Domes

DOME

Prof. A.M. Bhanderi

INTRODUCTION  A Dome is a thin shell generated by the revolution of a

regular geometrical curve about its vertical axis.  Revolution of a circular curve about the vertical diameter gives a Spherical Dome.

Prof. A.M. Bhanderi

Prof. A.M. Bhanderi

Prof. A.M. Bhanderi

Prof. A.M. Bhanderi

 Revolution of an elliptical curve about one of its axis gives

an Elliptical Dome.

Prof. A.M. Bhanderi

Prof. A.M. Bhanderi

 Revolution of a right angled triangle about one of the sides

gives a Conical Dome.

Prof. A.M. Bhanderi

Prof. A.M. Bhanderi

USE OF DOME

To cover any circular area

Prof. A.M. Bhanderi

To cover any circular area

Prof. A.M. Bhanderi

Circular tanks

Prof. A.M. Bhanderi

Circular tanks

Prof. A.M. Bhanderi

Biogas Circular tanks

Prof. A.M. Bhanderi

Prof. A.M. Bhanderi Exhibition hall, Stadium, Auditorium

Exhibition hall, Stadium, Auditorium

Prof. A.M. Bhanderi

Temple, Mosques

Prof. A.M. Bhanderi

Assembly hall

Prof. A.M. Bhanderi

 Mainly domes are used to cover large circular areas,

because they prove to be far economical with respect to material than any other types of roof.  In domes, the loads cause only direct stress ( compressive

or tensile) and bending moment and shear force are negligible.  Usual materials for construction of dome are steel,

masonry, timber and reinforced concrete. Due to development of tensile stress in large dome, the masonry domes become very heavy and expensive. Timber and steel domes are also expensive beside their usual shortcomings. Therefore concrete domes are most suitable for normal construction. Prof. A.M. Bhanderi

NATURE OF STRESS IN SPHERICAL DOMES  A spherical dome may be thought of as consisting of

circular rings of continuously reducing diameter placed one above the other.  The top point is called the crown of the dome.  Each ring supports the load of all the rings above it and

transfers the load to the ring immediately below it.  The joints between two rings being radial, the reaction

between them is tangential to the curved surface, giving rise to compression along the meridians.  This is termed as “Meridional Compression or Thrust” Prof. A.M. Bhanderi

Prof. A.M. Bhanderi

 Latitudes is the curve of a ring.  Meridian if the circles are drawn through the top and bottom

points diametrically opposite to each other, the circles are called of meridians.  Longitude A line corresponding to each circle of meridian is called longitude.  Meridional thrust (T) the direct compressive force acting along the meridians is called Meridional thrust or Meridional compression.  Hoop compression (H) the tendency of separation of any voussoir will be prevented because of its wedge shape giving rise to hoop compression (H) in each ring.

Prof. A.M. Bhanderi

Ring

Formation of spherical dome

Vertical section

Latitude Latitude Meridian

Plan of a Ring

Meridian

Prof. A.M. Bhanderi

Prof. A.M. Bhanderi

LOAD ACTING ON DOMES  The various types of load acting on the domes are: i.

Self weight of dome

ii.

Live load

iii.

Snow load

iv.

Wind load

Prof. A.M. Bhanderi

SPHERICAL DOME SUBJECTED TO UNIFORMLY DISTRIBUTED LOAD

Here, r = Radius of dome T = Intensity of Meridional thrust t = Thickness of dome shell w = U.D.L. inclusive of its self weight per unit area.Prof. A.M. Bhanderi

 Consider the equilibrium of a ring ABCD between two

horizontal planes AB and CD, the extremity of it makes an angle θ and θ+dθ with vertical at the centre respectively. So the ring subtends an angle dθ at the centre.  Force acting on unit length of the ring.

 The meridional thrust T, per unit length of the circle of

latitude AB, acting tangentially at B.  The reaction or thrust T+dT, per unit length of the circle

of latitude CD, acting tangentially at D.  The weight δW of the ring itself, acting vertically

downward.  Here, the meridional thrust T is caused by the weight of

the dome shell APB above the horizontal plane AB. Prof. A.M. Bhanderi

 Weight of the dome shell AB = 2πr x PQ x w

= 2πr (r-rcosθ) x w = 2πr2w(1-cosθ)  Now, this total load above the ring AB must be equal to the vertical components of T round the total periphery of the ring ABCD. T(2π x QB) sinθ = 2πr2w(1-cosθ) So, T sinθ (2πr sinθ) = 2πr2w(1-cosθ)

w r (1 - cos) T sin2

wr T 1  cos

…….Eq. of Meridional thrust, T.  Here meridional thrust T is acting on per unit length, so the Eq. for the meridional stress will be, w r (1 - cos)

