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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...
Structural design.ceng 501
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.
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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
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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
Structural design.ceng 501
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.
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Project on design of helical staircases 2
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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Project on design of helical staircases 3
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
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Project on design of helical staircases 4
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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Project on design of helical staircases 5
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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Project on design of helical staircases 6
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
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Project on design of helical staircases 7
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
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Project on design of helical staircases 8
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
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Project on design of helical staircases 9
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
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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Project on design of helical staircases 10
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
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=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
Then use 10mm c/c 280mm as the main reinforcement.
Asmin=0.5*b*d/ fyk =0.5*1000*122/300=203.33mm2
Spacing =b*as/As
=1000*78.5/203.33=386.129mm
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Project on design of helical staircases 11
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
Design load calculation
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
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Project on design of helical staircases 12
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
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Project on design of helical staircases 13
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
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Project on design of helical staircases 14
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
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Project on design of helical staircases 15
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
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Project on design of helical staircases 16
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
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Project on design of helical staircases 17
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
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Project on design of helical staircases 18
=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
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Project on design of helical staircases 19
=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
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Project on design of helical staircases 20
=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.
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Project on design of helical staircases 21
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
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Project on design of helical staircases 22
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
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Project on design of helical staircases 23
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
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=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
Then use 10mm 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
Structural design.ceng 501
Project on design of helical staircases 24
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
Structural design.ceng 501
Project on design of helical staircases 25
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)
Structural design.ceng 501
Project on design of helical staircases 26
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.
Structural design.ceng 501
Project on design of helical staircases 27
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
Structural design.ceng 501
Project on design of helical staircases 28
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
Structural design.ceng 501
Project on design of helical staircases 29
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.
Structural design.ceng 501
Project on design of helical staircases 30
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.
Structural design.ceng 501
Project on design of helical staircases 31
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.