Concrete Helical Stair Design

Structural design.ceng 501

Project on design of helical staircases i

Table of content Acknowledgement

Index .

1. Introduction..................................................................

2. Literature survey

2.1 Definition

2.2 Terms in stair...

2.3 Essential requirement..

2.4 Classification of stair ..

2.4.1 Based on material of construction.

2.4.2 Based on type.

2.5 helical 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 of spiral stair

4. Result and discussion 4.1 Design of geometric stair

4.2 Design of helical stair.

5. Recommendation and future work.

6. Reference...


Project on design of helical staircases ii

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.


Project on design of helical staircases iii

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

fcd. design compression strength of concrete

fyd design yield strength reinforcement

fyk .. 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

.............................................. Diameter of the reinforcing bars

s.. Partial factor of safety for steel

c partial factor of safety for concrete

unit weight

b effective geometrical ratio for reinforcement


Project on design of helical staircases iv

m .......................................... mess reinforcement diameter

2 subtended angle

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


Project on design of helical staircases 1

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.



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



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 45 and not less than 25 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



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.



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



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- work

Live load=5KN/m2 (for public building)

l =10mm

m =8mm

Cc=15mm for slab and 25mm for beam

Design constant using EBCS 1, 1995

fcd=0.85fck/c =13.4mpa

fyd=fyk/s =260.87mpa

m=fyd/(0.8*fcd) =24.335

c1=2.5/m =0.103

c2=0.4*fyd*m =2539.31

b= (0.8*Ec*fcd)/((Ec+Es)*fyd) =0.0299

=0.75*b =0.0224

Unit weight from EBCS 1, 1995

marble =27KN/m3 , 3cm thickness

concert =24KN/m3

screed =24KN/m3 , 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*(fyk/400))/ (le/a) use a =12 for span ratio 1:2.14

=106.25mm.

Over all depth (D) = d1 +cc+l/2+m

=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 = 5.304KN/m

Live load

5*1*0.15 =0.75KN/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

M=wl2/2 = 8.1*1.52/2 =9.113KN.m

F=wl =8.1*1.5 =12.15KN

Check depth for flexure

d2= (M/ (*b*fyd (1-0.4*b*m))) 0.5

= (M/ (b*4.569)) 0.5 =44.66mm

Since d1>d2 safe against deflection.

Actual depth (d) = D- Cc -l/2-m

=150-15-5-8

=122mm

Area of reinforcement using design chart of EBCS 1,1995

M=9.113KN/m

km =((M/b)0.5)/d =9.1130.5/0.122

=24.74

From the table ks = 3.975



Then the reinforcement will be

As = ks*M/d = 296.92mm2

S=b* as/As

=1000*78.5/296.92 where as =area of 10mm bar

=264.38mm

Then use 10mm c/c 260mm as the main reinforcement.

Minimum reinforcement for temperature and shrinkage.

Asmin=0.5*b*d/ fyk

=0.5*1000*122/300=203.33mm2

Spacing=b*as/As

=1000*78.5/203.33=386.129mm

Use 10mm 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


Pd= 1.3D.L+1.6L.L 7.557KN/m

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

1.5m



Moment and shear calculation

M=wl2/2 = 7.557*1.52/2 =8.5KN-m

F=wl = 7.557 *1.5 =11.34KN

Check depth for flexure

d2= (M/ (*b*fyd (1-0.4*b*m))) 0.5

= (M/ (b*4.569)) 0.5 =43.13mm


Actual depth (d) = D- Cc-l/2-m

=150-15-5-8

=122mm


M=8.5KN-m

km = ((M/b)0.5)/d =6.61570.5/0.122

=23.9

ks =3.97 As = ks*M/d =276.6mm2

Spacing =b*as/As =1000*78.5/276.6 =283.8mm


Asmin=0.5*b*d/ fyk =0.5*1000*122/300=203.33mm2

Spacing =b*as/As

=1000*78.5/203.33=386.129mm



Use 10mm c/c 380mm where minimum reinforcement required.

Design of the beam Design for flexure Determination of depth

d (0.4+0.6* fyk/400)*le/e = 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


D.L =24*0.2*0.25 =1.2KN/m

We have 11.34KN/m load coming from the

landing.