T

2 A.M. Bhanderi Prof. t  sin 

Hoop force (H)  The horizontal components of the meridional thrust will

produce the hoop stress along the periphery of the dome.  As the meridional thrust T increases to T+dT at the

bottom of the ring, this difference will cause hoop stress.  Let H be the hoop force per unit length of surface

measured on great circle arc.  As the breadth of ring = rdθ  The Hoop force = H x rdθ

…………….. (I)

 The horizontal component of T is T cosθ which produces

hoop tension. Prof. A.M. Bhanderi

 The magnitude of Hoop tension,

= T cosθ x Radius of ring AB = T cosθ x r sinθ = T r sinθ cosθ ……………………. (II)  Similarly, the horizontal components of the thrust T+dT will be (T+dT) cos(θ+dθ) and this horizontal component will cause hoop compression.  The magnitude of Hoop compression, = (T+dT) cos(θ+dθ) x Radius of ring CD = (T+dT) cos(θ+dθ) x r sin(θ+dθ) …………….. (III) The difference between (II) and (III) will cause the actual hoop stress. If (III) > (II), Hoop stress will be compressive Prof. A.M. Bhanderi If (III) < (II), Hoop stress will be tensile.

 So, the Hoop stress is due to change in the value of T

when θ increase by a small amount dθ, hence in the limiting case when dθ is extremely small, H x rdθ = d(T r sinθ cosθ)  by putting value of T and differentiating,

w r (cos2  cos - 1) H 1  cos  But, at the crown,

θ = 0 and also T = 0,

So, H = wr/2 H wr And Hoop Stress at crown =  t 2t This is the maximum value of hoop stress in compression. Prof. A.M. Bhanderi

 Now as the θ increase up to some value, the hoop stress

will go on decreasing and becomes tensile.

 To get the circle of zero hoop stress,

w r (cos2  cos - 1) H 0 1  cos Cos2θ + Cosθ – 1 = 0

Cosθ = 0.618 θ = 51.82° When θ = 51.82° , H = 0 When θ < 51.82°, H will be compressive (+ve) When θ > 51.82°, H will be Tensile (-ve) Prof. A.M. Bhanderi

Prof. A.M. Bhanderi

SPHERICAL DOME SUBJECTED TO CONCENTRATED LOAD AT THE CROWN

Prof. A.M. Bhanderi

 The sum of the vertical components of thrust T acting

along the circumference of the circle of latitude must be equal to the load W. T x 2πr sinθ x sinθ = W W T 2r  sin2 

OR

W T cosec2 2r

• Now, as the hoop stress developed in any horizontal ring

is due to the difference in the meridional thrust T and (T+dT), H x rdθ = d(T r sinθ cosθ) W H cosec2 2r Prof. A.M. Bhanderi

H W  cosec2  Hoop stress = t 2rt  The negative sign shows that the hoop stress developed in

the dome due to a concentrated load at the crown will always be tensile.  At crown θ = 0, hence, Hoop stress becomes infinite.

Therefore any concentrated load in the form of lantern or ornaments etc… should always be distributed over sufficient area, to reduce the hoop stress at the crown. It is also desirable to thicken the dome at the crown to spread the load over greater area.

Prof. A.M. Bhanderi

Prof. A.M. Bhanderi

Prof. A.M. Bhanderi

CONICAL DOME SUBJECTED TO UNIFORMLY DISTRIBUTED LOAD

Here, From the geometry, AB = 2y tanθ & Length AP = y/cosθ

Prof. A.M. Bhanderi

 The vertical component of the total meridional thrust at B

will evidently be equal to the load on the dome shell APB. w y  Load on dome shell APB     2y  tan  2 cos  Vertical component of total T, = T cosθ x (π x 2y tanθ)

w y T cos    2y  tan      2y  tan  2 cos W y  Meridional thrust, T   2 cos2  W y  Intensity of meridional stress   2t cos2  Prof. A.M. Bhanderi

 Now, Horizontal components of T will cause Hoop tension

at B, while the horizontal component of (T+dT) will cause hoop compression.  Magnitude of Hoop tension

H = T sinθ x Radius at ring AB

= T sinθ x y tanθ = T y x sin2θ/cosθ Magnitude of Hoop Compression, = (T +dT) sinθ x (y + dy) tanθ  The difference in these two horizontal components will give the value of Hoop force. Prof. A.M. Bhanderi

 Let H be the Hoop compression induced in the ring, per

unit breadth. Let ds be the breadth of the ring of height dy.

