Concrete-Helical-Stair-Design.pdf

Structural design………………………………………………………….ceng design………………………………………………………….ceng 501

Table of content Acknowledgement ……………………………………… Index ……………………………………………………. 1. Introduction......................................... Introduction.................................................................. ......................... 2. Literature survey………………………………………

2.1 Definition………………… Definition……………………………………………… …………………………… 2.2 Terms in stair…………………………………………... stair…………………………………………... 2.3 Essential requirement…………………… requirement………………………………….. …………….. 2.4 Classification of stair ………………………………….. 2.4.1 Based on material of construction………………. 2.4.2 Based on o n type……………………………………. type……………………………………. 2.5 helical stair……………………………………………. stair……………………………………………. 2.5.1 Geometric stair…………………………………. 2.5.2 Spiral stair………………………………………

3. Design procedures ……………………………………… 3.1 Design procedure of geometric stair…………………... 3.2 Design procedure procedu re of spiral stair…………………………

4. Result and discussion ……………………………………… 4.1 Design of geometric stair……………………………… stair……………………………… 4.2 Design of helical stair…………………………………. stair………………………………….

5. Recommendation and future work …………………………. …………………………. 6. Reference …………………………………………………...

Project on design of helical staircases

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Acknowledgement Of all, our deepest appreciation goes to our instructor Mr.Mebrehatom G. for getting us acquainted with civil engineering concepts through this mini project as part of structural design. And also we would like to thank our advisor Mr. Ashenafi and Mr.yonas for their unlimited support from the begging of the project to this final form. And of the last but not the least appreciation goes to our classmates for giving us some advice from the preparation till the edition of the project.


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Acknowledgement Of all, our deepest appreciation goes to our instructor Mr.Mebrehatom G. for getting us acquainted with civil engineering concepts through this mini project as part of structural design. And also we would like to thank our advisor Mr. Ashenafi and Mr.yonas for their unlimited support from the begging of the project to this final form. And of the last but not the least appreciation goes to our classmates for giving us some advice from the preparation till the edition of the project.


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Index as…........................................ ....

Area of standard diameter reinforcement

As………………………… …………………………….. …..

Total area of reinforcement calculated

Ac………………………… …………………………… …

area of concrete

B…………………………….

width of the strip

Cc………………………… ……………………………. ….

concrete cover

D………………………… …………………………….. …..

Over all depth

d1………………………… ……………………………. ….

minimum depth requirement for deflection

d2……………………………. …………………………….

depth required for flexure

dl………………………… ……………………………. ….

dead load

Ec……………………………

strain in concrete

Es………………………… …………………………… …

strain in steel

f cd ………………………….. cd……………………………

design compression strength of concrete

f yd yd……………………………

design yield strength reinforcement

f yk yk …………………………..

Characteristic yield strength of reinforcement

H ……………………………

room height

S ……………………………

spacing of reinforcing bars

le ………………………… …………………………… …

effective length

W………………………… …………………………… …

distributed load

ll ………………………… ……………………………. ….

live load

Pd……………………………

design load

Φ..............................................

Diameter of of the reinforcing bars

γs………………………… …………………………….. …..

Partial factor of safety for steel

γc …………………………… ……………………………

partial factor of safety for concrete

γ …………………………… ……………………………… …

unit weight

ρ b …………………………… ……………………………… … effective geometrical ratio for reinforcement


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Structural design………………………………………………………….ceng 501

Φm …..........................................

mess reinforcement diameter

2θ ……………………………… subtended angle α………………………………… inclination angle Aeff ……………………………… effective area heff ……………………………… effective height deff ……………………………… effective diameter Ueff ……………………………… effective perimeter Tc………………………………. Torsional capacity of concrete TRd……………………………… Torsional resistance Tsd……………………………... designed torsionsonal moment Vc ……………………………… shear capacity of concrete VRd……………………………... shear resistance Vsd …………………………….. Design shear force Bt……………………………….. Correction for torsion Bv………………………………..correction for shear


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1. Introduction Structural design is a very essential in many civil engineering works due to the result of safe design and economical of the structure. In structural design of reinforced concrete structures it involves the determination of the size and appropriate reinforcement so that the structure will serve for its purpose for the intended design period. So in this project we dealt about helical stairs which are astatically improved type of stairs that are used to link two floors in a public building. We hope that these two designs can be an alternative stair in the building where the obviously dog type stairs are inappropriate .we use Ethiopian building code (EBCS) in the limit state design of these helical stair. This project comprises



Literature survey



Design procedure



Result and discussion



Recommendation and future work

In the literature part it gives detailed information about stairs and helical stairs particularly. In the second part (body text), it sort out the procedure and approaches that should be followed in the design of helical stair case. In the third part there is detailed figural calculation using the procedures in the body text for the two stairs and there is a detailed reinforcement of both stair cases. In the last part we will recommend the feasibility of the design with respect to economy and site constraints.


