# Design of spherical dome to carry Load

## Design step of the dome

- With the help of the data given finalized the geometry of the dome.
- Calculate the various load coming on the structure.
- Calculate the maximum tensile & maximum compressive stress coming on the dome with the help of values of meridional and hoop stress.
- Tabulate the design table starting with the value of meridional & hoop stress with respect to the critical angle.
- Find the reinforcement required against the value of maximum tensile stress.

### Analysis of spherical dome

**a) Stress under UDL of ‘W’ per unit surface area**

T =

H = –

At crown = 0

H = 0.5

At the base for the hemispherical dome

= 90

H = –

For zero hoop stress, H = 0

0 = –

W r ( = 0

= 0

= 0.618

= 51.83

[br]

**b) Stress under concentrated load ‘W’ at the crown**

T = cosec 2

[br]

C' B'

0.9 m 0.9 m

D'

[br]

C 4.5 m 4.5 m D

[br]

B

G

H = – cosec 2

[br]

**c) Strees under combined UDL & concentrated load**

T = + cosec 2

H = – – cosec 2

### Q. The inside diameter of the circular room is 9m. design a spherical dome to carry a uniform distribution live load of 1.5 KN/m 2. It has to support a lantern of 18 KN as a point load. It has a circular opening of 1.8 m at crown use M-20 & Fe 415.

**Given**

Diameter of room = 9m

Live load = 1.5 KN/m 2

Cone load = 18 KN

Opening at crown = 1.8 m

M-20, Fe 415

**1) Geometry of dome (Rise, R, h, )**

i) Rise = = to

= and =

AB = 1.8 m

[br]

**ii) For R**

By chord property

CB x BD = AB x BO

4.5 x 4.5 = 1.8 x (2R – AB )

R = 6.525 m

[br]

**iii) For h**

By chord property

C’B’ x B’D’ = AB’ x B’O

0.9 x 0.9 = h x (2R – h )

0.81 = h x (13.05 – h )

0 = h 2 – 13.05h + 0.81

h = 12.98 m , 0.062 m

h = 0.062 m = 62 mm

[br]

**iv) For**

Sin = =

= 43.6

[br]

**v) For**

Sin = =

= 7.92

[br]

**2) Load calculation**

Let the thickness of dome be

t = 100 mm = 0.1 m

[br]

Total load per unit area = DL + LL

W = 25 x .1 + 1.5

W = 4 KN/m 2

Area of the opening C,A,D

= 2rh = 2 x 3.14 x 0.9 x 0.062 = 0.35 m 2

Load of the opening = 4 x 0.35 x 1.4 KN

Effective load at crown = 18 – 1.4 = 16.6 KN

[br]

**3) Stress calculation**

Meridional stress for comdined load

T = + cosec 2

T = + cosec 2

T = + 4. 05 cosec 2 KN/mm 2

A B

Hoop stress

H = cos – – cosec 2

H = cos – – cosec 2

H = 0.261 cos – – 0.004 cosec 2 N/mm 2

C D

[br]

**In absence of live load**

Load per unit area

W I = DL = 25 x 0.1 = 2.5 KN/m 2

Load of opening = 2.5 x area of opening

= 2.5 x 0.35 = 0.87 KN

Effective load of crown

W = 18 – 0.87 = 17.13 KN

[br]

H = cos – – cosec

[br]

H = cos – – cosec 2

H = 163 cos – – 4.18 cosec 2

H = 0.163 cos – – 0.00418 cosec 2 N/mm 2

E F

Maximum compressive stress = 0.3317 N/mm 2

Maximum temsile stress = 0.1311 N/mm 2

**4) Reinforcement**

Force = stress x area

Max. tensile force = 0.1311 x 100 x 1000 = 13110 N

Area of steel = = = 57 mm 2

Ast minimum = x 100 x 1000 = 120 mm

Total Ast = 120 + 57 = 177 mm 2

[br]

Using 8 mm dia of bars

No. of bar = = 3.52 4 bar

Spacing = x 1000 = 280 mm