So, ds  dy cos  T  y  sin 2   We have, H  ds  d   cos    H



d T  y  sin 2  dy



Substituting the value of T and differentiating, we get d w y 2  H   y  sin  2  dy  2 cos  

 

w sin 2  d 2 H   y 2 2 cos  dy

Prof. A.M. Bhanderi

 Hoop force , H = wy tan2θ

wy  Intensity of Hoop stress  tan2  t  Where, t = Thickness of dome slab

w = intensity of U.D.L. inclusive of self weight, per unit area of the dome. 2θ = angle of the apex

Prof. A.M. Bhanderi

CONICAL DOME SUBJECTED TO CONCENTRATED LOAD AT VERTEX

Here, From the geometry, AB = 2y tanθ & Length AP = y/cosθ

Prof. A.M. Bhanderi

 The sum of the vertical components of thrust T acting

along the circumference of the circle of latitude must be equal to the load W. T cosθ x π x 2y tanθ = W  So, Meridional thrust T,

W T 2  y  sin   Meridional stress,

W  2t  y  sin 

Prof. A.M. Bhanderi

 Now, Horizontal components of T will cause Hoop tension

at B, while the horizontal component of (T+dT) will cause hoop compression.  Magnitude of Hoop tension

H = T sinθ x Radius at ring AB

= T sinθ x y tanθ = T y x sin2θ/cosθ W sin 2   y 2  y  sin  cos  W H tan  2 Prof. A.M. Bhanderi

 Magnitude of Hoop Compression,

H + dH = (T +dT) sinθ x (y + dy) tanθ  So, Net Hoop compression on ring = H + dH – H = dH

And area of the ring /unit length = (ds x 1) = dy/cosθ  Now, Hoop compression /unit length

W But from, H  tan  2

dH  dy / cos 

It is clear that H is a constant quantity as w, π and θ are constant for particular dome. So, dH  0 dy Prof. A.M. Bhanderi

 dH = 0 shows that thrust on the bottom of the ring will

also produce a hoop compression equal to H itself.  Hence, net Hoop load on section = 0

Prof. A.M. Bhanderi

DESIGN OF R.C.C. DOMES  As compare to other structure, the required thickness

and % of steel in the R.C.C. dome is very less.  But as per the IS code, minimum thickness of 7.5 cm is

provided to protect steel. And a minimum steel provided is 0.15% for mild steel bars and 0.12% for HYSD bars, of the sectional area in each direction – meridionally as well as along the latitudes.

Prof. A.M. Bhanderi

Provision of Ring Beam  If the dome is not hemispherical, the meridional thrust at

the supporting circle of latitude will not be vertical. The inclined meridional thrust at the support will have horizontal component which will cause the supporting walls to burst outwards, causing its failure.  In order to bear this horizontal component of meridional thrust, a ring beam is provided at the base of the dome.  The reinforcement provided in the ring beam takes this hoop tension and transfer only vertical reaction to the supporting walls.  The tensile stress on the equivalent area of concrete on the ring beam section should not be exceed 1.2N/mm2. Prof. A.M. Bhanderi

Prof. A.M. Bhanderi

Prof. A.M. Bhanderi

Prof. A.M. Bhanderi

Prof. A.M. Bhanderi

Prof. A.M. Bhanderi

Prof. A.M. Bhanderi

Prof. A.M. Bhanderi

Placement of main reinforcement in Dome  A minimum reinforcement of 0.15% of area is provided

in both the direction of latitude as well as of the meridians.  If the reinforcement along the meridians is continued

upto crown, there will be congestion of steel there, hence from practical consideration, the meridional reinforcement is stopped at any latitude circle near crown, and a separate mesh is provided.

Prof. A.M. Bhanderi

Prof. A.M. Bhanderi

Provision of Opening  The opening is also provided in the dome as required

from other functional or architectural requirements. However sufficient trimming reinforcement should be provided all round the opening.  The meridional and hoop reinforcement reaching the

opening should be well anchored to the trimming reinforcement.  If there is an opening at the crown of the dome, and if

there is any concentrated load of lantern etc.. acting there a ring beam should be provided at the periphery of the opening. Prof. A.M. Bhanderi

Prof. A.M. Bhanderi

Prof. A.M. Bhanderi

Prof. A.M. Bhanderi

Prof. A.M. Bhanderi

Prof. A.M. Bhanderi