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Design of spherical dome to carry Load

Posted on May 26, 2021November 19, 2022 By YASH PAL No Comments on Design of spherical dome to carry Load

Table of Contents

  • Design step of the dome
    • Analysis of spherical dome
    • 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.

Design step of the dome

  1. With the help of the data given finalized the geometry of the dome.
  2. Calculate the various load coming on the structure.
  3. Calculate the maximum tensile & maximum compressive stress coming on the dome with the help of 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 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
Analysis and Design of spherical dome to carry Load
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
Analysis and Design of spherical dome to carry Load
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
civil engineering, engineering subjects Tags:civil engineering, engineering subjects

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