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2. Literature survey 2.1 Definition A stair may be defined as a set of steps leading from one floor to the other to afford a means of communications between the various floors of a building. The space in a building where the stair is located is called staircase for multi-storied flat buildings the staircase should be located either centrally or two staircases at two ends of the building.

2.2 Terms in stairs There are some important t technical terms used in connection with stairs are given below. Flight

- A continuous set of steps from floor to floor, floor to landing or Landing to landing is called a flight.

Landing – A platform at the end of the series of steps is called landing. Depending upon the arrangement of steps it may be half-landing Or quarter landing. Tread-

It is the horizontal part of a step on which the foot rests.

Rise –

It is the vertical distance between two consecutive treads or steps.

Riser –

The vertical member between two consecutive treads is known as Riser.

Nosing – It is the projected edge of a tread. Stringer –It is the sloping member in a stair which supports the steps. Newel – This is a post of a heavy section set at the two ends of a handrail. Balusters –These are intermediate vertical members supporting the handrail. Handrail –It is a rail of water or wood provided at the side of a stair for Safety and is fixed at about waist height parallel to the line of Nosing. Line of nosing -This is an imaginary line joining the noising points and it is


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Structural design………………………………………………………….ceng 501 Parallel to the slope of the stair.

2.3 Essential requirements Apart from structural design there are essential requirement for a stair. These are The width of a stair should be 0.9m for residential building and 1.35m for public buildings. The slope of the stair should not be greater than 45º and not less than 25º with the horizontal. The stair should be well-lighted and ventilated. All the risers and treads should be made uniform. The number of steps in a flight should not be more than 12. The head room in a stair should be at least 2m. The width of landing should be equal to the width of the stair. The width of tread should be from 225-275mm and the riser should not be more than 175mm and not less than 150mm. The nosing should not project beyond 15mm.

There are some thumb rules for proportioning steps. These are Rise + Tread=42.5 to 45cm 2 Rise +Tread=58.0 to 62cm Rise *Tread=420 to 460 sq.cm

2.4 Classification of stair case 2.4.1 –Based on material of construction Stairs are mostly made of concrete with reinforcement steel. In hilly areas and in forests where timber is readily available in plenty, wooden stairs are built in residential quarters and forest bungalows. For low-cost housing, pre cast light weight concrete steps are used. Stone steps are used very seldom. It is especially built in construction of temples.

2.4.2 –Based on type Straight flight or single flight

Segmental

Open well

circular

Quarter-turn

Dog-legged


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Structural design………………………………………………………….ceng 501 Half-turn open well

Bifurcated

Four flight

Octagonal

Hexagonal

Geometrical

Spiral

2.5 Helical stair case Since stairs renders access to various floors the choice of the stair best fit for a certain building depending on the aesthetic value needed for that building and the economy. But we will specifically deal with the economical choice between two types of helical stair case. Helical stairs can be modeled as geometric stair and spiral stair. These stairs mostly constructed in R.c.c, iron or stone.

2.5.1 Spiral (circular) stair –commonly provided at the back side of a building. In this form of stair all the steps radiate from the newel post or well hole, in the form of winders. These kinds of Rc stair cases can be analyzed as curved beam and there are internal stresses developed due to the specialty of the structure. The two critical moments are bending moment about the principal plane and the torsional moment and also shear force transferred to the circular column also create stress on the column.

2.5.2 Geometric stair – These type of stair have same principal like spiral one where the steps radiate from the shear wall and the forward and backward flight is curved and the change in direction is obtained through winder.


