Design of spherical dome to carry Load

In this tutorial, we will use the 5-step approach to design a spherical dome to carry a load.

Design step of the spherical dome

1. With the help of the data given finalized the geometry of the dome.
2. Calculate the various loads coming on the structure.
3. Calculate the maximum tensile and maximum compressive stress on the dome with the help of the values of meridional and hoop stress.
4. Tabulate the design table starting with the value of meridional & hoop stress with respect to the critical angle.
5. 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 of the hemispherical dome

= 90

H = –

For zero hoop stress, H = 0

0 = –

W r ( = 0

= 0

= 0.618

= 51.83

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

T = cosec 2

C’; B’;

0.9 m 0.9 m

D’;

C 4.5 m 4.5 m D

B

G

H = – cosec 2

c) Stress 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

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

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

iv) For

Sin = 43.6

v) For

Sin = 7.92

2) Load calculation

Let the thickness of the dome be

t = 100 mm = 0.1 m

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

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

In the absence of a 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

H = cos – – cosec

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

Using 8 mm dia of bars

No. of bar = = 3.52 4 bar

Spacing = x 1000 = 280 mm

Also, read

• Design step for cantilever retaining wall
• Design a rectangular beam