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
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) 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
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 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 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
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