Stair Cases Design

CHAPTER 9

STAIR CASES

9.1 GENERAL FEATURES

Stair cases are provided for connecting successive floors. It is comprised with flights of steps with inter mediate landings which provides rest to the user and support for the flight. A passage is provided at the start of staircase then for the vertical rise a flight is provided with rise and tread. Rise provided in the steps is normally 6 inch which conforms with the comfort of the user. Tread provided is 9.5 inch which can be more if the number of user is more depending on the type of building. The width of the stair can be between 3.5ft to 5 ft depending on the use. Generally public buildings should be provided with larger width. Going is the horizontal projection of the inclined flight between the first and the last riser. A flight is generally consist of two landings with going in between of 10 to 12 steps. Staircases can be designed in many forms as per the requirement of the user and the facility and space available in the construction. Design procedure of few types are discussed in this chapter.

9.2 TYPES OF STAIR CASES

Stair cases can be of varying geometrical shapes and structural behavior. Some of the most common types of staircases are shown is subsequent discussion.

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9.2.1 DOG LEGGED STAIR CASE

LANDING

PASSAGE

Figure 9.1 : Dog legged stair case

Most commonly used in buildings. It comprises with two flights and a landing or lobby in between. Normally the landing is provided at mid height. The landing acts as a support of the flight and landing is supported by beams or wall.

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9.2.2 OPEN WELL STAIR CASE

Open well

UPUP

Figure 9.2 : Open well stair case

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Generally adopted in public building where adequate space can be provided for staircases. It ahs quarter landings which provide more comfort to user. Moreover the open well provide adequate ventilation. The flights are consisted of lesser steps in comparison to dog legged staircases.

9.2.3 TREAD RISER STAIR CASE

This type of staircase is normally used for aesthetic beautification. No support for landing is provided. The tread and riser is constructed as folded plates. The construction of this types of staircase is costly as reinforcement required is more.

Riser

Tread

Figure 9.3 : Tread Riser Stair Case

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9.2.4 CANTILEVER STAIR CASE

Cantilever slab

Rise

Figure 9.4 : Cantilever Stair Case

In this type of staircase cantilever horizontal tread are projected from a wall or an inclined beam. This type of staircase needs complicated formwork and normally used for aesthetic beautification.

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9.3 DESIGN OF DOGLEGGED STAIR CASE

Step 1: General arrangement

PASSAGE

LANDING

Figure 9.5 : Dog legged stair case (general arrangement)

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The figure above shows the plan of the stair hall. Let the rise be 6 inch and trade be 9.5 inch. The width of each flight is 3.5 inch.

Height of each flight =2

10 = 5 ft.

No of risers required = 6125× = 10 risers in each flight.

No of tread in each flight = 10-1 = 9. Space occupied be trades = = 7.125 ft. 5.99×Width of landing =4.5 ft. Width of passage =4.5 ft. Size of stair hall = 7 ft ×16.125 ft.

Step 2: Design constants

For steel = 40,000 psi yfAnd for concrete = 3000 psi cf ′

Step 3: Determination of loading

The landing slab acts together with the going as a single slab. The bearing of the slab into the wall may be considered 6.5 inch.

Then the effective span = 17.1212

5.65.4125.7 =++ ft.

Considering one-way slab with both end continuous minimum thickness is 28l .

So, t = 622.528

17.1228

≈== inchesl inches.

• Self weight of the slab = 1150126

×× = 75 plf.

• Self weight of the steps =12

15012122

1 TreadRiserTread÷⎟

⎠⎞

⎜⎝⎛ ×××

=12

5.9150126

125.95.0 ÷⎟

⎠⎞

⎜⎝⎛ ×××

=37.5 plf.

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• Floor finish = 20 plf.

Total dead load =75+37.5+20=132.5. Live load = 100 plf. So, Design factored load = 1007.15.1324.1 ×+× =355.5 plf.

Step 4: Bending Moment Calculation

• Maximum Moment

60.658117.125.35581

82

2

max =××==wlM lb-ft =78.97 k-in.

