A First Course in Continuum Mechanics

Oscar Gonzalez, Andrew M. Stuart

Chapter 7 Isothermal Solid Mechanics - all with Video Answers

Educators

Chapter Questions

Problem 1

Let $B=\left\{\boldsymbol{X} \in \boldsymbol{E}^{3} \mid 0<X_{i}<1\right\}$ be the reference configuration of an elastic solid with stress response given by the St. Venant-Kirchhoff model with material constants $\lambda, \mu>0$. Consider a time-independent uniaxial deformation $\boldsymbol{x}=\varphi(\boldsymbol{X})$ of the form $x_{1}=X_{1}, x_{2}=q X_{2}, x_{3}=X_{3}$ where $q>0$ is a constant.
(a) Find the components of the deformation gradient $F$ and the Green-St. Venant strain tensor $\boldsymbol{G}=\frac{1}{2}(\boldsymbol{C}-\boldsymbol{I})$.
(b) Find the components of the two Piola-Kirchhoff stress fields $\boldsymbol{P}$ and $\boldsymbol{\Sigma}$, and the Cauchy stress field $\boldsymbol{S}_{m}$.
(c) Find the resultant force on each face of the deformed configuration $B^{\prime}=\varphi(B) .$ These are the forces required to maintain the deformation. Is there a non-zero force on each face for all $q \neq 1 ?$
(d) What happens to the forces from part (c) in the limits $q \rightarrow$ $\infty$ and $q \downarrow 0$ ? Can the body be compressed to zero volume with finite forces?

Narayan Hari

Numerade Educator

02:01

Problem 2

Consider an elastic body with reference configuration $B$ and elastic response function $\widehat{P}(\boldsymbol{F})$
(a) Show that any time-independent, homogeneous deformation $\boldsymbol{x}=\boldsymbol{\varphi}(\boldsymbol{X})$ satisfies the equilibrium equation (7.4) with zero body force.
(b) Show that the external surface force, per unit deformed area, required to maintain a homogeneous equilibrium configuration is of the form
$$
\boldsymbol{t}(\boldsymbol{x})=\boldsymbol{A n}(\boldsymbol{x})
$$
where $\boldsymbol{A}$ is a constant tensor depending on $\widehat{\boldsymbol{P}}$ and $\boldsymbol{\varphi}$, and $\boldsymbol{n}(\boldsymbol{x})$ is the unit outward normal field on the deformed body.

Akshaya Rs

Numerade Educator

01:23

Problem 3

Show that the three principal invariants $I_{i}(\boldsymbol{S})$ of a second-order tensor $\boldsymbol{S}$ are isotropic in the sense of Section $1.5$, namely
$$
I_{i}(\boldsymbol{S})=I_{i}\left(\boldsymbol{Q} \boldsymbol{S} \boldsymbol{Q}^{T}\right)
$$
for all $\boldsymbol{S} \in \mathcal{V}^{2}$ and all rotations $\boldsymbol{Q} \in \mathcal{V}^{2}$.

Raj Bala

Numerade Educator

08:53

Problem 4

Show that, if the three principal invariants of two second-order tensors $\boldsymbol{A}$ and $\boldsymbol{B}$ are equal, then $\boldsymbol{A}$ and $\boldsymbol{B}$ have the same set of eigenvalues.

Anthony Ramos

Numerade Educator

06:05

Problem 5

Show that a function $\bar{W}: \mathcal{V}^{2} \rightarrow \boldsymbol{R}$ has the property
$$
\bar{W}(\boldsymbol{C})=\bar{W}\left(\boldsymbol{Q} \boldsymbol{C} \boldsymbol{Q}^{T}\right)
$$
for all symmetric $\boldsymbol{C}$ and all rotations $\boldsymbol{Q}$, if and only if there is a function $\widehat{W}: \boldsymbol{R}^{3} \rightarrow \boldsymbol{R}$ such that
$$
\bar{W}(\boldsymbol{C})=\widehat{W}\left(\mathcal{I}_{C}\right)
$$
for all symmetric $\boldsymbol{C} .$ Here $\mathcal{I}_{C}$ denotes the three principal invariants of $C$.

Melvin Adkins

Numerade Educator

01:23

Problem 6

Let $I_{i}(\boldsymbol{S})$ denote the three principal invariants of a second-order tensor $\boldsymbol{S}$. Show that
$$
D I_{1}(\boldsymbol{S})=\boldsymbol{I} \quad \text { and } \quad D I_{2}(\boldsymbol{S})=\operatorname{tr}(\boldsymbol{S}) \boldsymbol{I}-\boldsymbol{S}^{T}
$$
Remark: By Result $2.22$ we have
$$
D I_{3}(\boldsymbol{S})=\operatorname{det}(\boldsymbol{S}) \boldsymbol{S}^{-T}
$$
Thus the derivatives of all three principal invariants of a secondorder tensor can be computed explicitly.

