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Given data: y=18 \mathrm{kN} / \mathrm{m}^{3}, \phi=35^{\circ} \quad c=0\begin{array}{c}D_{f}=1.5 \mathrm{~m} ; B=1.5 \mathrm{~m} ; L=2 \mathrm{~m} \\C_{L}=0.364 \mathrm{~m} \quad e_{B}=0.3 \mathrm{~m} \\\phi=5^{\circ} ; \quad N_{C}=46.12 ; N_{q}=33.30 ; N_{V}=48.03\end{array}Step is tind nl and B^{\prime}\frac{e B}{B}=\frac{0.3}{1.5}=0.2 \frac{1}{6} ; \frac{e L}{L}=\frac{0.364}{2}=182>\frac{1}{6}A=\frac{1}{2} B_{x} L_{x} \quad Case \quad \frac{e L}{L} \sum \frac{1}{6} ; \frac{e_{B}}{B} \geq \frac{1}{6}\begin{array}{l}B_{x}=B\left(1.5-\frac{3 e_{B}}{B}\right)=1.5(1.5-3 \times 0.2) \\B x=1.35 \mathrm{~m} \\L_{x}=L\left(1.5-\frac{3 e L}{L}\right)=2(1.5-3 \times 0.182) \\L_{x}=1.91 \mathrm{~m}\end{array}Now, A l=\frac{1}{2} B_{x} l_{x}=\frac{1}{2} \times 1.35 \times 1.91=1.29 \mathrm{~m}^{2}B^{\prime}=\frac{A^{\prime}}{L}=\frac{1.29}{2}=0.645 \mathrm{~m}Qult =q A^{\prime} N q F q s F q d+0.5 \mathrm{~V} B^{\prime} A^{\prime} N_{y} F_{r s} F v dB^{\prime}=0.645 \quad L^{\prime}=L=2 mN q, N y= Bearing Capacity FactorsFqs, Fys = shape factorsFad, Fyd = depth factors\begin{array}{l}N_{q}=33.30 ; N_{y}=48.03 \\F_{q s}=1+\left(\frac{B^{\prime}}{U}\right) \tan 35^{\circ}=1+\frac{0.645}{2} \tan 35^{\circ} \\F_{q s}=1.23 \\F_{y s}=1-0.4 \frac{B^{\prime}}{L}=1-0.4 \times \frac{0.645}{2}=0.871 \\\frac{D f}{B}=\frac{1.5}{1.5}=1 \Rightarrow F_{q d}=F_{y d}=1\end{array}Now,\begin{array}{l} Q= 18 \times 1.5 \times 1.29 \times 33.30 \times 1.23 \times 1+0.5 \times \\18 \times 0.645 \times 1.29 \times 48.03 \times 0.871 \times 1 \\Q= 1426.6+313.27=1739.871 . \mathrm{kN} \\Q=1739.874 \mathrm{~N}\end{array}Gross utimate load is 1739.87 \mathrm{kN} ...