A Method of Early Warning of Engineering Structure Failure
An engineering structure and finite element technology, applied in the direction of instrumentation, geometric CAD, calculation, etc., can solve problems such as failure warning of engineering structures, inapplicability of engineering structures, inability to apply engineering structures, etc., to improve work safety and stability, Effects of accident prevention and accurate failure prediction
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Embodiment 1
[0033] Embodiment 1: as Figure 1-2 Shown, a kind of engineering structure failure early warning method, the concrete steps of described engineering structure failure early warning method are as follows:
[0034] Step 1. According to the engineering structure design specification, obtain the design parameter Young's modulus E of the engineering structure 0 , and the design safety factor of the engineering structure [S];
[0035] Step 2, let the initial detection time node i=1;
[0036] Step 3. Calculate the corresponding average Young's modulus E of the engineering structure at the current time node i through finite element software i , where i=1, 2, 3...;
[0037] Step 4. Judgment E i Is it less than or equal to E 0 / [S];
[0038] If E i less than or equal to E 0 / [S], then give early warning to the engineering structure, otherwise according to E n Scattered distribution of (n=1,2,3...i), the function E=f(i) is fitted to predict the average Young's modulus E at the...
Embodiment 2
[0054] Embodiment 2: as Figure 1-2 Shown, a kind of engineering structure failure early warning method, the concrete steps of described engineering structure failure early warning method are as follows:
[0055] Step 1. According to the engineering structure design specification, obtain the design parameter Young's modulus E of the engineering structure 0 =10 10 Pa, and the design safety factor of the engineering structure [S] = 1.25;
[0056] Step 2, let the detection time node i=10;
[0057] Step 3. Calculate and obtain the corresponding average Young's modulus E of the engineering structure at the current time node i=10 through finite element software 10 ,E 10 =8.21×10 9 Pa;
[0058] Step 4. Judgment E 10 Is it less than or equal to E 0 / [S], where E 0 / [S]=8×10 9 Pa;
[0059] E. 10 greater than E 0 / [S], then according to E n (n=1,2,3...10) scattered point distribution, fitting the function E=E 0 -0.2×(i-1), predict the average Young's modulus E at the ...
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