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3. Design procedure 3.1 design procedure of geometric stair Step 1

depth determination

Step 2

design load calculation

Step 3

moment and shear calculation

Step 4

check depth for flexure

Step 5

area of reinforcement calculation

Step 6

reinforcement detail

Using the procedure above



Design of flight slab (as cantilever slab)



Design of landing



Design of beam for flexure and torsion



Design of shear wall

3.2 design procedure of spiral stair

Step 1

depth determination

Step 2

design load calculation

Step 3

moment and shear calculation

Step 4

check depth for flexure

Step 5

area of reinforcement calculation

Step 6

reinforcement detail

Using the procedure above



Design of the flight slab for flexure and torsion



Design of circular hallow column


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4. Result and discussion 4.1 design parameters Given data Room height (h)

=3.5m

Width of stair (b)

=1.5m

Use C-30 S-300 Class-І work 2

Live load=5KN/m

(for public building)

Φl =10mm Φm =8mm Cc=15mm for slab and 25mm for beam Design constant using EBCS 1, 1995 f cd=0.85f ck /γc

=13.4mpa

f yd=f yk /γs

=260.87mpa

m=f yd/(0.8*f cd)

=24.335

c1=2.5/m

=0.103

c2=0.4*f yd*m

=2539.31

ρ b= (0.8*Ec*f cd)/((Ec+Es)*f yd) =0.0299 ρ=0.75*ρ b

=0.0224

Unit weight from EBCS 1, 1995 3 ,

γ marble

=27KN/m

γ concert

=24KN/m

γscreed

=24KN/m

3cm thickness

3


3 ,

6

2cm thickness


4.2 design of geometric stair and shear wall. For the given room height considering the requirements of the stair we fix the dimension of trade and rise. Riser =159mm

22 risers

Trade =270mm

10 trade each flight

In order to analyze take a unit width strip as a cantilever and to calculate the design load start from minimum depth required from deflection.

Design of the stair slab. Minimum depth required for deflection

d1= (0.4+0.6*(f yk /400))/ (le/βa)

use βa =12 for span ratio 1:2.14

=106.25mm.

Over all depth (D) = d1 +cc+Φl/2+Φm =106.25mm+15mm+5mm+8mm=134.25mm Use D=150mm for construction convenience. Dead load

Due to waist slab=24*1*0.15

=3.6KN/m °

Due to floor finish =27*1*0.03*COS 30.5

=0.7KN/m °

Due to cement screed=24*1*0.02 *COS 30.5

=0.414KN/m °

Due to steps=24*(area of steps in 1 m) =24*0.0286*COS 30.5 =0.59KN/m Total Live load

5*1*0.15 =0.75KN/m


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= 5.304KN/m


Design load calculation

Pd=1.3D.L+1.6L.L =1.3*5.304+1.6*0.75KN/m

8.1KN/m

=8.1KN/m Moment and shear calculation 1.5m 2

2

M=wl /2 = 8.1*1.5 /2 =9.113KN.m

F=wl

=8.1*1.5

=12.15KN

Check depth for flexure 0.5

d2= (M/ (ρ*b*f yd (1-0.4*ρ b*m))) = (M/ (b*4.569))

0.5

=44.66mm

Since d1>d2 safe against deflection.

Actual depth (d) = D- Cc -Φl/2-Φm =150-15-5-8 =122mm

Area of reinforcement using design chart of EBCS 1,1995

M=9.113KN/m 0.5

k m =((M/b) )/d

0.5

=9.113 /0.122 =24.74

From the table

k s

= 3.975


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Then the reinforcement will be As = k s*M/d

2

= 296.92mm

S=b* as/As =1000*78.5/296.92

where as =area of Φ10mm bar

=264.38mm

Then use Φ10mm c/c 260mm as the main reinforcement. Minimum reinforcement for temperature and shrinkage.

Asmin=0.5*b*d/ f yk 2

=0.5*1000*122/300=203.33mm Spacing=b*as/As

=1000*78.5/203.33=386.129mm Use Φ10mm c/c 380mm where minimum reinforcement required.

Design of landing Take maximum diameter as a unit strip. Due to the uniformity of the structure, depth requires is the same.

Total dead load (due to depth of lading, cement screed, floor finish) =24*0.15+24*0.02+27*0.03 =4.89KN/m Live load

=0.75KN/m

Design load calculation

Pd= 1.3D.L+1.6L.L

7.557KN/m

=1.3*4.89+1.6*0.75 =7.557KN/m 1.5m


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Structural design………………………………………………………….ceng 501 Moment and shear calculation 2

2

M=wl /2 = 7.557*1.5 /2 =8.5KN-m

F=wl

= 7.557 *1.5

=11.34KN

Check depth for flexure d2= (M/ (ρ*b*f yd (1-0.4*ρ b*m))) = (M/ (b*4.569))

0.5

0.5

=43.13mm


Actual depth (d) = D- Cc-Φl/2-Φm =150-15-5-8 =122mm

Area of reinforcement using design chart of EBCS 1,1995

M=8.5KN-m 0.5

k m = ((M/b) )/d

0.5

=6.6157 /0.122 =23.9 →

k s =3.97

As = k s*M/d

Spacing =b*as/As =1000*78.5/276.6 =283.8mm Then use Φ10mm c/c 280mm as the main reinforcement. Asmin=0.5*b*d/ f yk 2

=0.5*1000*122/300=203.33mm Spacing =b*as/As =1000*78.5/203.33=386.129mm


10

2

=276.6mm


Use Φ10mm c/c 380mm where minimum reinforcement required.

