23条数据的回归结果

Thesis

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一、初始刚度

$$k_0^{model}=\frac{0.22(1+6.82 \psi_{axis})}{(1-1.048 \rho_l+10.097 \rho_t)(1+\frac{3.68}{\phi_{asp}})}$$

或者

$$k_0^{model}=\frac{0.35(1+7.09 \psi_{axis})}{(1+15.10 \rho_t)(1+\frac{4.02}{\phi_{asp}})}$$

或者

$$k_0^{model}=\frac{0.40(1+19.11 \psi_{axis})}{(1+29.71 \rho_t)(1+\frac{13.92}{\phi_{asp}})}+0.18\rho_l$$

二、峰值割线刚度

$$k_u^{model}=\frac{0.204(1+6.80 \psi_{axis})}{(1+17.68 \rho_t)(1+\frac{2.62}{\phi_{asp}})}$$

或者

$$k_u^{model}=\frac{0.168(1+12.18 \psi_{axis})}{(1+28.48 \rho_t)(1+\frac{4.08}{\phi_{asp}})}+0.092\rho_l$$

三、峰值点恢复力

$$v_u^{model}=0.25 \rho_t +\frac{0.120 \sqrt{1+0.77 \psi_{axis}}}{\phi_{asp}}$$

四、开裂点恢复力

$$v_y^{model}=0.22 \rho_t +\frac{0.071 \sqrt{1+1.28 \psi_{axis}}}{\phi_{asp}}$$

五、下降段刚度$\gamma$

用刚度的预测公式进行预测

$$-\gamma k_0^{model} = \frac{0.0045\phi_{asp}(1+6.53 \psi_{axis})(1+6.16\rho_l)}{(1+9.12 \rho_t)}$$

6、卸载刚度和涅拢系数

卸载刚度系数取为$\theta = -0.51$，涅拢系数取为$\xi = 0.18$