Pd = (1.3*1.2) + 11.34

=12.9KN/m

L =1.5m

M = wl2/2 =12.9*1.52/2 =14.51KN.m

F =wl =12.9*1.5 = 19.35KN

d= (M/ (b*4.569)) 0.5 = (14.51E6/ (250*4.569)) 0.5

=112.71mm



We have 11.34KN/m load coming from the

landing.

Pd = (1.3*2.4) + 11.34

=14.46KN/m

L =1.5m

M = wl2/2 =14.46*1.52/2 =16.27KN.m

F =wl =14.46*1.5 = 21.69KN

d= (M/ (b*4.569)) 0.5 = (16.27E6/ (250*4.569)) 0.5

=119.4mm 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)

=59.4E3 mm2

h ef =def /5 = (b-2d) /5

=26mm < A/U = (400*250)/ ((250+400)*2) = 76.92mm)

Tc = 1.2*fctd*Aef*hef

=1.2*1.165*59.4E3*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



Since Tc < Tsd =12.75KN.m we need a reinforcement

Section capacity against diagonal compression

TRd =0.8*fcd* Aef*hef

=0.8*13.4*59.4 E3 *26 =16.55KN.m > Tsd =12.75KN.m

Vc =0.25*K1*K2* fctd*bW*d

=0.25*(1+50*0.002)*(1.6-0.357)*1.165*250*357 =35.54KN

Vsd / (1.5-0.357) = 21.69/1.5

Vsd =16.53KN

VRd =0.25* fcd* bW*d

=0.25*13.4*250*357 =298.99KN

The limiting value for torsion and shear are

TRd,cor =t * TRd and VRd,cor = v* VRd

t =1/(1+(( Vsd / VRd)/( Tsd / TRd ))2)0.5 v =1/(1+(( Tsd / TRd)/( Vsd / VRd))2)0.5 By using the above results we will get

t =0.997 , v =0.072

12.75KN.m

0.357m 1.5m-0.357m=1.143m

21.69KN

1.5m



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 btc = 1/(1+(( Vsd / Vc)/( Tsd/ Tc))2)0.5 , bvc = 1/(1+(( Tsd / Tc)/( Vsd/Vc))2)0.5

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*fyd)

= ((12.75-3.05)*1020)/ (2*59.4E3*260.87)

=319.25mm2

Then use 2 16 longitudinal bars

Transversal bars

S (2* Aeff*fyd *Astr)/ ( Tsd - Tc, cor)

= (2*59400*260.87*((*102)/4))/ (12.75-3.03)

=250.42mm by using 10

10 C/C 250mm

And

S (2* Aeff*fyd *Astr)/( Tsd - Tc,cor)

= (2*59400*260.87*((*82)/4))/ (12.75-3.03)

=160.27mm in using 8

8 C/C 160mm

So use 8 C/C 160mm



Shear design

Vc, cor =3.91KN , Vsd =16.53KN , VRd, cor =21.53KN

16.53/1.5 =3.91/X

X =0.355mm

Therefore from X=0 to x =0.355mm, provide minimum reinforcement with spacing

(Asv*fyk)/ (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*fyd )*(d-dc)/ Vsd

Let us assume dc =35mm

S=2*((*62)/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.

0.357mX

16.53KN

1.5-0.357m

3.91KN



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)

Where etot =total eccentricity

K=relative eccentricity ratio

=is the function of relative normal force.

To get the relative eccentricity, first lets 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



a y C2 =a2 + y2

= 2.72 + 1.752

=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



=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

Mb =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



=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

= Nsd / (Ac *fcd)

= (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

= Msd / (Ac *fcd *b)

=35.71E-03/ (13.4*0.2*2.7*0.2)

=0.025

Having =0.025 and =0.03

Then w=0.015

The total reinforcement is

As = w*Ac *fcd/260.87

=0.015*13.4*0.2*2.7/260.87



=416.07

For vertical reinforcement in walls the code gives the upper and lower limits as follows Asmax = 0.04 * Ac =0.04*0.2*2.7 =0.0216m2 =21600mm2

Asmin =0.004* Ac =0.004*0.2*2.7 =2160mm2

Then use the minimum longitudinal reinforcement

As =2160mm2

Use 20 12

Check for shear The resistance of a section is obtained from

VRd =0.25 * fcd *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

As,hor =0.5*2160mm2

=1080 mm2

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.