Raj Bala

Numerade Educator

07:28

Problem 7

Show that, if the strain energy density for a hyperelastic solid takes the isotropic form
$$
W(\boldsymbol{F})=\widehat{W}\left(\mathcal{I}_{C}\right)
$$
then the stress response function $\widehat{\boldsymbol{P}}(\boldsymbol{F})$ takes the isotropic form
$$
\widehat{\boldsymbol{P}}(\boldsymbol{F})=\boldsymbol{F}\left[\gamma_{0}\left(\mathcal{I}_{C}\right) \boldsymbol{I}+\gamma_{1}\left(\mathcal{I}_{C}\right) \boldsymbol{C}+\gamma_{2}\left(\mathcal{I}_{C}\right) \boldsymbol{C}^{-1}\right]
$$
for appropriate functions $\gamma_{0}, \gamma_{1}$ and $\gamma_{2}$ which you should determine.

Dr. Rajveer Singh

Numerade Educator

06:45

Problem 8

Consider a hyperelastic body with reference configuration $B$ undergoing a motion $\varphi$ in the absence of body forces, and suppose the motion satisfies the boundary condition $\varphi(\boldsymbol{X}, t)=\boldsymbol{X}$ for all $\boldsymbol{X} \in \partial B$ and $t \geq 0$. Show that the sum of kinetic and strain energy is constant in the sense that
$$
\frac{d}{d t}\left(K\left[B_{t}\right]+U\left[B_{t}\right]\right)=0, \quad \forall t \geq 0
$$

Lucas Finney

Numerade Educator

02:03

Problem 9

Consider a hyperelastic body under the same conditions as in Exercise 8 and let
$$
U^{*}\left[B_{t}\right]=\int_{B} W^{*}(F(\boldsymbol{X}, t)) d V_{\boldsymbol{X}}
$$
for some $W^{*}: \mathcal{V}^{2} \rightarrow \mathbb{R}$ with the property that
$$
\nabla^{X} \cdot\left[D W^{*}(\boldsymbol{F}(\boldsymbol{X}, t))\right]=\mathbf{0}, \quad \forall \boldsymbol{X} \in B, t \geq 0
$$
Show that
$$
\frac{d}{d t} U^{*}\left[B_{t}\right]=0, \quad \forall t \geq 0
$$
Remark: The strain energy density $W^{*}$ can be referred to as a null energy density since it does not upset the energy balance established in Exercise 8. In particular, the modified energy $K\left[B_{t}\right]+U\left[B_{t}\right]+\alpha U^{*}\left[B_{t}\right]$ is also conserved for any scalar constant $\alpha .$

Manik Pulyani

Numerade Educator

01:59

Problem 10

Complete the proof of Result $7.9 .$ In particular, show $\mathbf{B}=\mathbf{C}$, where $\mathrm{B}$ and $\mathbf{C}$ are the elasticity tensors associated with the second Piola-Kirchhoff and Cauchy stress response functions $\widehat{\boldsymbol{\Sigma}}(\boldsymbol{F})$ and $\widehat{\boldsymbol{S}}(\boldsymbol{F})$

Tanishq Gupta

Numerade Educator

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Problem 11

Show that the components of the fourth-order elasticity tensor C for an isotropic, linear elastic material are
$$
\mathrm{C}_{i j k l}=\lambda \delta_{i j} \delta_{k l}+\mu\left(\delta_{i k} \delta_{j l}+\delta_{j k} \delta_{i l}\right)
$$

Victor Salazar

Numerade Educator

01:21

Problem 12

Show that a linear elastic solid is hyperelastic if and only if the fourth-order elasticity tensor $\mathbf{C}$ has major symmetry.

Dominador Tan

Numerade Educator

04:20

Problem 13

Consider an isotropic, linear elastic body under the same conditions as in Exercise $8 .$
(a) Show that the result in Exercise 8 implies
$$
\frac{1}{2} \frac{d}{d t} \int_{B}\left[\rho_{0}|\dot{\boldsymbol{u}}|^{2}+\nabla^{x} \boldsymbol{u}: \mathbf{C}\left(\nabla^{X} \boldsymbol{u}\right)\right] d V_{\boldsymbol{X}}=0, \quad \forall t \geq 0
$$
where $\boldsymbol{u}(\boldsymbol{X}, t)=\boldsymbol{\varphi}(\boldsymbol{X}, t)-\boldsymbol{X}$ is the displacement, $\rho_{0}$ is the reference mass density and $\mathbf{C}(\boldsymbol{H})=\lambda(\operatorname{tr} \boldsymbol{H}) \boldsymbol{I}+\mu\left(\boldsymbol{H}+\boldsymbol{H}^{T}\right)$ is the elasticity tensor.
(b) Supposing $\lambda=-\mu$, use the balance of linear momentum equation (with zero body force) $\rho_{0} \ddot{\boldsymbol{u}}=\nabla^{x} \cdot\left[\mathbf{C}\left(\nabla^{x} \boldsymbol{u}\right)\right]$ to show
$$
\frac{1}{2} \frac{d}{d t} \int_{B}\left[\rho_{0}|\dot{\boldsymbol{u}}|^{2}+\mu \nabla^{x} \boldsymbol{u}: \nabla^{x} \boldsymbol{u}\right] d V_{\boldsymbol{X}}=0, \quad \forall t \geq 0
$$
(c) Use the results from (a) and (b) to conclude that $B$ possesses a null energy density $W^{*}(\boldsymbol{F})$, which up to multiplicative and additive constants, can be written as
$$
W^{*}(\boldsymbol{F})=-\mu\left(\operatorname{tr}\left(\nabla^{x} \boldsymbol{u}\right)\right)^{2}+\mu \nabla^{x} \boldsymbol{u}: \nabla^{x} \boldsymbol{u}^{T}
$$
where $\nabla^{x} \boldsymbol{u}=\boldsymbol{F}-\boldsymbol{I}$.

