Thread: My anti seismic systems View Single Post
#130 seismic Registered User Join Date: Feb 2010 Location: ISLAND OF IOS CYCLADES Posts: 124
Experimental findings

EXPERIMENT MEASUREMENTS https://www.youtube.com/watch?v=RoM5pEy7n9Q Acceleration measurement. I did various micro-scale experiments with an essay scale 1 to 7, mass 900kg with double grid reinforcement 5X5 cm Φ / 1.5mm, with micro-scale concrete material. I used Sand with cement in a ratio of 6 to 1 Oscillation width 0.15m Displacement 0.30m Full oscillation 0.60m Frequency 2 Hz Acceleration to (g) a = (- (2 * π * 2) ^ 2 * 0,15) / 9.81 a = 3.14x2 = 6.28x2 = 12.56X12.56 = 157.754X0.15 = 23.6631 / 9.81 = 2.41g of natural earthquake. Two floors 900 kg 450 kg each Inertia and shear base measurement Acceleration 2.41g X 900kg = inertia and base cutting 2169 kg Each of the 4 walls has a cutting base 2169kg / 4 = cutting. base 542.24kg Rollover torque measurement To find the tipping moment of each wall we must first find the lateral force received by each wall per floor. So we say 2169 kg / 2 = 1084.5 kg of inertia per floor / 4 walls = 271.2 kg of inertia force is received by each wall per floor. To find the overturning moment of the height of the walls we add all the heights of the floors, ie (0.67 + 1.34) = 2.01m and we multiply them by the kg of inertia that each floor wall receives per floor which is 271.2kg The result is 2.01X271.02 = Rolling torque for each wall 545.1kg If we want to create a reciprocal moment of stability so that the wall does not tip over, it must be greater than> 545.1kg Each wall gets a tipping torque of 545.1kg and the 4 walls together get a torque of 2180 kg The weight of the construction in immobility is 900 kg That is, a two-storey specimen weighing 900 kg with an acceleration of 2.41g creates a tipping torque of 2180/900 = 2.42 times its weight.
Here in this video you can see what happens when we disconnect the construction ground connection.