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 (2)tana Where h = room height

2 = subscribed angle by the curved beam =180=p a= tan-1 (h/(R*2 ) =39.53 =400

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



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*(fyk/400))/(le/a) use a =12 =106.25mm. Over all depth (D) = d1 +cc+l/2+m =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 =0.701KN/m Due to cement screed=24*1*0.02*COS 30 =0.415KN/m Due to steps=24*(area of steps in 1 m)=24*0.06936*COS 30 =1.44KN/m Total =6.156 KN/m Live load 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 M=wl2/2 = 9.203 *1.52/2 =10.35KN-m

1.5m



F=wl =9.203 *1.5 =13.80KN Check depth for flexure d2= (M/ (*b*fyd (1-0.4*b*m))) 0.5

= (M/ (b*4.569)) 0.5 =47.60mm


Actual depth (d) = D-Cc-l/2-m

=150-15-5-8

=122mm


M=10.35KN-m

km = ((M/b)0.5)/d =10.350.5/0.122

=26.37

From the chart ks =3.98

As = ks*M/d= 3.98*10.35/0.122= 337.65mm2

S=b* as/As

=1000*78.5/337.65 where as =area of 10mm bar

=232.5mm


Design for torsion

Since there is no clear understanding to formulate the

exact torsional stress in a spiral concrete plate, different



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-ofcos =r (1-cos)

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 above

it,then we have



T*2p*sf sin Q =2p*r*w (1-cosQ)

Or T*2p*r sinQ* sinQ=2p* r2 (1-cosQ)

T=Wr((1-cos Q)/ sin2Q . 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 T=Wr ((1-cos Q)/ sin2Q

H=W*r*(d/dQ)*(( cosQ/ sinQ)-( cos2Q/ sinQ) differentiating the

hoop force is H=W*r (cos2Q+ cosQ-1)/(1+ cosQ)

and the stress=H/thickness of the dome .

Stress=horizontal force/thickness of the plate.

=H/t where H=Wr ( cos2+ cos-1)/(1+ cos)

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

43.35 9.203 2.1 19.07 (compression)

50 9.203 2.1 4.387 (compression)



51.48 9.203 2.1 0

60 9.203 2.1 -21.4 (tension)

65 9.203 2.1 -36.41 (tension)

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)

Maximum tension produced -164.6 at 900 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/fyd

=268.84*1000/260.87 =1030.58mm2

S=1500*78.5/1030.58=114.25mm Therefore use 10mm 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

=tan-1(h/r*2)

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.



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



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

Mb=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*fcd) =90.137/(13.4*0.15*1.885) =23.8*E-3 =0.024

from table 4.1 of EBCS-2,1995 we get

=0.6

eeq= etot(1+k)=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

= Nsd/(Ac*fcd*b) =80.71/(13.4*0.15*1.885) =0.142, n=0.024

then w=0.3



the total reinforcement is

As = (w* Ac*fcd)/fyd

= (0.3*0.15*1.885*13.4)/260.87

=4357.17mm2

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

As(max)=0.04*0.15*1.885=11310mm2

As(min)=0.004*0.15*1.885=1131mm2

Then use As=4357.17mm2

Check for shear The resistance of the section is obtained from

VRd=0.25*fcd*bw*d

Where bw=is the minimum width of the web=150mm

d1/h=0.05 =d1/1885

d1 =94.25 =94mm

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.

That is As=0.5*4357.17mm2=2178.58 mm2

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



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 dont 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.



6. Reference 1. Sushil Kumar, treasure of Rcc designs, 15th edition, India, standard book house, 2003.

2. P.dayartnam,design of reinforcing concrete structures3rd edition, India , oxford and ibh

publishing co.pvt.ltd,1998.

3. EBCS 2, 1995,structural use of concrete, Ethiopia, ministry of works & urban

development,1995.

4. EBCS 1, 1995 , ,Ethiopia, ministry of works & urban

development,1995.

5. Reinforcing concrete text book.

Concrete Helical Stair Design

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