egzamin nosnosc lawy


$${tg(45 + \frac{\varphi}{2}) = {\frac{H}{\frac{B}{2}}\ \ \rightarrow}\ H = {\frac{B}{2}tg(45 + {\frac{\varphi}{2}) = {\frac{B}{2}\sqrt{K_{p}}}}}}{\sigma_{\text{hp}} = \gamma_{D} \bullet D \bullet K_{p} + \gamma_{B} \bullet z \bullet K_{p}}{P_{p} = H \bullet \gamma_{D} \bullet D \bullet K_{p} + \frac{1}{2}\ \bullet \ \gamma_{B} \bullet H \bullet K_{p} = \frac{B}{2}K_{p}^{\frac{3}{2}} \bullet \gamma_{D} \bullet D + \frac{1}{2} \bullet K_{p}^{2} \bullet \frac{B^{2}}{4} \bullet \gamma_{B}}{\sigma_{\text{ha}} = q_{\text{gr}} \cdot K_{a} + \gamma_{B} \cdot z \cdot K_{a}}{P_{a} = H \cdot q_{\text{gr}} \cdot \frac{1}{K_{p}} + \frac{1}{2}H \cdot \gamma_{B} \cdot H \cdot \frac{1}{K_{p}} = q_{\text{gr}} - \frac{B}{2} \cdot \frac{\sqrt{K_{p}}}{K_{p}} + \ \frac{1}{8} \cdot B^{2} \cdot \gamma_{B}}{q_{\text{gr}} = \frac{B}{2} \cdot \frac{\sqrt{K_{p}}}{K_{p}} + \frac{1}{8} \cdot B^{2} \cdot \gamma_{B} \cdot \frac{B}{2} + \frac{1}{8}K_{p}^{2} \cdot B^{2} \cdot \gamma_{B}}{q_{\text{gr}} = \ \frac{2}{B}\sqrt{K_{p}}(K_{p}^{\frac{3}{2}} \cdot D \cdot \gamma_{D} \cdot \frac{B}{2} + \frac{1}{2} \cdot K_{p}^{2} \cdot B^{2} \cdot \gamma_{B} - \frac{1}{8} \cdot B^{2} \cdot \gamma_{B})}{q_{\text{gr}} = K_{p}^{2} \cdot D \cdot \gamma_{D} + \frac{1}{4}\left( K_{P}^{\frac{5}{2}} - K_{p}^{\frac{1}{2}} \right) \cdot B \cdot \gamma_{B}}{\mathbf{q}_{\mathbf{\text{gr}}}\mathbf{=}\mathbf{N}_{\mathbf{D}}\left( \mathbf{\varphi} \right)\mathbf{D \cdot}\mathbf{\rho}_{\mathbf{D}}\mathbf{\cdot g +}\mathbf{N}_{\mathbf{B}}\left( \mathbf{\varphi} \right)\mathbf{B \cdot}\mathbf{\rho}_{\mathbf{B}}\mathbf{\cdot g}}$$