Design of the beam Design for flexure Determination of depth d≥ (0.4+0.6* f yk /400)*le/βe = 0.85*1650/12 =116.875mm take 117mm Assuming

∅l =20mm,

b =250mm

Over all depth D=d+c+∅l/2 + ∅s =117+25+10+8 =160mm Take D=200mm Design load calculation

12.9KN/m

D.L =24*0.2*0.25 =1.2KN/m We have 11.34KN/m load coming from the

1.5m

landing. Pd = (1.3*1.2) + 11.34 =12.9KN/m L =1.5m 2

2

M = wl /2 =12.9*1.5 /2 =14.51KN.m

F =wl =12.9*1.5 = 19.35KN 0.5

d= (M/ (b*4.569))

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0.5

= (14.51E / (250*4.569))

=112.71mm <117mm which is safe! K m = (M/b)

0.5

0.5

/d = (14.51/0.25)

/.117 =65.11

Since it will be double reinforcement let us increase the depth to make it singly reinforced. Take D=300mm d=357mm

14.46KN/m

D.L =24*0.4*0.25 =2.4KN/m 1.5m


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Structural design………………………………………………………….ceng 501 We have 11.34KN/m load coming from the landing. Pd = (1.3*2.4) + 11.34 =14.46KN/m L =1.5m 2

2

M = wl /2 =14.46*1.5 /2 =16.27KN.m

F =wl =14.46*1.5 = 21.69KN 0.5

6

d= (M/ (b*4.569))

0.5

= (16.27E / (250*4.569))

=119.4mm <357mm which is safe! K m = (M/b)

0.5

0.5

/d = (16.27/0.25)

/0.357 =21.51

K s = Then

As =K s *M/d = For

∅14

/153.94 =

take 2

Use 2 ∅14 for flexural design in

∅14

250*300mm section

Let us take total depth D=400>300mm for decreasing the reinforcement needed for torsion and shear.

Design for combination action (torsion and shear) ’

’

Aef = (D-2d )*(b-2d ) = (400-2*35)*(250-2*35) 3

2

=59.4E mm ’

h ef =def /5 = (b-2d ) /5 =26mm < A/U = (400*250)/ ((250+400)*2) = 76.92mm) Tc = 1.2*f ctd*Aef *hef 3

=1.2*1.165*59.4E *26 = 3.05KN.m By drawing the torsion diagram we will have a torsion of Tsd =8.5KN.m/m*1.5m =12.75KN.m


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12.75KN.m

1.5m

Since Tc < Tsd =12.75KN.m

we need a reinforcement

Section capacity against diagonal compression TRd =0.8*f cd* Aef *hef 3

=0.8*13.4*59.4 E *26 =16.55KN.m > Tsd =12.75KN.m Vc =0.25*K 1*K 2* f ctd*bW*d =0.25*(1+50*0.002)*(1.6-0.357)*1.165*250*357 =35.54KN

21.69KN

0.357m

1.5m-0.357m=1.143m

Vsd / (1.5-0.357)

= 21.69/1.5

Vsd =16.53KN VRd =0.25* f cd* bW*d =0.25*13.4*250*357 =298.99KN

The limiting value for torsion and shear are

TRd,cor =βt * TRd

and

VRd,cor = βv* VRd

2 0.5

2 0.5

βt =1/(1+(( Vsd / VRd)/( Tsd / TRd )) )

βv =1/(1+(( Tsd / TRd)/( Vsd / VRd)) )

By using the above results we will get βt =0.997

,


βv =0.072

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Therefore TRd, cor =16.5KN.m

VRd, cor =21.53KN

And the torsion and shear resistance of the concrete Tc, cor =btc* Tc

Vc, cor = bvc * Vc 2 0.5

btc = 1/(1+(( Vsd / Vc)/( Tsd/ Tc)) )

2 0.5

,

bvc = 1/(1+(( Tsd / Tc)/( Vsd/Vc)) )

By substituting the above results we will have btc = 0.994

, bvc =0.110

Therefore, Tc, cor =3.03KN.m

,

Vc, cor = 3.91KN.m

Tsd < TRd,cor

Vsd < VRd,cor

12.75KN.m < 16.5KN.m

16.53KN < 21.53KN

Reinforcement ’

