Question

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.

Short answer

Isotropic stress implies equal principal stresses and strains. The bulk modulus relates to the material's incompressibility.

Step-by-step solution

Understanding Isotropic Stress

In an isotropic stress state, all principal stresses are equal: \( \sigma_1 = \sigma_2 = \sigma_3 = p \), where \( p \) is the pressure. Thus, the stress in all directions is the same, leading to an isotropic condition.

Analyzing Principal Strains

The principal strains \( \varepsilon_1, \varepsilon_2, \varepsilon_3 \) in a solid subjected to isotropic stresses are equal to each other and related to the dilation \( \Delta \) by \( \varepsilon_1 = \varepsilon_2 = \varepsilon_3 = \frac{1}{3} \Delta \). This shows each strain component as one third of the total dilatation.

Deriving the Bulk Modulus Expression

By summing equations for isotropic stress and strain, we find \( p = \left(\frac{3\lambda + 2G}{3}\right) \Delta = K \Delta = \frac{1}{\beta} \Delta \). Here, \( K \) is the bulk modulus and \( \beta \) is compressibility.

Volume and Density Conservation

Using mass conservation, \( \delta(\rho V) = 0 \) implies \( \rho \delta V + V \delta \rho = 0 \), which rearranges to \( \frac{-\delta V}{V} = \Delta = \frac{\delta \rho}{\rho} \). This equates fractional changes in volume and density.

Relation Between Density Change and Pressure

Equation \((3-50)\) and \((3-53)\) result in \( \delta \rho = \rho \beta p \), allowing us to compute the density change at depth under isotropic pressure.

Rewriting Bulk Modulus

Using elasticity equations \((3-11)\) to \((3-13)\), the bulk modulus is rewritten as \( K = \frac{1}{\beta} = \frac{E}{3(1-2u)} \). Here \( E \) is Young's modulus and \( u \) is Poisson's ratio.

Behavior of Bulk Modulus with Poisson's Ratio

As \( u \) approaches \( 1/2 \), indicating an incompressible material, \( K \) tends towards infinity, showing that incompressibility is related to an infinite bulk modulus.

Key concepts

Principal Stresses

In solid mechanics, principal stresses are the normal stresses acting on particular planes where shear stress is zero. These planes are perpendicular to each other, and the stresses acting on them are known as principal stresses. When all the principal stresses are equal, the state of stress is referred to as isotropic stress. This simply means that the material experiences the same amount of stress in all directions, and the value of each principal stress is equal to the pressure \( p \). Isotropic stress conditions are common in many practical situations such as fluid under pressure and materials subjected to uniform expansion or contraction due to temperature changes.

Bulk Modulus

The bulk modulus, denoted by \( K \), is a measure of a material's resistance to compression under isotropic stress conditions. It is the ratio of the infinitesimal change in pressure to the relative change in volume. In mathematical terms, it is defined as \( K = \frac{1}{\beta} \), where \( \beta \) is the material compressibility.

A higher bulk modulus means a material is less compressible, hence more resistant to volume changes under uniform pressure. The relationship can also be expressed in terms of Young's modulus \( E \) and Poisson's ratio \( v \) as \( K = \frac{E}{3(1-2v)} \). This indicates that as Poisson's ratio approaches \( 1/2 \), which corresponds to an incompressible material, the bulk modulus tends towards infinity.

Density Change

Density change in materials subjected to isotropic stress is linked to changes in volume. When a material's volume changes due to stress, its density must adjust since mass is conserved. The relationship is given by \( \frac{-\delta V}{V} = \Delta = \frac{\delta \rho}{\rho} \). This equation equates the fractional change in volume to the fractional change in density.

For practical applications, this can be used to calculate changes in density with depth in the Earth's crust, where pressure changes due to overlaying materials. The change in density \( \delta \rho \) can be related to the pressure \( p \) using \( \delta \rho = \rho \beta p \), reflecting how compacted a material might become under increased pressure conditions.

Poisson's Ratio

Poisson's ratio \( v \) is a material property that expresses the ratio of transverse strain to axial strain. It provides insight into how a material deforms in the lateral direction when it is stretched or compressed along one axis. The typical range for most materials is between 0 and 0.5, with values close to 0.5 indicating near incompressibility due to their resistant nature to volume change.

Poisson’s ratio plays a critical role in determining material stability and ease of deformation, influencing the computation of other mechanical properties like the bulk modulus. Thus, as \( v \) approaches \( 0.5 \), the bulk modulus becomes significantly large, reflecting the material's capacity to resist deformation.