$${tg(45 + \frac{\varphi}{2}) = {\frac{H}{\frac{B}{2}}\ \ \rightarrow}\ H = {\frac{B}{2}tg(45 + {\frac{\varphi}{2}) = {\frac{B}{2}\sqrt{K_{p}}}}}}{\sigma_{\text{hp}} = \gamma_{D} \bullet D \bullet K_{p} + \gamma_{B} \bullet z \bullet K_{p}}{P_{p} = H \bullet \gamma_{D} \bullet D \bullet K_{p} + \frac{1}{2}\ \bullet \ \gamma_{B} \bullet H \bullet K_{p} = \frac{B}{2}K_{p}^{\frac{3}{2}} \bullet \gamma_{D} \bullet D + \frac{1}{2} \bullet K_{p}^{2} \bullet \frac{B^{2}}{4} \bullet \gamma_{B}}{\sigma_{\text{ha}} = q_{\text{gr}} \cdot K_{a} + \gamma_{B} \cdot z \cdot K_{a}}{P_{a} = H \cdot q_{\text{gr}} \cdot \frac{1}{K_{p}} + \frac{1}{2}H \cdot \gamma_{B} \cdot H \cdot \frac{1}{K_{p}} = q_{\text{gr}} - \frac{B}{2} \cdot \frac{\sqrt{K_{p}}}{K_{p}} + \ \frac{1}{8} \cdot B^{2} \cdot \gamma_{B}}{q_{\text{gr}} = \frac{B}{2} \cdot \frac{\sqrt{K_{p}}}{K_{p}} + \frac{1}{8} \cdot B^{2} \cdot \gamma_{B} \cdot \frac{B}{2} + \frac{1}{8}K_{p}^{2} \cdot B^{2} \cdot \gamma_{B}}{q_{\text{gr}} = \ \frac{2}{B}\sqrt{K_{p}}(K_{p}^{\frac{3}{2}} \cdot D \cdot \gamma_{D} \cdot \frac{B}{2} + \frac{1}{2} \cdot K_{p}^{2} \cdot B^{2} \cdot \gamma_{B} - \frac{1}{8} \cdot B^{2} \cdot \gamma_{B})}{q_{\text{gr}} = K_{p}^{2} \cdot D \cdot \gamma_{D} + \frac{1}{4}\left( K_{P}^{\frac{5}{2}} - K_{p}^{\frac{1}{2}} \right) \cdot B \cdot \gamma_{B}}{\mathbf{q}_{\mathbf{\text{gr}}}\mathbf{=}\mathbf{N}_{\mathbf{D}}\left( \mathbf{\varphi} \right)\mathbf{D \cdot}\mathbf{\rho}_{\mathbf{D}}\mathbf{\cdot g +}\mathbf{N}_{\mathbf{B}}\left( \mathbf{\varphi} \right)\mathbf{B \cdot}\mathbf{\rho}_{\mathbf{B}}\mathbf{\cdot g}}$$


$${tg(45 + \frac{\varphi}{2}) = {\frac{H}{\frac{B}{2}}\ \ \rightarrow}\ H = {\frac{B}{2}tg(45 + {\frac{\varphi}{2}) = {\frac{B}{2}\sqrt{K_{p}}}}}}{\sigma_{\text{hp}} = \gamma_{D} \bullet D \bullet K_{p} + \gamma_{B} \bullet z \bullet K_{p}}{P_{p} = H \bullet \gamma_{D} \bullet D \bullet K_{p} + \frac{1}{2}\ \bullet \ \gamma_{B} \bullet H \bullet K_{p} = \frac{B}{2}K_{p}^{\frac{3}{2}} \bullet \gamma_{D} \bullet D + \frac{1}{2} \bullet K_{p}^{2} \bullet \frac{B^{2}}{4} \bullet \gamma_{B}}{\sigma_{\text{ha}} = q_{\text{gr}} \cdot K_{a} + \gamma_{B} \cdot z \cdot K_{a}}{P_{a} = H \cdot q_{\text{gr}} \cdot \frac{1}{K_{p}} + \frac{1}{2}H \cdot \gamma_{B} \cdot H \cdot \frac{1}{K_{p}} = q_{\text{gr}} - \frac{B}{2} \cdot \frac{\sqrt{K_{p}}}{K_{p}} + \ \frac{1}{8} \cdot B^{2} \cdot \gamma_{B}}{q_{\text{gr}} = \frac{B}{2} \cdot \frac{\sqrt{K_{p}}}{K_{p}} + \frac{1}{8} \cdot B^{2} \cdot \gamma_{B} \cdot \frac{B}{2} + \frac{1}{8}K_{p}^{2} \cdot B^{2} \cdot \gamma_{B}}{q_{\text{gr}} = \ \frac{2}{B}\sqrt{K_{p}}(K_{p}^{\frac{3}{2}} \cdot D \cdot \gamma_{D} \cdot \frac{B}{2} + \frac{1}{2} \cdot K_{p}^{2} \cdot B^{2} \cdot \gamma_{B} - \frac{1}{8} \cdot B^{2} \cdot \gamma_{B})}{q_{\text{gr}} = K_{p}^{2} \cdot D \cdot \gamma_{D} + \frac{1}{4}\left( K_{P}^{\frac{5}{2}} - K_{p}^{\frac{1}{2}} \right) \cdot B \cdot \gamma_{B}}{\mathbf{q}_{\mathbf{\text{gr}}}\mathbf{=}\mathbf{N}_{\mathbf{D}}\left( \mathbf{\varphi} \right)\mathbf{D \cdot}\mathbf{\rho}_{\mathbf{D}}\mathbf{\cdot g +}\mathbf{N}_{\mathbf{B}}\left( \mathbf{\varphi} \right)\mathbf{B \cdot}\mathbf{\rho}_{\mathbf{B}}\mathbf{\cdot g}}$$


