A seismic survey was carried out for a large project to determine the nature of the substrata. The results of the survey are given in Fig. Ex 9.12 in the form of a graph. Determine the depths of the strata.

Question

Accepted Answer

Two methods may be used

1. Use of Eq (9.35)

[latex]H_{1}=\frac{d_{1}}{2} \sqrt{\frac{V_{2}-V_{1}}{V_{2}+V_{1}}}[/latex] (9.35a)

[latex]H_{2}=0.85 H_{1}+\frac{d_{2}}{2} \sqrt{\frac{V_{3}-V_{2}}{V_{3}+V_{2}}}[/latex] (9.35b)

2. Use of Eqs (9.36) and (9.37)

[latex]H_{1}=\frac{t_{1} V_{1}}{2 \cos \alpha}[/latex] (9.36)

[latex]H_{2}=\frac{t_{2} V_{2}}{2 \cos \beta}[/latex] (9.37)

First we have to determine the velocities in each stratum (Fig. Ex. 9.12).

[latex]V_{1}=\frac{ ext { distance }}{A \bar{B}}=\frac{2.188}{12.75 imes 10^{-3}}=172 m / sec[/latex]

[latex]V_{2}=\frac{d_{2}}{A C_{2}-A B_{1}}=\frac{22.5}{(7.75-1.75) 5 imes 10^{-3}}=750 m / sec[/latex]

In the same way, the velocity in the third stata can be determined. The velocity obtained is [latex]V_{3}=2250[/latex] m/sec

Method 1

From Eq (9.35 a), the thickness [latex]H_{1}[/latex] of the top layer is

[latex]H_{1}=\frac{d_{1}}{2} \sqrt{\frac{V_{2}-V_{1}}{V_{2}+V_{1}}}=\frac{2.188}{2} \sqrt{\frac{750-172}{1000}}=0.83 m[/latex]

From Eq (9.35b) the thickness [latex]H_{2}[/latex] is

[latex]H_{2}=0.85 H_{1}+\frac{d_{2}}{2} \sqrt{\frac{V_{s}-V_{2}}{V_{s}+V_{2}}}[/latex]

[latex]\begin{aligned}H_{2} &=0.85 imes 0.83+\frac{22.5}{2} \sqrt{\frac{2250-750}{3000}} \&=0.71+7.955=8.67 m\end{aligned}[/latex]

Method 2

From Eq (9.36)

[latex]H_{1}=\frac{t_{1} V_{1}}{2 \cos \alpha}[/latex]

[latex]t_{1}=A B_{1}=1.75 imes 5 imes 10^{-3}[/latex] sec(Fig.Ex.9.12)

[latex]\alpha=\sin ^{-1} \frac{V_{1}}{V_{2}}=\sin ^{-1} \frac{172}{750}=13.26^{\circ}[/latex]

[latex]\cos \alpha=0.9733[/latex]

[latex]H_{1}=\frac{12.75 imes 10^{-3} imes 172}{2 imes 0.9737}=1.13 m[/latex]

From Eq (9.37)

[latex]H_{2}=\frac{t_{2} V_{2}}{2 \cos \beta}[/latex]

[latex]t_{2}=5 imes 5 imes 10^{-3} sec[/latex]

[latex]\beta=\sin ^{-1} \frac{750}{2250}=19.47^{\circ} ; \cos \beta=0.9428[/latex]

[latex]H_{2}=\frac{5 imes 5 imes 10^{-3} imes 750}{2 imes 0.9428}=9.94 m[/latex]

Question 9.12: A seismic survey was carried out for a large project to dete...

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