Effect of the movement across a surface breaking, inclined, locked, finite strike-slip fault in visco-elastic medium of Burger’s rheology

Abstract

Seismically active regions are associated with fault systems, and the movement of these faults influences the nature of stress accumulation/release in that region. For in-depth analysis of the effect of such movement, a mathematical model has been assumed by considering a surface breaking, inclined, finite strike-slip fault in visco-elastic medium of Burger’s rheology and considered the deformation caused in the medium. Analytical expressions for displacements, stresses and strains have been obtained by using Green’s function technique and Laplace transform. The variations of displacements, stresses and strains due to locked fault have been studied. These variations are graphically depicted for different inclinations of the fault with the free surface and for different slip magnitude using MATLAB. Such studies may throw some light on the nature of stress and strain accumulation/ release due to fault movement in Burger medium.

Appendices

Appendix 1.

Before fault movement

To solve the boundary value problem, it is customary to assume that

(16)

Taking Laplace transform on all constitutive equations, boundary conditions and then using Eq. (16) (following Kundu et al. (2021a), Kundu et al. 2021b), it is found that

$$ {\displaystyle \begin{array}{c}{A}_1=\frac{\tau_L}{2\ {s}^2\ \left({q}_1+{q}_2s\right)}\\ {}{B}_1={A}_2=\frac{1+{p}_1+{p}_2{s}^2}{2s\left({q}_1+{q}_2s\right)}\left[{\tau}_{\infty }(s)-\frac{\left({p}_1+{p}_2s\right){\tau}_{\infty }(0)}{1+{p}_1+{p}_2{s}^2}\right]\end{array}} $$

and C₂ = B₂ = C₃ = B₃ = 0

Therefore, Eq. (16) becomes

$$ {\displaystyle \begin{array}{c}\overline{u_1}=\frac{{\left({u}_1\right)}_0}{s}+\frac{\tau_L{y}_1}{2{s}^2\left({q}_1+{q}_2s\right)}+\frac{1+{p}_1+{p}_2{s}^2}{2s\left({q}_1+{q}_2s\right)}\left[{\tau}_{\infty }(s)-\frac{\left({p}_1+{p}_2s\right){\tau}_{\infty }(0)}{1+{p}_1+{p}_2{s}^2}\right]{y}_2\\ {}\overline{u_2}=\frac{{\left({u}_2\right)}_0}{s}+\frac{1+{p}_1+{p}_2{s}^2}{2s\left({q}_1+{q}_2s\right)}\left[{\tau}_{\infty }(s)-\frac{\left({p}_1+{p}_2s\right){\tau}_{\infty }(0)}{1+{p}_1+{p}_2{s}^2}\right]{y}_1\\ {}\overline{u_3}=\frac{{\left({u}_3\right)}_0}{s}\end{array}} $$

whose inverse Laplace transform results in the following:

$$ {\displaystyle \begin{array}{c}{u}_1={\left({u}_1\right)}_0+\frac{\tau_L{y}_1}{2{q}_1}\left[t-\frac{q_2}{q_1}\left(1-{e}^{-\frac{q_1t}{q_2}}\right)\right]+\frac{\tau_{\infty }(0)}{2{q}_1}\left[t-\frac{q_2}{q_1}\left(1-{e}^{-\frac{q_1t}{q_2}}\right)+k\left\{\frac{t^2}{2}+\frac{q_2}{q_1}\left(\frac{q_2}{q_1}\left(1-{e}^{-\frac{q_1t}{q_2}}\right)-t\right)+{p}_1\left(t-\frac{q_2}{q_1}\left(1-{e}^{-\frac{q_1t}{q_2}}\right)\right)+{p}_2\left(1-{e}^{-\frac{q_1t}{q_2}}\right)\right\}\right]{y}_2\\ {}\begin{array}{c}{u}_2={\left({u}_2\right)}_0+\frac{\tau_{\infty }(0)}{2{q}_1}\left[t-\frac{q_2}{q_1}\left(1-{e}^{-\frac{q_1t}{q_2}}\right)+k\left\{\frac{t^2}{2}+\frac{q_2}{q_1}\left(\frac{q_2}{q_1}\left(1-{e}^{-\frac{q_1t}{q_2}}\right)-t\right)+{p}_1\left(t-\frac{q_2}{q_1}\left(1-{e}^{-\frac{q_1t}{q_2}}\right)\right)+{p}_2\left(1-{e}^{-\frac{q_1t}{q_2}}\right)\right\}\right]{y}_1\\ {}{u}_3={\left({u}_3\right)}_0\end{array}\end{array}} $$

