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The Amazon River basin in Brazil has a width of \(400 \mathrm{~km}\). Assuming that the basin is caused by a line load at its center and that the elastic lithosphere is not broken, determine the corresponding thickness of the elastic lithosphere. Assume \(E=70\) \(\mathrm{GPa}, v=0.25,\) and \(\rho_{m}-\rho_{s}=700 \mathrm{~kg} \mathrm{~m}^{-3}\)

Short Answer

Expert verified
The lithosphere thickness is found using the flexural rigidity equation with the width of the basin.

Step by step solution

01

Understand the Problem

The problem asks us to find the thickness of the elastic lithosphere under a line load at the center of a basin. We are given the width of the basin, the Young's modulus \(E\), Poisson's ratio \(v\), and the density difference between the mantle and the sediment \(\rho_m - \rho_s\).
02

Identify the Key Equation

We need to use the flexural rigidity equation for a lithospheric plate: \( D = \frac{E h^3}{12(1-v^2)} \), where \(D\) is the flexural rigidity and \(h\) is the thickness of the lithosphere.
03

Find the Relationship Between Flexural Rigidity and Basin Width

For a line load, the characteristic wavelength \(\lambda\) is related to flexural rigidity \(D\) and load parameters: \( \lambda = 4\pi \sqrt{\frac{D}{{(\rho_m - \rho_s) g}}} \), where \(g = 9.81 \mathrm{~m/s^2}\) is the acceleration due to gravity. The given basin width is \(400 \mathrm{~km}\), so the half-width \(L = 200 \mathrm{~km} = 200,000 \mathrm{~m}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Flexural Rigidity
Flexural rigidity is an important concept in geology and geophysics. It describes a lithospheric plate's ability to resist bending or flexing under a load. This rigidity is quantified by the parameter \( D \), and it is a function of the thickness, elastic properties, and geometric configuration of the plate.

In mathematical terms, flexural rigidity is expressed as:\[D = \frac{E h^3}{12(1-v^2)}\]Where:
  • \( E \) is the Young's modulus, representing the elasticity of the material.

  • \( h \) is the thickness of the lithosphere.

  • \( v \) is Poisson's ratio, a measure of the material's ability to undergo volume changes.
This equation shows how thickness impacts rigidity dramatically; the cubic power implies that even small changes in thickness can greatly affect rigidity.
Understanding flexural rigidity helps us grasp how various forces, like those from tectonic activities, are distributed and balanced across the lithosphere, influencing surface structures and depressions such as basins.
Line Load
The concept of a line load refers to a force applied along a linear path, which is especially relevant in geological scenarios like the formation of basins due to tectonic or surface loads. This type of load can create depressions or bending in the Earth's lithosphere over time.

In the case of the Amazon River basin, the line load at the center affects the surrounding lithosphere, leading to the characteristic geological feature. The width of this bending or basin, noted to be \( 400 \) km, provides insight into how the lithosphere responds to such loads.
The behavior of the lithosphere under line loads can be predicted using the expression for the characteristic wavelength \( \lambda \), which is connected to the flexural rigidity \( D \) and other geological parameters:
\[\lambda = 4\pi \sqrt{\frac{D}{{\rho_m - \rho_s}} g}\]This equation shows how line loads result in particular wavelengths or widths of bending in the lithosphere, depending on its flexural rigidity and the density difference between layers.
Density Difference
Density difference is a crucial factor when studying lithospheric bending under loads like those forming basins. It refers to the disparity between the densities of the mantle (\( \rho_m \)) and the sediment (\( \rho_s \)) layers.

This difference (\( \rho_m - \rho_s \)), influences how deep or prominently the lithosphere can bend or flex under applied pressures such as line loads. A higher density difference generally corresponds to increased gravitational forces acting on the lithosphere, which dictates the degree of bending.