$${tg(45 + \frac{\varphi}{2}) = {\frac{H}{\frac{B}{2}}\ \ \rightarrow}\ H = {\frac{B}{2}tg(45 + {\frac{\varphi}{2}) = {\frac{B}{2}\sqrt{K_{p}}}}}}{\sigma_{\text{hp}} = \gamma_{D} \bullet D \bullet K_{p} + \gamma_{B} \bullet z \bullet K_{p}}{P_{p} = H \bullet \gamma_{D} \bullet D \bullet K_{p} + \frac{1}{2}\ \bullet \ \gamma_{B} \bullet H \bullet K_{p} = \frac{B}{2}K_{p}^{\frac{3}{2}} \bullet \gamma_{D} \bullet D + \frac{1}{2} \bullet K_{p}^{2} \bullet \frac{B^{2}}{4} \bullet \gamma_{B}}{\sigma_{\text{ha}} = q_{\text{gr}} \cdot K_{a} + \gamma_{B} \cdot z \cdot K_{a}}{P_{a} = H \cdot q_{\text{gr}} \cdot \frac{1}{K_{p}} + \frac{1}{2}H \cdot \gamma_{B} \cdot H \cdot \frac{1}{K_{p}} = q_{\text{gr}} - \frac{B}{2} \cdot \frac{\sqrt{K_{p}}}{K_{p}} + \ \frac{1}{8} \cdot B^{2} \cdot \gamma_{B}}{q_{\text{gr}} = \frac{B}{2} \cdot \frac{\sqrt{K_{p}}}{K_{p}} + \frac{1}{8} \cdot B^{2} \cdot \gamma_{B} \cdot \frac{B}{2} + \frac{1}{8}K_{p}^{2} \cdot B^{2} \cdot \gamma_{B}}{q_{\text{gr}} = \ \frac{2}{B}\sqrt{K_{p}}(K_{p}^{\frac{3}{2}} \cdot D \cdot \gamma_{D} \cdot \frac{B}{2} + \frac{1}{2} \cdot K_{p}^{2} \cdot B^{2} \cdot \gamma_{B} - \frac{1}{8} \cdot B^{2} \cdot \gamma_{B})}{q_{\text{gr}} = K_{p}^{2} \cdot D \cdot \gamma_{D} + \frac{1}{4}\left( K_{P}^{\frac{5}{2}} - K_{p}^{\frac{1}{2}} \right) \cdot B \cdot \gamma_{B}}{\mathbf{q}_{\mathbf{\text{gr}}}\mathbf{=}\mathbf{N}_{\mathbf{D}}\left( \mathbf{\varphi} \right)\mathbf{D \cdot}\mathbf{\rho}_{\mathbf{D}}\mathbf{\cdot g +}\mathbf{N}_{\mathbf{B}}\left( \mathbf{\varphi} \right)\mathbf{B \cdot}\mathbf{\rho}_{\mathbf{B}}\mathbf{\cdot g}}$$