’

Ueff =2(b-2d +D-2d ) =2(250-2*35+400-2*35) =1020mm Asl= ((Tsd - Tc)* Ueff )/ (2*Aeff *f yd) 3

= ((12.75-3.05)*1020)/ (2*59.4E *260.87) 2

=319.25mm Then use 2

∅16

longitudinal bars

Transversal bars S≤ (2* Aeff *f yd *Astr )/ ( Tsd - Tc, cor ) 2

= (2*59400*260.87*((π*10 )/4))/ (12.75-3.03) =250.42mm by using ∅10

∅10

C/C 250mm

And S≤ (2* Aeff *f yd *Astr )/( Tsd - Tc,cor ) 2

= (2*59400*260.87*((π*8 )/4))/ (12.75-3.03) =160.27mm in using ∅8

→ So use

∅8

C/C 160mm ∅8

C/C 160mm


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Structural design………………………………………………………….ceng 501 Shear design Vc, cor =3.91KN , Vsd =16.53KN , VRd, cor =21.53KN 3.91KN 16.53KN

X 0.357m 1.5-0.357m

16.53/1.5 =3.91/X X =0.355mm Therefore from X=0 to x =0.355mm, provide minimum reinforcement with spacing

(Asv*f yk )/ (0.4*bw) = (2*28*300)/ (0.4*250) =168mm Smax

d=357mm 800

Use

∅6

C/C 168mm

take C/C 160mm

And from X=0.355mm to X=1.143mm S = (Asv*f yd )*(d-dc)/ Vsd Let us assume dc =35mm 2

S=2*((π*6 )/4))*260.87*(357-35)/16.53 =510mm then we will take

∅6

C/C 510mm

But since the spacing for the minimum reinforcement is greater than the spacing for Vsd, we will use the minimum one which is


∅8

C/C 160mm.

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Design of shear wall The walls are designed as isolated sway element of the frame using the second order theory of column given in section 4.4 of EBCS 2, 1995 following the approximate method (section 4.4.15.3) the wall shall be designed for uniaxial bending with an equivalent eccentricity of load along the axis parallel to the larger relative eccentricity.

eequ=etot(1 + k α)

Where

etot =total eccentricity K=relative eccentricity ratio α=is the function of relative normal force.

To get the relative eccentricity, first let’s calculate total eccentricity etot =ee+ea+e2 ee =is equivalent constant first order eccentricity of the design axial load ea=eccentricity due to imperfection =le/300>20mm e2=is the second order eccentricity, where it is applied only for non sway frame as per EBCS 2.1995 in section 4.4.10.3 Le=effective buckling depth.

Dimension

l-direction l=2700mm b=200mm l-direction l

b

c


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Structural design………………………………………………………….ceng 501 a y 2+ 2 C =a y 2

2

2

= 2.7 + 1.75 =3.217m

Moment and axial load transferred from stair slab. F1 =21.69*2 =43.38KN

M1 =25.05KN.m/m

F2 =12.15KN/m

M2 =9.113KN.m/m

M3 =25.05KN.m/m Then the concentrated load will be F2 =12.15*3.217 =39.09KN M2 =9.113*3.217 =29.32KN/m M3 = -29.32KN/m For the total height of the critical force and moment is at the bottom of the column. Fl =2*39.09 + 43.38 =121.56KN Ml =25.05KN/m

→ Axial

load due to self weight =24 * 0.2* 4* 2.7

=51.84KN

Then axial load, including self weight at the bottom of the wall will be Nsd

=121.56 + (1.3 *51.84) =188.95KN

Assuming the wall behaves as a cantilever in this direction, the effective buckling length in this plane is Le =2 * L

=2*4000 =8000mm

First order eccentricity and the additional eccentricity are ee= Ml / Nsd

=25.05/188.95


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Structural design………………………………………………………….ceng 501 =0.13m =130mm ea = Le /300

= 8000/300 =26.67mm

Total eccentricity etot = ee + ea

= 26.67+130 =156.67mm

Then

Msd

=188.95*0.15667 =29.6KN.m

Therefore the relative eccentricity erel =156.67/2700 =0.058

b –direction Pd =43.38KN is the force transferred from the beam Total axial load including the self weight at the bottom of the wall Nsd = 188.95KN M b =29.32KN.m

