Intro Doc
$\underline{Case\space 4-\space Anchored\space TERS\space in\space cohesionless\space soil}$ $\\$ $\\$ This application determines the the capacity of a TERS with a single anchor. This application will find the required section modulus bsed on a safety factor of one. The required depth will then be found based on a user determined factor of safety. $\\$ This application requires the following inputs: $\phi`$, Surcharge Load (psf), $\gamma$ (pcf), $\gamma$ saturated (pcf), height of dredge depth (ft), and anchor depth (ft). $\\$ First active and passive pressure coefficients are calculated. \begin{equation}\label{eq1} Ka=tan(45-\frac{\phi`}{2})^2 \end{equation} $\\$ \begin{equation}\label{eq2} Kp=tan(45+\frac{\phi`}{2})^2 \end{equation} $\\$ Next, active and passive pressures are determined. $\\$ \begin{equation}\label{eq3} \sigma\space a\space sur=Ka\cdot Surcharge \end{equation} $\\$ \begin{equation}\label{eq4} \sigma\space a\space aw=\sigma\space a\space sur+Ka\cdot height\cdot \gamma \end{equation} $\\$ \begin{equation}\label{eq5} \sigma\space a\space bw=\sigma\space a\space aw+Ka\cdot height\cdot \gamma\space sub \end{equation} $\\$ \begin{equation}\label{eq6} \sigma\space p\space sur=Kp\cdot depth\cdot \gamma\space sub \end{equation} $\\$ Where: $\\$ \begin{equation}\label{eq7} \gamma sub=\gamma sat-\gamma water \end{equation} $\\$ These values are then used to find the active forces at seven locations, and the passive force at a single location, along the TERS. The force due to water pressure is also found at this time. All lengths start at the anchor depth. La6 and La7 directed toward the surface of the soil and all the remaining lengths going deeper into the soil. $\\$ The locations of interest are as follows: $\\$ \begin{equation}\label{eq8} La1=\frac {1}{2}\cdot Hz+\frac {1}{2}\cdot D \end{equation} $\\$ \begin{equation}\label{eq9} La2=\frac {1}{2}\cdot Hz \end{equation} $\\$ \begin{equation}\label{eq10} La3=\frac{2}{3}\cdot Hz \end{equation} $\\$ \begin{equation}\label{eq11} La4=Hz+\frac {1}{2}\cdot D \end{equation} $\\$ \begin{equation}\label{eq12} La5=Hz+\frac{2}{3}\cdot D \end{equation} $\\$ \begin{equation}\label{eq13} La6=\frac {1}{2}\cdot anchor\space depth \end{equation} $\\$ \begin{equation}\label{eq14} La7=\frac{1}{3}\cdot D \end{equation} $\\$ \begin{equation}\label{eq15} Law=La5 \end{equation} $\\$ \begin{equation}\label{eq16} Lp1=Hz+\frac{2}{3}\cdot D \end{equation} $\\$ \begin{equation}\label{eq17} Lpw=Lp1 \end{equation} $\\$ Where: $\\$ \begin{equation}\label{eq18} Hz=height-anchor\space depth \end{equation} $\\$ The corresponding forces are: $\\$ \begin{equation}\label{eq19} Pa1=Ka\cdot Surcharge\cdot (Hz+D)\cdot \frac {1}{2}\cdot anchor\space depth \end{equation} $\\$ \begin{equation}\label{eq20} Pa2=Ka\cdot anchor\space depth\cdot \gamma\cdot Hz\cdot \frac {1}{2}\cdot anchor\space depth \end{equation} $\\$ \begin{equation}\label{eq21} Pa3=Ka\cdot \frac {1}{2}\cdot \gamma\cdot Hz^2\cdot \frac {1}{2}\cdot anchor\space depth \end{equation} $\\$ \begin{equation}\label{eq22} Pa4=Ka\cdot height\cdot \gamma\cdot D\cdot anchor\space depth\cdot \frac {1}{2} \end{equation} $\\$ \begin{equation}\label{eq23} Pa5=Ka\cdot \frac {1}{2}\cdot \gamma\space sub\cdot D^2\cdot \frac {1}{2}\cdot anchor\space depth \end{equation} $\\$ \begin{equation}\label{eq24} Pa6=Ka\cdot Surcharge\cdot anchor\space depth \end{equation} $\\$ \begin{equation}\label{eq25} Pa7=Ka\cdot \frac {1}{2}\cdot anchor\space depth^2\cdot \gamma \end{equation} $\\$ \begin{equation}\label{eq26} Paw=\frac {1}{2}\cdot \gamma\space w\cdot D^2 \end{equation} $\\$ \begin{equation}\label{eq27} Pp1=\frac {1}{2}\cdot Kp\cdot \gamma\space sub\cdot D^2\cdot \frac {1}{2}\cdot anchor\space depth \end{equation} $\\$ \begin{equation}\label{eq28} Ppw=\frac {1}{2}\cdot \gamma\space w\cdot D^2 \end{equation} $\\$ Using these values, the disturbing and restoring moments are found. $\\$ \begin{equation}\label{eq29} Mdist=Pa1\cdot La1+Pa2\cdot La2+Pa3\cdot La3+Pa4\cdot La4+Pa5\cdot La5+Paw\cdot Law \end{equation} $\\$ \begin{equation}\label{eq30} Mres=Pp1\cdot Lp1+Pa6\cdot La6+Pa7\cdot La7+Ppw\cdot Lpw \end{equation} $\\$ The required depth to be driven to below the dredge line is found by iteration until the safety factor met: \begin{equation}\label{eq31} \frac{Mres}{Mdist}=1 \end{equation} $\\$ The load in the anchor is then calculated: $\\$ \begin{equation}\label{eq32} T=Active-Passive \end{equation} $\\$ The total moment is calculated based on these forces. It is used to calculate a minimum section modulus of the piling. \begin{equation}\label{eq33} S=M/fs \end{equation} $\\$ Where fs is assumed to be 25000psi. $\\$ $\\$ The process of finding the locations and forces is then repeated to find a final depth to be driven to below the dredge line based on the user specified safety factor.