Stopa fundamentowa – wypadkowej obciążenia schemat 1

{N_{r1}} = {\rm{ }}{P_{r1}} + {G_r} = {\rm{ }}745 + 191,42{\rm{ }} = {\rm{ }}936,42{\rm{ }}kN

{{M_{ry}}^I = {\rm{ }}{M_y}^I - {H_x}^I \times h{\rm{ }} = {\rm{ }}250,00 + 30,00 \times 0,80{\rm{ }} = {\rm{ }}274,00{\rm{ }}kNm}\\

{{M_{rx}}^I = {\rm{ }}{M_x}^I + {H_y}^I \times h{\rm{ }} = {\rm{ }}0{\rm{ }}kNm}

{{e_L}^I = {\rm{ }}{e_{rx}}^I = \left( {\frac{{{M_{ry}}^I}}{{{N_r}^I}}} \right){\rm{ }} = \left( {\frac{{274,00}}{{936,42}}} \right){\rm{ }} = {\rm{ }}0,293{\rm{ }}m{\rm{ }} < {\rm{ }}\left( {\frac{L}{6}} \right){\rm{ }} = {\rm{ }}\left( {\frac{{2,55}}{6}} \right){\rm{ }} = 0,425{\rm{ }}m}\\

{{e_B}^I = {\rm{ }}{e_y} = \left( {\frac{{{M_{rx}}^I}}{{{N_r}^I}}} \right){\rm{ }} = {\rm{ }}0,00{\rm{ }}m}\\

\frac{{{q_{max}}}}{{{q_{min}}}} = {\rm{ }}2 \Leftrightarrow \left( {1 + \frac{{6{e_L}}}{L}} \right) \times {\left( {1 - \frac{{6{e_L}}}{L}} \right)^{ - 1}} = {\rm{ }}2\; =  > {\rm{ }}{e_L} \approx 0,15{\rm{ }}m

{{e_{xs}}^I = {\rm{ }}0,2m{\rm{ }} =  > {\rm{ }}{M_{ry}}^I = {\rm{ }}{M_y}^I - {H_x}^I \times h - {P_r}^I \times {e_{xs}} }\\

{{e_{xs}}^I = {\rm{ }}250 + 30 \times 0,8 - 745 \times 0,2{\rm{ }} = {\rm{ }}125{\rm{ }}kNm}

{{e_L}^I = {\rm{ }}\frac{{{M_{ry}}^I}}{{{N_r}^I}} = {\rm{ }}\frac{{125,00}}{{936,42}}{\rm{ }} = {\rm{ }}0,133{\rm{ }}m{\rm{ }} =  > {\rm{ }}\frac{{{q_{max}}}}{{{q_{min}}}} = {\rm{ }}1,91}