Le =8000mm Eccentricities ee= Ml / Nsd

=29.32/188.95 =0.155m =155mm

ea = Le /300

= 8000/300 =26.67mm

Total eccentricity etot = ee + ea

= 26.67+155 =181.67mm

Msd = Nsd * etot

=188.95*0.18167


18


=34.33KN.m

The relative eccentricity erel =181.67 / 200

= 0.908

Hence, relatively eccentricity ratio K = (smaller relative eccentricity)/ (larger relative eccentricity) = 0.058/0.908 =0.064 ν = Nsd /

(Ac *f cd) = (188.95E-03) / (13.4 *(2.7*0.2) =0.03

From table 4.1 of EBCS-2, 1995 we get a=0.63

eeq

= etot*(1+ (K*a)) =181.67*(1+ (0.064*0.63)) =189mm

The equivalent moment becomes Msd

=0.189*188.95 =35.71KN.m

From chart No.6 (uniaxial )of EBCS -2 1995, part 2 ,for

µ= Msd / (Ac *f cd *b) =35.71E-03/ (13.4*0.2*2.7*0.2) =0.025 Having µ=0.025 Then

and

ν=0.03

w=0.015

The total reinforcement is As = w*Ac *f cd/260.87 =0.015*13.4*0.2*2.7/260.87 Project on design of helical staircases

19


=416.07 ►For

vertical reinforcement in walls the code gives the upper and lower limits as follows 2

2

Asmax = 0.04 * Ac =0.04*0.2*2.7 =0.0216m =21600mm 2

Asmin =0.004* Ac =0.004*0.2*2.7 =2160mm Then use the minimum longitudinal reinforcement 2

As =2160mm

Use 20

∅12

Check for shear The resistance of a section is obtained from VRd =0.25 * f cd *bw *d Where bw =200mm ’

d /l = d’/2700=0.05 d’ =135mm d=2700-135 =2565mm The shear resistance becomes VRd = 0.25*13.4*200*2565 =1718.55KN > Vsd =0 because we neglect transversal load (like wind load etc) But according to section 7.2.5 of EBCS-2, 1995, the area of horizontal reinforcement shall not be less than one-half of that of the vertical reinforcement. That is 2

As,hor =0.5*2160mm 2

=1080 mm Use

∅8

C/C 300 through out the wall height

►The horizontal reinforcement shall enclose and be tied to the vertical bars so as to form a rigid mat.


20


4.3 design of spiral stair For the given room height considering the requirements of the stair we fix the dimension of trade and rise. Use radius of circular column(r)

=0.6

Radius from column center to slab center(R) =1.35 h=R (2ϑ)tana Where

h = room height 2ϑ = subscribed angle by the curved beam =180=p

-1

a= tan (h/(R*2 ϑ) 0

=39.53 =40

Perimeter of the concrete plate

At column edge

= p*0.6=1.885

At the middle of the plate

= p*1.35=4.241

At the free end

= p*2.1=6.597

Assume riser

=160mm

Number of riser =3.5/0.16=21.87 =22 risers and actual dimension =159mm Number of trade= Number of riser-1 = 22-1 = 21 Number of trade

Trade width=total length/number of width At edge

=1.885/21 =89.7 =90mm


21

Structural design………………………………………………………….ceng 501 At middle =4.241/21 =201.9 =202mm At end

=6.597/21 =314.14=315mm

So t he dimension requirement for the comfort of transportation is adequate.

Design of the stair slab. Design of the spiral stair plate must consider the two dominant stresses and reinforcement must be applied to resist the tensile stress in the plate. These stresses are

1.due to bending along the radial line 2. due to torsion

Design for flexure. Minimum depth required for deflection d1= (0.4+0.6*(f yk /400))/(le/βa)

use βa =12

=106.25mm. Over all depth (D) = d1 +cc+Φl/2+Φm =106.25mm+15mm+5mm+8mm=134.25mm Use D=150mm for construction convenience Dead load Due to waist slab=24*1*0.15 ° Due to floor finish =27*1*0.03* COS 30 ° Due to cement screed=24*1*0.02*COS 30 ° Due to steps=24*(area of steps in 1 m)=24*0.06936*COS 30 Total Live load

=3.6KN/m =0.701KN/m =0.415KN/m =1.44KN/m =6.156 KN/m

5*1*0.15=0.75KN/m

9.203KN/m Design load calculation Pd=1.3dl+1.6ll =1.3*6.156+1.6*0.75 =9.203N/m 1.5m Moment and shear calculation 2

2

M=wl /2 = 9.203 *1.5 /2 =10.35KN-m


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F=wl

=9.203 *1.5

=13.80KN

Check depth for flexure 0.5

d2= (M/ (ρ*b*f yd (1-0.4*ρ b*m))) = (M/ (b*4.569))

0.5

=47.60mm


Actual depth (d) = D-Cc-Φl/2-Φm =150-15-5-8 =122mm

Area of reinforcement using design chart of EBCS 1,1995 M=10.35KN-m 0.5

k m = ((M/b) )/d

0.5

=10.35 /0.122 =26.37

From the chart k s

=3.98 2

As = k s*M/d= 3.98*10.35/0.122= 337.65mm

S=b* as/As =1000*78.5/337.65

where as =area of Φ10mm bar

=232.5mm

Then use Φ10mm c/c 230mm as the main reinforcement.

