Concrete Dome Seismic Analysis

  • November 16, 2011, 1:34am UTC
  • by Dr. Arnold Wilson

Example

Dome Diameter = 110’
Height of Dome = 37’
Thickness = 3” @ top and 8” @ bottom
Ref: Billington 1985 Ed., p55

Pz= pcos <b> sin <a>
P<a> = p<b> = 0

Membrane forces

N<a> = apk1 cos<b>
N<ab> = apk2 sin<b>
N<b> =apk3 cos<b>

Seismic Force (UBC 1985 Edition)

V= ZSICKW (Formula for the total design lateral force)
Z=1.0 (Zone IV – Seismic Zone Factor)
CS=.14
I=1.5 (Importance Factor = Hospital)
K=2.0 (Unusual building such as dome – very conservative

Therefore: V = (1.0) (1.5) (.14) (2.0) W = .42OW – Note: V =.14W for normal shear wall building

V=(.42) (100) = 42.0 psf – one square foot of shell 8”thick weighs 100 lbs.

The value of p=V=42.0 psf

For demonstration purposes assume p = 60 psf. This represents earthquake forces in excess of the most severe code requirement by a factor of 1.4

Maximum stress due to N,b. is -64.8 psi; n<a> is 70.6 psi. Maximum bending moment is 909.3 lbs – ft/ft.

For a vertical live load of 40 psf in addition to the dead load of the shell the following stresses and moment are obtained. Maximum stress due to N<a> = -82.5 psi; N,b> = -70.7 psi or 146.5 psi. The maximum bending moment is 1,588.0 lbs – ft/ft.

The maximum allowable compressive force in the concrete is: fc = 1.33 (.45) (4000psi)=-2.394 psi. This is many times greater than the 70.6 psi needed.

Conclusion

The forces caused by a major earthquake are considerably less than normal provided for when a dome is designed for nominal vertical loads.

Filed in: Durability
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