Problem 6
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 $$ \begin{array}{l} \varepsilon_{1}=\frac{(1+v)}{E}\left\\{\sigma_{1}(1-v)-v \sigma_{2}\right\\} \\\ \varepsilon_{2}=\frac{(1+v)}{E}\left\\{\sigma_{2}(1-v)-v \sigma_{1}\right\\} \end{array} $$
Problem 8
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.
Problem 13
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?
Problem 22
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 \(\rho_{m}-\rho_{s}=700 \mathrm{~kg} \mathrm{~m}^{-3}\)