Design for torsion

Since there is no clear understanding to formulate the exact torsional stress in a spiral concrete plate, different


23

Structural design………………………………………………………….ceng 501 investigators develop their own approximation to find the area of reinforcement to resist torsional stress. In this project we design for the stress caused by torsion as that of dome structure since they have nearly same structural feature. And also the analysis of the dome is done dividing the structure in to horizontal strips which gives us curved beams as that of the spiral stair. In dome structure there are meridonal thrust and hoop stress, where the torsional effects create as that of hoop stress in the concrete plate. Force act on unit length of the ring are

1. Meridonal trust, T, per unit of the circle of latitude ef acting tangentially or at right angleto the radial line of.

2. The reaction of trust T+dT per unit length of the circle of latitude gh acting at right angle to the radial line og. 3. The weight of the ring

So T=2pr*rs

where rs=or-os =or-ofcosϑ =r (1-cosϑ)

Since the sum of the vertical component of the trust T, acting along the along the Circumference of the circle must be equal to the weight of the portion the dome it,then we have


24

above

Structural design………………………………………………………….ceng 501 T*2p*sf sin Q =2p*r*w (1-cosQ) 2

Or T*2p*r sinQ* sinQ=2p* r (1-cosQ) 2

T=Wr((1-cos Q)/ sin Q

. On the account of the difference in the

magnitude

of T and

dT there is the rise of hoop force

Let H be the hoop force per unit length, the breadth of the ring rdQ hoop force= H* rdQ Horizontal component of T is T cosQ produces hoop tension The magnitude of hoop tension = T cosQ*radius of the ring = T cosQ*r sinQ T+dT also produce the same effect but in opposite direction resulting hoop compression in the ring.

= (T+dQ) cos (Q+dQ)*r*(sinQ+dQ) Hence the difference of the horizontal components of the two trusts causes the actual hoop compression or tension. The hoop force on the ring in illustration is due to the change in the magnitude of T,when Q increases by a very small dT when this increases tends to zero,e have

H*r* dQ=d (T cosQ*r sinQ)

from the above 2

H=W*r*(d/dQ)*(( cosQ/ sinQ)-( cos Q/ sinQ) 2

hoop force is H=W*r (cos Q+ cosQ-1)/(1+ cosQ) and the stress=H/thickness of the dome . Stress=horizontal force/thickness of the plate. =H/t

0

Θ ( )

2 where H=Wr ( cos ϑ+ cosϑ-1)/(1+ cosϑ)

W(KN/m) r (m) Stress (H/t) in KN/m

43.35 9.203

2.1

19.07

50

2.1

4.387

9.203

(compression) (compression)


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2

T=Wr ((1-cos Q)/ sin Q

differentiating the

Structural design………………………………………………………….ceng 501 51.48 9.203

2.1

0

60

9.203

2.1

-21.4

65

9.203

2.1

-36.41

70

9.203

2.1

-51.9 (tension)

75

9.203

2.1

-53.63 (tension)

80

9.203

2.1

-87.367 (tension)

85

9.203

2.1

-107.25 (tension)

90

9.203

2.1

-128.02 (tension)

(tension) (tension)

0

Maximum tension produced -164.6 at 90

at the free end of the slab then, hoop tension

tending to rapture the plate per meter length =-128.02*2.1=268.84KN The area of the steel required

= maximum tension/f yd 2

=268.84*1000/260.87 =1030.58mm S=1500*78.5/1030.58=114.25mm Therefore use Φ10mm c/c 110mm.

Design or circular column Design of the column is done like shear wall design in the previous section.

b direction Dimensions. L=1885mm B=150mm -1

α=tan (h/r*2Ө) we design the loaded part of the circular column as a shear wall .we approximate the curved shape and we consider the eccentricity coming from changing curved shape in to rectangular shape.