• Check for depth

0278.014087

8740385.085.075.075.0max =

+××××== bρρ

⎟⎟⎠

⎞⎜⎜⎝

⎛−

=

c

yy f

fbf

Md

ρφρ 59.01

max2

= ⎟⎠⎞

⎜⎝⎛ ××−××××

30400278.059.0112400278.09.0

97.78

∴ d = 2.9 inch And t = 2.9+1=3.9 inch (with 1 inch clear cover) t =3.9 inch < 6 inch (Ok)

availabled = 6-1 = 5 inch

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Step 5: Reinforcement Calculation

• Distribution Bar. Minimum reinforcement is provided as temperature and shrinkage reinforcement. Temperature and shrinkage reinforcement,

ftintbAst /144.0612002.0002.0 2=××=××= # 3 bar can be used. The spacing will be,

S = =×

144.01211.0 9 inch c/c.

• Longitudinal Steel.

This is selected by trial. Trial No

Assumed ‘a’ (inch) Steel Area,

⎟⎠⎞

⎜⎝⎛ −

=

2adf

MAy

s

φ

(inch2)

bf

fAa

c

ys

×=

′85.0

(inch)

Comments

Trial-1

a=1.0 49.0

215409.0

97.78=

⎟⎠⎞

⎜⎝⎛ −×

=sA 64.012385.0

4049.0=

×××

=a

Not OK

Trial-2

a=0.6 47.0

26.05409.0

97.78=

⎟⎠⎞

⎜⎝⎛ −×

=sA 61.012385.0

4047.0=

×××

=a OK

So, is provided. 247.0 inchAs = It can be furnished by using # 4 bar.

Spacing = 662.547.0

1222.0≅=

× inch center to center.

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Step 6: Detailing

The following points are to be remembered in detailing:

• The main reinforcement should be bent to follow the bottom profile of the stair. • Near the landing the reinforcement should be taken straight up and then bent in the compression zone of landing. • For tensile stress in the landing zone separate set of bars should be used as shown in the detailing. • The length of each type of bar on either side of the crossing should be at least equal to 2 ft 2 inches. • All the bars of the tensile reinforcement should be taken into the supports and anchorage and development length requirement must be fulfilled. • Distribution bars should be used parallel to the width of the steps.

# 4 bar @ 6 inch c/c


PASSAGE


LANDING


Figure 9.6: Detailing of Stair Case

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PASSAGE

LANDING

Figure 9.6.: Detailing of Stair Case (continued)

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9.4 DESIGN OF OPEN WELL STAIR CASE

4.79 ft

4.5 ft

4 ft 4.5 ft

4.5 ft

7.67 ft

13.79 ftFigure 9.7 : Plan View of Open Well Stair Case

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Step 1: General Arrangements

Width of steps= 4.5 ft Height of first flight =4.5 ft Height of 2nd flight = 3 ft Total height between the floors = 15 ft First landing= 4.5 ft 2nd landing = 4.5 ft Riser raised in first flight = 9 Riser = 6 inch Tread = 11.5 inch The size of stair hall= 13.79 ft ×16.17 ft

Step 2: Design Constants

Let for steel psi 40000=yfAnd for concrete psi 3000=′cf

Step 3: Design of First Flight

The bearing of the landing slab into the wall is 6.5 inch.

Therefore the effective span = ft71.1212

5.65.467.7 =++

Considering one-way slab with both end continuous minimum thickness is 28l

So, inchesincheslt 5.545.528

71.1228

≈===

• Self weight of the slab = plf75.68115012

5.5=××

• Self weigh of the steps = 12

15012122

1 treadrisertread÷⎟

⎠⎞

⎜⎝⎛ ×××

= plf5.3712

5.11150126

125.115.0 =÷⎟

⎠⎞

⎜⎝⎛ ×××

• Floor finish=20 plf • Live load = 100 plf [can be determined by table 1.1 of ACI code]

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Total Dead load =68.75+37.5+20=126.25 plf Design factored load = 1.4×126.25+1.7×100=346.75 plf

Bending Moment Calculation:

• 93.700171.1275.34681

82

2

max =××==wlM lb-ft =84.02 k-in

• Check for the depth

⎟⎠⎞

⎜⎝⎛ ××−××××

=

⎟⎟⎠

⎞⎜⎜⎝

⎛−

=

=+

××××==

′3400278.059.0112400278.09.0

02.84

59.01

0278.04087

874385.085.075.075.0

max2

max

c

yy

b

ff

bf

Md

ρφρ

ρρ

d = 2.99 inch 3 inch ≈ Provide 1inch clear cover t=3+1=4 inch<6 inch or 5.5 inch

So, Design is OK. Available d =5.5-1=4.5 inch

Reinforcement Calculation:

• Distribution Bar

Only minimum reinforcement is provided as temperature and shrinkage reinforcement. Temperature and shrinkage reinforcement

ftinbtAst /132.05.512002.0002.0 2=××=×= So # 3 Bar can be used.

