hence F2

=1

Allowable

  Span  Effective Depth  = Basic  

Allowable

  Span   = 26 × 1.17 × 1.15 = 34.98 Effective Depth  



Span



Span

  Span  Effective Depth  × F1 × F2  

 4862.5  = 145 = 33.53 Effective Depth  

Actual 

   Span  < Allowable    Effective Depth   Effective Depth 

Actual 

Hence deflection check is satisfied

Deflection is Ok.


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Reference

Calculation

Out Put

Checks for Shear

Shear is critical at support B. V = 27.34kN

Clause

Maximum possible shear , lesser of

3.4.5.2

0.8 f cu or 5N/mm

2

0.8f cu = 0.8 × 25 = 4N/mm2

Clause 3.4.5.2

v=

V

=

bv d

27.34 × 10

3

1000 × 145

Maximum shear

= 0.19N/mm2 < 4N/mm2

Ok

Hence applied shear stress is less tan maximum allowable. Table 3.8

1

1

 100A s  3  400  4 1  f cu  v c = 0.79 ×  × ×   ×  δm  25  b d d    v 

1

3

 100As   100 × 524  = 0.361 < 3  =  × b d 1000 145   v   1

1

 400  4  400  4  d  =  145  = 1.29 > 1 δm = 1.25

1 1 1 30   3 ×   = 0.62N/mm2 v c = 0.79 × [0.361] 3 × 1.29 × 1.25  25 

v < v c Hence shear check is satisfied.


Shear is Ok

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Staircase Design to BS 8110-1:1997

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