土の応力ーひずみ関係

これを使うと

$ \mathrm d\sigma_m'=K\mathrm d{\rm tr}\bm\varepsilon=0

$ \iff \mathrm d\sigma_m'=0

$ \therefore\mathrm du=\mathrm d(\sigma_m-\sigma_m')=\mathrm d\sigma_m

$ \mathrm du=B\mathrm d\sigma_m

$ \mathrm d{\rm tr}\bm\varepsilon=\frac1K\mathrm d\sigma_m'

$ {\rm tr}\bm\varepsilon\simeq\varepsilon_V=\frac{e_0-e}{1+e_0}

$ \mathrm d\varepsilon_V=-\frac{\mathrm de}{1+e_0}

$ V_a=(1-S_r)\frac{e}{1+e}V=:n_aV

$ \implies n_a=(1-S_r)\frac{e}{1+e}

$ \varepsilon_{aV}=\frac{V_{a0}-V_{a}}{V_{a0}}=\frac{n_{a0}-n_a}{n_{a0}}=\frac{e_0-e}{(1-S_{r0})e_0}=\frac{\varepsilon_V}{n_{a0}}

土全体の圧縮量は間隙空気の圧縮量と等しい

$ V_0-V=(e_0-e)V_s

$ V_{a0}-V_a=n_{a0}V_0-n_aV

$ =(n_{a0}(1+e_0)-n_a(1+e))V_s

$ =((1-S_{r0})e_0-(1-S_r)e)V_s

$ =(e_0-e)V_s+(S_{r0}e_0-S_re)V_s

$ =(e_0-e)V_s

$ \because \mathrm d(S_re)=0

$ \mathrm d\varepsilon_{aV}=-\frac{\mathrm dn_a}{n_{a0}}=-\frac{\mathrm de}{(1+e_0)n_{a0}}=\frac{\mathrm d\varepsilon_V}{n_{a0}}

$ \mathrm d\varepsilon_{aV}=\frac{\mathrm d\varepsilon_V}{n_{a0}}=K_a\mathrm du

$ K_a=-\frac{\mathrm dv}{v}\frac{1}{\mathrm du}=\frac 1u

$ \therefore \mathrm dv=-\frac vu\mathrm du

$ \therefore \mathrm d\varepsilon_V=\frac{n_{a0}}{u}\mathrm du

$ Kn_{a0}\mathrm du=u\mathrm d\sigma_m-u\mathrm du

$ \iff \mathrm du=\frac{u}{Kn_{a0}+u}\mathrm d\sigma_m

$ =\frac1{n_{a0}\frac{K}{u}+1}\mathrm d\sigma_m

$ =:B\mathrm d\sigma_m

$ B=\frac1{n_{a0}\frac{K}{u}+1}=\frac1{\frac{K}{u}(1-S_{r0})\frac{e_0}{1+e_0}+1}

$ \mathrm d{\rm tr}\bm\varepsilon=\frac1K\mathrm d\sigma_m'-3D\mathrm d\frac13(\sigma_1'-\sigma_3')

剪断応力に比例して体積膨張する

$ n_{a0}K\mathrm du=u(\mathrm d\sigma_m-\mathrm du)-uKD\mathrm d(\sigma_1-\sigma_3)

$ \iff (n_{a0}K+u)\mathrm du=\frac13u\mathrm d(\sigma_1+2\sigma_3)-uKD\mathrm d(\sigma_1-\sigma_3)

$ = u\mathrm d\sigma_3+\mathrm d\left(\frac13u(\sigma_1-\sigma_3)-uKD(\sigma_1-\sigma_3)\right)

$ \iff \mathrm du=B\mathrm d\sigma_3+B(1-3KD)\mathrm d\frac13(\sigma_1-\sigma_3)

$ =:B\mathrm d\sigma_3+3BA\mathrm d\frac13(\sigma_1-\sigma_3)

$ A=\frac13(1-3KD)

$ \mathrm du=B\mathrm d\sigma_m-BKD\mathrm d(\sigma_1-\sigma_3)

$ = B\mathrm d\sigma_m-B(1-3A)\mathrm d\frac13(\sigma_1-\sigma_3)

$ \mathrm de=-\frac{C_c}{\ln10}\frac{\mathrm d\sigma_m'}{\sigma_m'}-\mu(1+e_0)\mathrm d\left(\frac{\tau_{oct}}{\sigma_m'}\right)

$ \mathrm d e=-\frac{C_c}{\ln10}\frac{\mathrm d\sigma_m'}{\sigma_m'}+3D(1+e_0)\mathrm d\frac13(\sigma_1'-\sigma_3')

$ = -\frac{C_c}{\ln10}\frac{\mathrm d\sigma_m'}{\sigma_m'}+\frac{3}{\sqrt2}D(1+e_0)\mathrm d\tau_{oct}

$ \mathrm du= B\mathrm d\sigma_m-\frac{3}{\sqrt2}B(1-3A)\mathrm d\tau_{oct}

地盤中の有効応力がサンプリング後も変化しないと仮定する