Therefore,

$$ {\displaystyle \begin{array}{c}{\tau}_{11}=\frac{{\left({\tau}_{11}\right)}_0}{A}\left[\left({p}_1-{p}_2{r}_1\right){e}^{-{r}_1t}-\left({p}_1-{p}_2{r}_2\right){e}^{-{r}_2t}\right]+\left[{\tau}_L-\frac{\tau_L}{A}\left\{\left({p}_1-{p}_2{r}_1\right){e}^{-{r}_1t}-\left({p}_1-{p}_2{r}_2\right){e}^{-{r}_2t}\right\}\right]\\ {}{\tau}_{12}=\frac{{\left({\tau}_{12}\right)}_0}{A}\left[\left({p}_1-{p}_2{r}_1\right){e}^{-{r}_1t}-\left({p}_1-{p}_2{r}_2\right){e}^{-{r}_2t}\right]+\left[{\tau}_{\infty }(t)-\frac{{\left({\tau}_{\infty}\right)}_0}{A}\left\{\left({p}_1-{p}_2{r}_1\right){e}^{-{r}_1t}-\left({p}_1-{p}_2{r}_2\right){e}^{-{r}_2t}\right\}\right]\\ {}\begin{array}{c}{\tau}_{13}=\frac{{\left({\tau}_{13}\right)}_0}{A}\left[\left({p}_1-{p}_2{r}_1\right){e}^{-{r}_1t}-\left({p}_1-{p}_2{r}_2\right){e}^{-{r}_2t}\right]\\ {}{\tau}_{23}=\frac{{\left({\tau}_{23}\right)}_0}{A}\left[\left({p}_1-{p}_2{r}_1\right){e}^{-{r}_1t}-\left({p}_1-{p}_2{r}_2\right){e}^{-{r}_2t}\right]\\ {}\begin{array}{c}{\tau}_{22}=\frac{{\left({\tau}_{22}\right)}_0}{A}\left[\left({p}_1-{p}_2{r}_1\right){e}^{-{r}_1t}-\left({p}_1-{p}_2{r}_2\right){e}^{-{r}_2t}\right]\\ {}\begin{array}{c}{\tau}_{33}=\frac{{\left({\tau}_{33}\right)}_0}{A}\left[\left({p}_1-{p}_2{r}_1\right){e}^{-{r}_1t}-\left({p}_1-{p}_2{r}_2\right){e}^{-{r}_2t}\right]\\ {}{e}_{11}=\frac{1}{2}{\left({e}_{11}\right)}_0\\ {}\begin{array}{c}{e}_{12}=\frac{1}{2}{\left({e}_{12}\right)}_0+\frac{\tau_{\infty }(0)}{4{q}_1}\left[t-\frac{q_2}{q_1}\left(1-{e}^{-\frac{q_1t}{q_2}}\right)+k\left\{\frac{t^2}{2}+\frac{q_2}{q_1}\left(\frac{q_2}{q_1}\left(1-{e}^{-\frac{q_1t}{q_2}}\right)-t\right)+{p}_1\left(t-\frac{q_2}{q_1}\left(1-{e}^{-\frac{q_1t}{q_2}}\right)\right)+{p}_2\left(1-{e}^{-\frac{q_1t}{q_2}}\right)\right\}\right]\\ {}{e}_{13}=\frac{1}{2}{\left({e}_{13}\right)}_0\end{array}\end{array}\end{array}\end{array}\end{array}} $$

Appendix 2.

After fault movement

After the seismic event when the fault becomes locked, the displacements, stresses and strains have been found in the form given in Eq. (13), where (u_i)₁, (τ_ij)₁, (e_ij)₁ (i, j = 1, 2, 3) are given in Appendix 1. (u_i)₂, (τ_ij)₂, (e_ij)₂ (i, j = 1, 2, 3) satisfy all the constitutive equations and boundary conditions. These boundary value problems involving (u_i)₂, (τ_ij)₂, (e_ij)₂ (i, j = 1, 2, 3) can be solved by using modify Green’s function technique developed by Maruyama (Kundu et al. 2021a; Kundu et al. 2021b) and Rybicki (Pal et al. 1979; Rosenman and Singh 1973) and the corresponding principle. According to them, it is found that

$$ \overline{{\left({u}_1\right)}_2}(Q)={\iint}_F\left[\overline{{\left({u}_1\right)}_2}(s)\right]G\left(P,Q\right)d{x}_3d{x}_1 $$

where Q(y₁, y₂, y₃) is field point in the half-space, P(x₁, x₂, x₃) is any point on the fault F, [(u₁)₂(p)] is the magnitude of discontinuity of u₁ across F and G is the Green’s function.

