 Problems for Chapter 2

Governing Equations

2.4.  Work and Energy, the Principle of Virtual Work

2.4.1.      A solid with volume V is subjected to a distribution of traction ${t}_{i}$ on its surface. Assume that the solid is in static equilibrium. By considering a virtual velocity of the form $\delta {v}_{i}={A}_{ij}{y}_{j}$, where ${A}_{ij}$ is a constant symmetric tensor, use the principle of virtual work to show that the average stress in a solid can be computed from the shape of the solid and the tractions acting on its surface using the expression

$\frac{1}{V}\underset{V}{\int }{\sigma }_{ij}dV=\frac{1}{V}\underset{S}{\int }\frac{1}{2}\left({t}_{i}{y}_{j}+{t}_{j}{y}_{i}\right)dA$ 2.4.2.      The figure shows a cantilever beam that is subjected to surface loading $q\left({x}_{1}\right)$ per unit length.  The state of stress in the beam can be approximated by ${\sigma }_{11}=M\left({x}_{1}\right){x}_{2}/I$, where $I=\underset{A}{\int }{x}_{2}^{2}dA$ is the area moment of inertia of the beam’s cross section and $M\left({x}_{1}\right)$ is an arbitrary function (all other stress components are zero).  By considering a virtual velocity field of the form

$\delta {v}_{1}=-\frac{w\left({x}_{1}\right)}{d{x}_{1}}{x}_{2}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\delta {v}_{2}=w\left({x}_{1}\right)$

where $w\left({x}_{1}\right)$ is an arbitrary function satisfying $w=0$ at ${x}_{1}=0$, show that the beam is in static equilibrium if

$\underset{0}{\overset{L}{\int }}M\left({x}_{1}\right)\frac{{d}^{2}w}{d{x}_{1}^{2}}d{x}_{1}+\underset{0}{\overset{L}{\int }}q\left({x}_{1}\right)wd{x}_{1}=0$

By integrating the first integral expression by parts twice, show that the equilibrium equation and boundary conditions for $M\left({x}_{1}\right)$ are

$\frac{{d}^{2}M}{d{x}_{1}^{2}}+q\left({x}_{1}\right)=0\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}M\left({x}_{1}\right)=\frac{dM\left({x}_{1}\right)}{d{x}_{1}}=0\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{x}_{1}=L$ 2.4.3.      The figure shows a plate with a clamped edge that is subjected to a pressure $p\left({x}_{1},{x}_{2}\right)$ on its surface.  The state of stress in the plate can be approximated by

${\sigma }_{\alpha \beta }={M}_{\alpha \beta }\left({x}_{1},{x}_{2}\right){x}_{3}/3{h}^{3}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\sigma }_{33}={\sigma }_{3\alpha }=0$

where the subscripts $\alpha ,\beta$ can have values 1 or 2, and ${M}_{\alpha \beta }\left({x}_{1},{x}_{2}\right)$ is a tensor valued function.  By considering a virtual velocity of the form

$\delta {v}_{\alpha }=-\frac{\partial w}{\partial {x}_{\alpha }}{x}_{3}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\delta {v}_{3}=w\left({x}_{1},{x}_{2}\right)$

where $w\left({x}_{1},{x}_{2}\right)$ is an arbitrary function satisfying $w=0$ on the edge of the plate, show that the beam is in static equilibrium if

$\underset{A}{\overset{}{\int }}{M}_{\alpha \beta }\left({x}_{1}\right)\frac{{\partial }^{2}w}{\partial {x}_{\alpha }\partial {x}_{\beta }}dA+\underset{A}{\overset{}{\int }}p\left({x}_{1},{x}_{2}\right)wdA=0$

By applying the divergence theorem appropriately, show that the governing equation for ${M}_{\alpha \beta }\left({x}_{1},{x}_{2}\right)$ is

$\frac{{\partial }^{2}{M}_{\alpha \beta }}{\partial {x}_{\alpha }\partial {x}_{\beta }}+p=0$ 2.4.4.      The shell shown in the figure is subjected to a radial body force $b=b\left(R\right){e}_{R}$, and a radial pressure ${p}_{a},{p}_{b}$ acting on the surfaces at $R=a$ and $R=b$. The loading induces a spherically symmetric state of stress in the shell, which can be expressed in terms of its components in a spherical-polar coordinate system as ${\sigma }_{RR}{e}_{R}\otimes {e}_{R}+{\sigma }_{\theta \theta }{e}_{\theta }\otimes {e}_{\theta }+{\sigma }_{\varphi \varphi }{e}_{\varphi }\otimes {e}_{\varphi }$.   By considering a virtual velocity of the form $\delta v=w\left(R\right){e}_{R}$, show that the stress state is in static equilibrium if

$\begin{array}{l}\underset{a}{\overset{b}{\int }}\left\{{\sigma }_{RR}\frac{dw}{dR}+\left({\sigma }_{\theta \theta }+{\sigma }_{\varphi \varphi }\right)\frac{w}{R}\right\}4\pi {R}^{2}dR-\underset{a}{\overset{b}{\int }}b\left(R\right)w\left(R\right)4\pi {R}^{2}dR\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-4\pi {a}^{2}{p}_{a}w\left(a\right)+4\pi {b}^{2}{p}_{b}w\left(b\right)=0\end{array}$

for all w(R).  Hence, show that the stress state must satisfy

$\frac{d{\sigma }_{RR}}{dR}+\frac{1}{R}\left(2{\sigma }_{RR}-{\sigma }_{\theta \theta }-{\sigma }_{\varphi \varphi }\right)+b=0\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\sigma }_{RR}=-{p}_{a}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(R=a\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\sigma }_{RR}=-{p}_{b}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(R=b\right)\text{\hspace{0.17em}}$ 2.4.5.      In this problem, we consider the internal forces in the polymer specimen described in Problem 2.1.29 and 2.3.5 (you will need to solve 2.1.29 and 2.3.5 before you can attempt this one). Suppose that the specimen is homogeneous, has mass density $\rho$ in the reference configuration, and may be idealized as a viscous fluid, in which the Kirchhoff stress is related to stretch rate by

$\tau =\mu D+pI$

where p is an indeterminate hydrostatic pressure and $\mu$ is the viscosity.

2.4.5.1.            Calculate the rate of external work done by the torque acting on the rotating exterior cyclinder

2.4.5.2.            Calculate the rate of internal dissipation in the solid as a function of r.

2.4.5.3.            Show that the total internal dissipation is equal to the external work done on the specimen.