Moment and axial force transferred from the stair slab. Project on design of helical staircases

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Structural design………………………………………………………….ceng 501 F=18.8KN/m M=10.35KN-M The concentrated load will be F=13.8*3.975

=54.85KN

M=10.35*3.975

=41.14KN-M

For the total height the critical force and moment is at the bottom of the column. Axial load due to self weight. =gc*t*h*l=24*0.15*4*1.885=27.144KN Then the total axial load including self weight (Nsd) Nsd

=54.85+1.3*27.144=90.137KN

Assuming the wall behaves as a cantilever in this direction, the effective buckling length is in this plane is Le=2*L=8000mm Eccentricities ee =Mu/Nsd

=41.14/90.137 =0.456m

=456mm

ea=Le/300

=8000/300

=26.67mm

eo=(2/3)*r

=(2/3)*0.6

=0.4m

=400mm

when we align the loaded part (which is half circle) in to the rectangular form ,we know that eo is zero but when we deal with the half circle the load is at the centroid of the half circle. since we want to design the full circle, we place the concentrated load at the centroid of the full circle so by doing this, there is the emergence of the moment with eccentricity eo . Total eccentricity(e tot ) e tot = ee + ea +eo =26.67+456+400 =882.67mm


27

Structural design………………………………………………………….ceng 501 Then Msd=90.137*0.88267=79.55KN-M

There fore the relative eccentricity(erel) erel =882.67/150=5.88

l direction Total axial load including self weightat the bottom of the wall Nsd =90.137KN M b=0 Le=8000mm Eccentricities ee =Mu/Nsd

=0/90.137

=0

ea=Le/300

=8000/300

=26.67mm

etot=26.67mm Msd=90.137*0.02667=2.4KN-M There fore the relative eccentricity (erel) erel =26.67/1885=0.0141 Hence the relative eccentricity ratio(k) K=smaller relative essentricity/larger relative essentricity =0.0141/5.88

=0.0024

n= Nsd/(Ac*f cd) =90.137/(13.4*0.15*1.885)

-3

=23.8*E

=0.024

from table 4.1 of EBCS-2,1995 we get α=0.6 eeq= etot(1+k α)=882.67*(1+0.024*0.6) =895.4mm The equivalent moment becomes Msd=0.8954*90.137=80.71KN-M From chart no 6(uniaxial of EBCS -2,1995 PART 2,for µ= Nsd/(Ac*f cd*b) =80.71/(13.4*0.15*1.885) =0.142, n=0.024

then w=0.3


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Structural design………………………………………………………….ceng 501 the total reinforcement is As

= (w* Ac*f cd)/f yd = (0.3*0.15*1.885*13.4)/260.87 2

=4357.17mm

For vertical reinforcement in walls, the code gives the upper and lower limit as follows 2

As(max)=0.04*0.15*1.885=11310mm

2

As(min)=0.004*0.15*1.885=1131mm 2

Then use As=4357.17mm

Check for shear The resistance of the section is obtained from VRd=0.25*f cd*bw*d Where bw=is the minimum width of the web=150mm 1

=d /1885

1

=94mm

d /h=0.05 d =94.25

1

d=1885-94 =1791mm shear resistance becomes VRd=0.25*13.4*150*1791 =899.55>vsd=0 assuming that there is no transversal load. But according to section 7.2.5of EBCS 2,1995, the area or the horizontal reinforcement shall not be less than one half of that of the vertical reinforcement. 2

2

That is As=0.5*4357.17mm =2178.58 mm

the spiral reinforcement shall enclose and tied to the vertical bars so as to form a riged mat.


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5. Recommendations and future work

In conclusion to these designs, spiral and geometric as part of helical stair, we hope that they can help as an alternative structures beside other staircases in which most building comprise. The best example is doglegged staircase. The best alternative to apply these two designs where when we don’t have any room for stair case installation in the building , so that they can be construct out side the building. The type of the building also have significant effect on the selection of stair cases so,One should specify the use of the building. Since due to the aesthetic out put spiral Rc designs can be applied for a places like cafeterias, trade centers. Site condition should be investigated properly because each design needs safe and sufficient foundation. Height of the construction should be specified to select with respect to economy and workability. Since reinforcement placement and form work construction of the spiral stair is a little bit complicated than other types there is a need for skilled person.

And finally we would like to inform for any one who reads this project work, it is possible to know the best fit and economical structure by laying out a take off sheet for the volume of the concrete, the total form work per meter square and bar schedule for the total weight of the reinforcing bar and by providing their respective unit price then preparing bill of quantity will tell as which is the most economical.


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Concrete-Helical-Stair-Design.pdf

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