where $G\left(P,Q\right)=\frac{\partial }{\partial {x}_2}{G}_1\left(P,Q\right)$

$$ {G}_1\left(P,Q\right)=\frac{1}{{\left[{\left({y}_1-{x}_1\right)}^2+{\left({y}_2-{x}_2\right)}^2+{\left({y}_3-{x}_3\right)}^2\right]}^{\frac{1}{2}}}-\kern0.5em \frac{1}{{\left[{\left({y}_1+{x}_1\right)}^2+{\left({y}_2-{x}_2\right)}^2+{\left({y}_3-{x}_3\right)}^2\right]}^{\frac{1}{2}}} $$

Then $G\left(P,Q\right)==\frac{y_2-{x}_2}{{\left[{\left({y}_1-{x}_1\right)}^2+{\left({y}_2-{x}_2\right)}^2+{\left({y}_3-{x}_3\right)}^2\right]}^{\frac{3}{2}}}-\kern0.5em \frac{y_2-{x}_2}{{\left[{\left({y}_1+{x}_1\right)}^2+{\left({y}_2-{x}_2\right)}^2+{\left({y}_3-{x}_3\right)}^2\right]}^{\frac{3}{2}}}$

P(x₁, x₂, x₃) being a point on the fault F. Since the fault is inclined at an angle θ and depth is d from the free surface, therefore, 0 ≤ x₂ ≤ D cos θ, 0 ≤ x₃ ≤ D sin θ and x₂ = x₃ cot θ. A change in coordinate from (x₁, x₂, x₃) to $\left({x}_1^{\prime },{x}_2^{\prime },{x}_3^{\prime}\right)$ is considered which is connected by the relations:

$$ {x}_1={x}_1^{\prime },\kern1em {x}_2={x}_2^{\prime}\sin \theta +{x}_3^{\prime}\cos \theta, \kern0.5em {x}_3=-{x}_2^{\prime}\cos \theta {x}_3^{\prime}\sin \theta $$

From x₂ = x₃ cot θ, it is found that ${x}_2^{\prime }=0$

Then ${x}_1={x}_1^{\prime },\kern0.5em {x}_2={x}_3^{\prime}\cos \theta, \kern0.75em {x}_3={x}_3^{\prime}\sin \theta$

and $d{x}_1=d{x}_1^{\prime },\kern0.75em d{x}_2^{\prime }=0,\kern0.75em d{x}_3=\sin \theta d{x}_3^{\prime }$

$$ \overline{{\left({u}_1\right)}_2}(Q)={\iint}_F\left[\overline{{\left({u}_1\right)}_2}(s)\right]G\left(P,Q\right)d{x}_3d{x}_1 $$

Since the coordinate of the end points of the fault are taken w.r.t prime coordinate system as A (−L, 0, 0) and B (L, 0, 0) respectively then the limit of the integration of ${x}_1^{\prime }$is −L to L. Also w.r.t prime coordinate system the width of the fault is D which is in ${x}_3^{\prime }$ direction; then the limit of ${x}_3^{\prime }$ is taken as 0 to D.

Then $\overline{{\left({u}_1\right)}_2}\ (Q)$ = ${\int}_{-L}^L{\int}_0^D\frac{U_1}{s}f\left({x}_1^{\prime },{x}_3^{\prime}\right)G\left(P,Q\right)\sin \theta d{x}_3^{\prime }d{x}_1^{\prime }$

= U₁ /s ϕ(y₁, y₂, y₃)

Taking inverse Laplace transform we get,

${\left({u}_1\right)}_2(Q)=\frac{U_1}{2\pi}\phi \left({y}_1,{y}_2,{y}_3\right),$where t₁ = t − T₁ and

$$ \phi \left({y}_1,{y}_2,{y}_3\right)={\int}_{-L}^L{\int}_0^Df\left({x}_1^{\prime },{x}_3^{\prime}\right)G\left(P,Q\right)\sin \theta\ {d}^{\prime }{x}_3{d}^{\prime }{x}_1={\int}_{-L}^L{\int}_0^D\left\{\frac{y_2-{x}_3^{\prime}\cos \theta }{{\left[{\left({y}_1-{x}_1^{\prime}\right)}^2+{\left({y}_2-{x}_3^{\prime}\cos \theta \right)}^2+{\left({y}_3-{x}_3^{\prime}\sin \theta \right)}^2\right]}^{\frac{3}{2}}}-\frac{y_2-{x}_3^{\prime}\cos \theta }{{\left[{\left({y}_1+{x}_1^{\prime}\right)}^2+{\left({y}_2-{x}_3^{\prime}\cos \theta \right)}^2+{\left({y}_3-{x}_3^{\prime}\sin \theta \right)}^2\right]}^{\frac{3}{2}}}\right\}f\left({x}_1^{\prime },{x}_3^{\prime}\right)\sin \theta d{x}_3^{\prime }d{x}_1^{\prime } $$