Spacing = 10132.0

1211.0=

× inch c/c.

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• Longitudinal bar

This is selected through trials. Trial No


⎟⎠⎞

⎜⎝⎛ −

=

2adf

MAy

s

φ

(inch2)

bf

fAa

c

ys

×=

′85.0

(inch)

Comments

Trial-1

a=1.0 58.0

215.4409.0

02.84=

⎟⎠⎞

⎜⎝⎛ −×

=sA 76.012385.0

4058.0=

×××

=a

Not OK

Trial-2

a=0.7 56.0

27.05.4409.0

02.84=

⎟⎠⎞

⎜⎝⎛ −×

=sA

73.012385.0

4056.0=

×××

=a

OK

So, can be provided. It can be furnished by using # 4 bar. 256.0 inAs =

Spacing= 571.456.0

1222.0≈=

× inch c/c.

Step 4: Design of Second Flight

Let the bearing of the landing slab into the wall is 6.5 inch.

The effective span= ft87.1412

5.65.479.45.412

5.6=++++

Considering one way slab with both end continuous minimum thickness is 28l

So, t= inchl 5.628

87.1428

==

• Self weight of the slab = 25.81115012

5.6=×× plf

• Self weight of the steps =12

15012122

1 treadrisertread÷⎟

⎠⎞

⎜⎝⎛ ×××

=12

5.11150126

125.115.0 ÷⎟

⎠⎞

⎜⎝⎛ ×××

=37.5 plf

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• Floor finish=20 plf

Total dead load =81.25+37.5+20=138.75 Live load = 100 plf So, designed factored load=1.4 138.75+1.7× ×100=364.25 plf

Bending Moment Calculation:

• 73.1006787.1425.36481

82

2

max =××==wlM lb-ft = 120.81 k-in.

• Check for depth:

⎟⎠⎞

⎜⎝⎛ ××−××××

=

⎟⎟⎠

⎞⎜⎜⎝

⎛−

=

=+

××××==

′3400278.059.0112400278.09.0

81.120

59.01

0278.04087

874385.085.075.075.0

max2

max

c

yy

b

ff

bf

Md

ρφρ

ρρ

d=3.59 4 inch ≈ Provide 1-inch clear cover t=4+1=5 inch <6.5 inch So, design is Ok inchdavailable 5.515.6 =−=

Reinforcement Calculation:

• Distribution Bar

Only minimum reinforcement is provided as temperature and shrinkage reinforcement. Temperature and shrinkage reinforcement

ftinAst /156.05.612002.0 2=××= If # 3 bar is used as distribution reinforcement

Spacing= 846.8156.0

1211.0≈=

× inch c/c.

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• Longitudinal bar This is selected through trials.

Trial No


⎟⎠⎞

⎜⎝⎛ −

=

2adf

MAy

s

φ

(inch2)

bf

fAa

c

ys

×=

′85.0

(inch)

Comments

Trial-1

a=1.0 67.0

215.5409.0

81.120=

⎟⎠⎞

⎜⎝⎛ −×

=sA 88.012385.0

4067.0=

×××

=a

Not OK

Trial-2

a=0.85 66.0

285.05.5409.0

81.120=

⎟⎠⎞

⎜⎝⎛ −×

=sA

86.012385.0

4066.0=

×××

=a

OK

So, can be provided. It can be furnished by using # 4 bar. 266.0 inAs =

Required spacing=66.0

1222.0 × = 4 inch c/c.

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Step 5: Detailing

4 ft7.67 ft4.5 ft

# 4 bar @5 inch c\c

# 3 Bar @10 inch c/c

# 4 bar @5 inch c/c 11.5 inch

6 inch

Figure 9.8: First Flight

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7.67 ft

11.5 inch

6 inch

Figure 9.9: Second Flight

4.5 ft 4.5 ft

# 4 bar @ 4 inch c\c


# 3 Bar @ 8 inch c/c

# 3 Bar @ 8 inch c/c

Stair Cases Design

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