$${tg(45 + \frac{\varphi}{2}) = {\frac{H}{\frac{B}{2}}\ \ \rightarrow}\ H = {\frac{B}{2}tg(45 + {\frac{\varphi}{2}) = {\frac{B}{2}\sqrt{K_{p}}}}}}{\sigma_{\text{hp}} = \gamma_{D} \bullet D \bullet K_{p} + \gamma_{B} \bullet z \bullet K_{p}}{P_{p} = H \bullet \gamma_{D} \bullet D \bullet K_{p} + \frac{1}{2}\ \bullet \ \gamma_{B} \bullet H \bullet K_{p} = \frac{B}{2}K_{p}^{\frac{3}{2}} \bullet \gamma_{D} \bullet D + \frac{1}{2} \bullet K_{p}^{2} \bullet \frac{B^{2}}{4} \bullet \gamma_{B}}{\sigma_{\text{ha}} = q_{\text{gr}} \cdot K_{a} + \gamma_{B} \cdot z \cdot K_{a}}{P_{a} = H \cdot q_{\text{gr}} \cdot \frac{1}{K_{p}} + \frac{1}{2}H \cdot \gamma_{B} \cdot H \cdot \frac{1}{K_{p}} = q_{\text{gr}} - \frac{B}{2} \cdot \frac{\sqrt{K_{p}}}{K_{p}} + \ \frac{1}{8} \cdot B^{2} \cdot \gamma_{B}}{q_{\text{gr}} = \frac{B}{2} \cdot \frac{\sqrt{K_{p}}}{K_{p}} + \frac{1}{8} \cdot B^{2} \cdot \gamma_{B} \cdot \frac{B}{2} + \frac{1}{8}K_{p}^{2} \cdot B^{2} \cdot \gamma_{B}}{q_{\text{gr}} = \ \frac{2}{B}\sqrt{K_{p}}(K_{p}^{\frac{3}{2}} \cdot D \cdot \gamma_{D} \cdot \frac{B}{2} + \frac{1}{2} \cdot K_{p}^{2} \cdot B^{2} \cdot \gamma_{B} - \frac{1}{8} \cdot B^{2} \cdot \gamma_{B})}{q_{\text{gr}} = K_{p}^{2} \cdot D \cdot \gamma_{D} + \frac{1}{4}\left( K_{P}^{\frac{5}{2}} - K_{p}^{\frac{1}{2}} \right) \cdot B \cdot \gamma_{B}}{\mathbf{q}_{\mathbf{\text{gr}}}\mathbf{=}\mathbf{N}_{\mathbf{D}}\left( \mathbf{\varphi} \right)\mathbf{D \cdot}\mathbf{\rho}_{\mathbf{D}}\mathbf{\cdot g +}\mathbf{N}_{\mathbf{B}}\left( \mathbf{\varphi} \right)\mathbf{B \cdot}\mathbf{\rho}_{\mathbf{B}}\mathbf{\cdot g}}$$


Wyszukiwarka

Podobne podstrony:
SPRAWDZENIE NOŚNOŚCI ŁAWY FUNDAMENTOWEJ W POZIOMIE POSADOWIENIA
Nośność i odkształenie, Materiały na egzamin mechanika gruntów
Egzamin zaoczne
Pytania egzaminacyjneIM
ANALIZA WYNIKÓW EGZAMINU GIMNAZJALNEGO DLA UCZNIÓW KLAS III
zadania egzaminacyjne
Egzamin 2008 2009
Egzamin poprawkowy I 2009 2010
Egzamin II ze statystyki luty 2007
14 Nośnośc Graniczna Przekroju Poprzecznego
312[01] 01 122 Arkusz egzaminac Nieznany (2)
Egzamin praktyczny Zadanie Nr 4
konta egzaminacyjne id 246765 Nieznany
EGZAMIN PKM2 pytania2011
na co nalezy zwrocic uwage przygotowujac uczniow do nowego ustnego egzaminu maturalnego

więcej podobnych podstron