In practical scenarios, the density difference helps determine the characteristic wavelength \( \lambda \), by contributing to the gravitational component in the equation:\[\lambda = 4\pi \sqrt{\frac{D}{{\rho_m - \rho_s}} g}\]This underlines the significant interplay between density differences, gravitational forces, and lithospheric rigidity—in shaping observable geological structures like basins and troughs.

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Most popular questions from this chapter

Plane Strain In the case of plane strain, \(\varepsilon_{3}=0,\) for example, and \(\varepsilon_{1}\) and \(\varepsilon_{2}\) are nonzero. Figure \(3-7\) illustrates a plane strain situation. A long bar is rigidly confined between supports so that it cannot expand or contract parallel to its length. In addition, the stresses \(\sigma_{1}\) and \(\sigma_{2}\) are applied uniformly along the length of the bar. Equations \((3-1)\) to \((3-3)\) reduce to $$ \begin{aligned} \sigma_{1} &=(\lambda+2 G) \varepsilon_{1}+\lambda \varepsilon_{2} \\ \sigma_{2} &=\lambda \varepsilon_{1}+(\lambda+2 G) \varepsilon_{2} \\ \sigma_{3} &=\lambda\left(\varepsilon_{1}+\varepsilon_{2}\right) \end{aligned} $$. From Equation \((3-6)\) it is obvious that $$ \sigma_{3}=v\left(\sigma_{1}+\sigma_{2}\right) $$ This can be used together with Equations \((3-4)\) and (3-5) to find $$ \begin{array}{l} \varepsilon_{1}=\frac{(1+v)}{E}\left\\{\sigma_{1}(1-v)-v \sigma_{2}\right\\} \\\ \varepsilon_{2}=\frac{(1+v)}{E}\left\\{\sigma_{2}(1-v)-v \sigma_{1}\right\\} \end{array} $$

Find the deflection of a uniformly loaded beam pinned at the ends, \(x=0, L\). Where is the maximum bending moment? What is the maximum bending stress?

If all the principal stresses are equal \(\sigma_{1}=\sigma_{2}=\sigma_{3} \equiv p\), then the state of stress is isotropic, and the principal stresses are equal to the pressure. The principal strains in a solid subjected to isotropic stresses are also equal \(\varepsilon_{1}=\varepsilon_{2}=\varepsilon_{3}=\frac{1}{3} \Delta ;\) each component of strain is equal to one-third of the dilatation. By adding Equations \((3-1)\) to \((3-3),\) we find $$ p=\left(\frac{3 \lambda+2 G}{3}\right) \Delta \equiv K \Delta \equiv \frac{1}{\beta} \Delta . $$ The quantity \(K\) is the bulk modulus, and its reciprocal is \(\beta,\) the compressibility. The ratio of \(p\) to the bulk modulus gives the fractional volume change that occurs under isotropic compression. Because the mass of a solid element with volume \(V\) and density \(\rho\) must be conserved, any change in volume \(\delta V\) of the element must be accompanied by a change in its density \(\delta \rho .\) The fractional change in density can be related to the fractional change in volume, the dilatation, by rearranging the equation of mass conservation $$ \delta(\rho V)=0 $$ which gives $$ \rho \delta V+V \delta \rho=0 $$ or $$ \frac{-\delta V}{V}=\Delta=\frac{\delta \rho}{\rho} $$ Equation \((3-53)\) of course assumes \(\Delta\) to be small. The combination of Equations \((3-50)\) and \((3-53)\) gives $$ \delta \rho=\rho \beta p $$ This relationship can be used to determine the increase in density with depth in the earth. Using Equations \((3-11)\) to \((3-13),\) we can rewrite the formula for \(K\) given in Equation \((3-50)\) as $$ K=\frac{1}{\beta}=\frac{E}{3(1-2 v)} $$ Thus as \(v\) tends toward \(1 / 2,\) that is, as a material becomes more and more incompressible, its bulk modulus tends to infinity.

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