To determine the Lame parameters $\lambda$ and $\mu$ for the Poisson solid, we can use the relationship between the elastic properties of the material and its velocity, density, and Poisson's ratio.

Given:

• P-wave velocity $V_P = 5.5 \, \text{km/s} = 5500 \, \text{m/s}$
• Density $\rho = 2.6 \, \text{Mg/m}^3 = 2600 \, \text{kg/m}^3$
• The material is a Poisson solid, so it obeys the usual elastic relationships for isotropic media.

Step 1: Relating the P-wave velocity to $\lambda$ and $\mu$

For a Poisson solid, the P-wave velocity $V_P$ is related to the Lame parameters $\lambda$ and $\mu$ by the formula:

$V_P = \sqrt{\frac{\lambda + 2\mu}{\rho}}$

Rearranging this equation to solve for $\lambda + 2\mu$:

$\lambda + 2\mu = V_P^2 \cdot \rho$

Substitute the known values:

$\lambda + 2\mu = (5500)^2 \cdot 2600 = 30.25 \times 10^9 \, \text{m}^2/\text{s}^2 \cdot \text{kg/m}^3$

$\lambda + 2\mu = 78.65 \times 10^{12} \, \text{Pa}$

Step 2: Relating the S-wave velocity to $\lambda$ and $\mu$

The shear wave (S-wave) velocity $V_S$ is related to the shear modulus $\mu$ and density by:

$V_S = \sqrt{\frac{\mu}{\rho}}$

However, we don't have the S-wave velocity directly given. To proceed, we can use the Poisson's ratio ($\nu$) to estimate the value of $\mu$. Poisson's ratio for rocks like granite is typically around 0.25 (this is a typical estimate for granite).

Step 3: Estimating $\mu$ using Poisson's ratio

The relationship between $\mu$, $\lambda$, and Poisson’s ratio $\nu$ is:

$\mu = \frac{\lambda}{2(1 + \nu)}$

Now, using $\nu = 0.25$:

$\mu = \frac{\lambda}{2(1 + 0.25)} = \frac{\lambda}{2.5}$

Step 4: Substituting and solving

We already know:

$\lambda + 2\mu = 78.65 \times 10^{12} \, \text{Pa}$

Substitute $\mu = \frac{\lambda}{2.5}$ into this equation:

$\lambda + 2 \left(\frac{\lambda}{2.5}\right) = 78.65 \times 10^{12}$

Simplify:

$\lambda + \frac{2\lambda}{2.5} = 78.65 \times 10^{12}$

$\lambda + 0.8\lambda = 78.65 \times 10^{12}$

$1.8\lambda = 78.65 \times 10^{12}$

Solving for $\lambda$:

$\lambda = \frac{78.65 \times 10^{12}}{1.8} = 43.7 \times 10^{12} \, \text{Pa}$

Now, substitute this back to find $\mu$:

$\mu = \frac{43.7 \times 10^{12}}{2.5} = 17.48 \times 10^{12} \, \text{Pa}$

Final Answers:

• Lame parameter $\lambda$ = $43.7 \times 10^{12} \, \text{Pa}$
• Lame parameter $\mu$ = $17.48 \times 10^{12} \, \text{Pa}$

Further clarification:

Would you like to dive deeper into how these parameters are related to material behavior or how to calculate other properties of the granite, such as the S-wave velocity?

Additional questions for exploration:

1. How does the Poisson’s ratio affect the values of $\lambda$ and $\mu$?
2. What are the typical values of the S-wave velocity in granite?
3. What is the relationship between elastic modulus and Lame parameters?
4. How can we use these parameters in seismic studies or modeling?
5. How would the Lame parameters change if the material were a different type of rock?

Tip:

When estimating properties of materials like rocks, it's helpful to refer to standard values for Poisson's ratio and typical velocities, as they can provide a reasonable approximation even when specific data is missing.