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Changes to markdown file

Co-authored-by: Ben Spencer <benjamin.spencer@inl.gov>

doc/content/source/materials/DamagePlasticityStressUpdate.md

@@ -43,10 +43,8 @@ sections.
 The yield function, $\mathfrak{F}$ is a function of $\boldsymbol{\sigma}$, the strength of the material in uniaxial tension, $f_t$, and the strength of the material in uniaxial compression, $f_c$. It was used to describe the admissible stress space. For this implementation, the yield function in stress space is defined as follows
 \begin{equation} 
 \label{yf}
-\begin{gathered}
     \mathfrak{F}\left(\boldsymbol{\sigma},f_t,f_c\right) = \frac{1}{1-\alpha} 
     \left(\alpha I_1 + \sqrt{3J_2} + \beta\left(\boldsymbol{\kappa}\right)<{\hat{\boldsymbol{\sigma}}_{max}}>\right) - f_c\left(\boldsymbol{\kappa}\right)    
-\end{gathered}
 \end{equation}
 where $I_1$ and $J_2$ is first invariant of stress and second invariant of the deviatoric
 component of the stress, respectively, $<x> =\frac{x+|x|}{2}$ is the Macaulay bracket function, ${\hat{\boldsymbol{\sigma}}_{max}}$ is algebraically maximum principal
@@ -57,7 +55,7 @@ relates tensile, $f_t\left(\boldsymbol{\kappa}\right)$, and compressive, $f_c\le
 function of a vector of damage variable, $\boldsymbol{\kappa} = \{\kappa_t, \kappa_c\}$ and $\kappa_t$
 and $\kappa_c$ are the damage variables in tension and compression, respectively.
 
-The implementation first solves the given problem in the effective stress space and then transform the effective stress to stress space using [sigma_def2]. Thus, the yield strength of the concrete under uniaxial loading is expressed as effective yield strength as follows
+The implementation first solves the given problem in the effective stress space and then transforms the effective stress to stress space using [sigma_def2]. Thus, the yield strength of the concrete under uniaxial loading is expressed as effective yield strength as follows
 \begin{equation}
     \label{ft}
     f_t\left(\boldsymbol{\kappa}\right) = \left(1-D_t \left(\kappa_t\right)\right)f_{t}^{e}\left(\kappa_t\right)    
@@ -200,7 +198,7 @@ Thus, $a_t$ was obtained using [bt]-[slope] as follows
     \label{a_t}
      a_t = \sqrt{\frac{9}{4}+\frac{2\frac{G_t}{l_t} \left(\frac{d\sigma}{d\varepsilon^p}\right)_{\varepsilon^p=0}}{f_{t0}^2}} 
 \end{equation}
-To obtain a real value of $a_t$, the quantity inside the square root must be $\geq$ 0. Therefore, the minimum possible slope of the $\sigma$ versus $\varepsilon^p$ curve is
+To obtain a real value of $a_t$, the quantity inside the square root must be $\geq$ 0. Therefore, the minimum admissible slope of the $\sigma$ versus $\varepsilon^p$ curve is
 $\left(\left(\frac{d\sigma}{d\varepsilon^p}\right)_{\varepsilon^p=0}\right)_{min}=
 -\frac{9}{8}\frac{f_{t0}^2}{\frac{G_t}{l_t}}$, which is a function of the characteristic length in tension.
 Therefore, a mesh independent slope parameter $\omega\in\left(0,1\right)$, is defined such that
@@ -307,11 +305,11 @@ During the plastic corrector step, the returned effective stress should satisfy
         \mathfrak{F}\left(\boldsymbol{\sigma}^e,f_t^e,f_c^e\right) = 0
     \end{split}
 \end{equation}
-As per flow rule in [flowRule], the plastic corrector step, i.e., [plasticCorrector] can be rewritten as
+Per the flow rule in [flowRule], the plastic corrector step, i.e., [plasticCorrector] can be rewritten as
 \begin{equation}
 \boldsymbol{\sigma^e}_{n+1} = \boldsymbol{\sigma}_{n+1}^{e^{tr}}-\dot{\gamma}\left(2G\frac{\boldsymbol{s}_{n+1}^e}{\|\boldsymbol{s}_{n+1}^e\|} + 3K\alpha_p\boldsymbol{I}\right) \label{returnMap1}
 \end{equation}
-where $G$ is shear modulus and $K$ is bulk modulus. After separating the volumetric and deviatoric components from [returnMap1] following relations can be obtained
+where $G$ is the shear modulus and $K$ is the bulk modulus. After separating the volumetric and deviatoric components from [returnMap1] the following relations can be obtained
 \begin{equation}
     I_{1|n+1} = I_{1|n+1}^{e^{tr}} - 9K\alpha \alpha_p \dot{\gamma} \label{stressRelation1}
 \end{equation}