# Thread: I could save the world from earthquakes ... but .. no one listens

1. Would you like to discuss a new method of reducing displacements ?

The simplest description of the method I can do is that if we join the nodes of the highest level with foundation ground, it will withstand larger lateral overturning forces than another wall that simply rests on on the ground. If we stop the primary torque of the wall overturning by this method we have stopped the displacement of the structure. By controlling the displacement of the structure, you also control the failures .
I just did experiments and I measured them. I have my own theory on earthquake design. All this is under investigation.
Theory
By controlling the displacement of the structure, you also control the tensions ... responsible for the inelastic deformations of the structure
With the method of designing, clamping the top-level nodes with the ground, I hope to divert the lateral inertial stresses of the earthquake into more powerful areas of the structure than those currently driven. These strong areas have the ability to absorb these tensions (preventing and relieving the relative displacements (ie drifts) and thus the tension that develops throughout the vector is limited) and returning them to the soil from where they came by subtracting in this way, great tensions and failures over the load-bearing structure of the building while ensuring a stronger bearing capacity of the foundation soil. Tensioning between of the upper edge of the walls with the earth reduces the displacements responsible for all the stresses that develop on the structural carrier. The patent is stacked into the ground to draw from it a force that transfers it to the upper end of the wall in order to apply a counterbalance to the torque of the wall
Τhe patent achieves the following
1)The consolidation of the nodes of highest level of the walls with the ground, using the mechanism of the invention, deflects the upward tensions created by the wall overturning torque transporting them freely and directly from the roof into the ground and in this way stops the displacements responsible for all growing tensions on the body of the bearing elements which they cause inelastic bending deformations and failures in a major earthquake. 2 ) Also the mechanism and method of anchoring provides very strong foundation in soft soils 3)The wall receives only compressive stresses at both ends a) at the upper end b) and the facing lower end near the base. Does not exist anymore tensile strength. This means that there are no longer torques in the nodes Does not exist anymore mechanism of concentric forces failure The floor mechanism (soft floor) does not exist anymore 4) Does not exist anymore coordination because the whole construction is shifted with the same frequency and the same oscillation amplitude 5) The wall also receives horizontal shear forces. Apply tension at all edges of the wall with the patent mechanism increases the ability to horizontal shear forces.

My own experiment. The model in this experiment From 2.45 minutes to 2.50 minutes, that is, within 5 seconds, made 20 journeys of 25 cm ... so in 20 seconds made 80 journeys with 25 cm of oscillation width. These oscillations from one end to the other measure, and their respective time in sec. Frequency (Hz) is the fraction: v = number of such paths / corresponding time. 80/20 = 4Hz ...9.81 is the acceleration of the earth and we divide it with the acceleration we found to find the g. That is, how many times the acceleration is accelerated by a body that falls on the earth. In a natural earthquake I did this experiment with a 0.25 cm oscillating amplitude and a frequency of 4 Hz we have an acceleration ... a = (- (2 * π * 4) ^ 2 * 0.22) / 9.81 3,14x2 = 6,28x4 = 25,12X25,12 = 631,0144X0,22 = 157,754 / 9,81 = 16 g acceleration
The specimen in the experiment had a general mass weighing 880 kg. The second floor because of the inverted beam it carries is more pounds than half so I would say it is about 450kg and the ground floor is 430kg So to find the inertia force F first on the ground floor we say ... F = m.a 430 Χ 157,754 = 67834,22 Newton or 68 kN. and the first floor 450 Χ 157,754 = 70989 Newton or 71 kN. Total force F (Inertia) 68 + 71 = 139 kN Moment of inertia Strength X Height ^ 2 Ground floor 68Χ0,67Χ0,67= 30,53 kN First floor 71Χ1,35Χ1,35 = 129,4 kN Total Inertia Torque 30,53+129,4 = 160 kN
More.. https://file.scirp.org/Html/6-1880388_59888.htm
Most popular papers in Open Journal of Civil Engineering https://www.scirp.org/journal/Hottes...?JournalID=788

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