Joseph Liao

Numerade Educator

01:01

Problem 14

Consider a general linear elastic body with fourth-order elasticity tensor $\mathbf{C}$.
(a) Show that $\mathbf{C}$ can be represented by a matrix $[[\mathbf{C}]] \in \mathbb{R}^{6 \times 6}$. (Hint: Use the left and right minor symmetries to show that the transformation $\mathbf{C}: \mathcal{V}^{2} \rightarrow \mathcal{V}^{2}$ reduces to $\mathbf{C}: \operatorname{sym}\left(\mathcal{V}^{2}\right) \rightarrow \operatorname{sym}\left(\mathcal{V}^{2}\right)$, where $\operatorname{sym}\left(\mathcal{V}^{2}\right) \subset \mathcal{V}^{2}$ is the subspace of symmetric second-order tensors.)
(b) Each $\boldsymbol{A} \in \operatorname{sym}\left(\mathcal{V}^{2}\right)$ can be uniquely represented by a column vector $[[\boldsymbol{A}]] \in \mathbb{R}^{6}$ in a number of different ways. Use the representation
$$
[[\boldsymbol{A}]]=\left(A_{11}, A_{22}, A_{33}, \sqrt{2} A_{12}, \sqrt{2} A_{13}, \sqrt{2} A_{23}\right)^{T}
$$
to show that major symmetry of $\mathbf{C}$ is equivalent to symmetry of its matrix representation $[[\mathbf{C}]]$
(c) Use the results from (a) and (b) to show that a general linear elastic material model is defined by 36 independent elastic constants (components of $\mathbf{C})$. Moreover, if this model has major symmetry (thus hyperelastic), then the number of independent elastic constants is reduced to 21 .

Raj Bala

Numerade Educator

02:54

Problem 15

Consider the linear momentum equation for an isotropic, linear elastic body with constant mass density $\rho_{0}>0$, namely
$$
\rho_{0} \ddot{\boldsymbol{u}}=\mu \Delta^{x} \boldsymbol{u}+(\mu+\lambda) \nabla^{x}\left(\nabla^{x} \cdot \boldsymbol{u}\right)
$$
where $\boldsymbol{u}(\boldsymbol{X}, t)$ is the displacement field and $\lambda, \mu$ are the Lamé constants. Here we study plane or progressive wave solutions of the form
$$
\boldsymbol{u}(\boldsymbol{X}, t)=\sigma \boldsymbol{a} \phi(\gamma \boldsymbol{k} \cdot \boldsymbol{X}-\omega t)
$$
where $\phi: \mathbb{R} \rightarrow \mathbb{R}$ is an arbitrary function which defines the profile of the wave (we assume $\phi^{\prime \prime}$ is not identically zero), $a$ and $k$ are unit vectors called the direction of displacement and propagation, and $\sigma, \gamma$ and $\omega$ are constants that determine the amplitude, spatial scaling and speed of the wave. In particular, the wave speed is given by $\omega / \gamma$
(a) Show that $(7.41)$ satisfies $(7.40)$ if and only if $\boldsymbol{a}, \boldsymbol{k}, \sigma, \gamma, \omega$ satisfy
$$
\sigma \gamma^{2}[\mu \boldsymbol{a}+(\mu+\lambda)(\boldsymbol{a} \cdot \boldsymbol{k}) \boldsymbol{k}]=\sigma \rho_{0} \omega^{2} \boldsymbol{a}
$$
Notice that these equations are satisfied by the trivial wave defined by $\sigma=0$
(b) Suppose $\lambda+\mu \neq 0 .$ For any given $k, \sigma \neq 0$ and $\gamma \neq 0$ show that the only independent solutions $(\boldsymbol{a}, \omega)$ to $(7.42)$ are:
(1) $\boldsymbol{a}=\pm \boldsymbol{k}, \quad \omega^{2}=\gamma^{2}(\lambda+2 \mu) / \rho_{0}$
(2) $\boldsymbol{a} \cdot \boldsymbol{k}=0, \quad \omega^{2}=\gamma^{2} \mu / \rho_{0}$
What happens in the case when $\lambda+\mu=0 ?$
Remark: The above results show that in an isotropic, linear elastic solid with $\lambda+\mu \neq 0$ only two types of plane waves are possible for any given direction $\boldsymbol{k}$ : longitudinal waves with $a$ parallel to $k$ as in (1), and transverse waves with $a$ perpendicular to $k$ as in (2). The speeds of these two types of waves are different. Notice that these speeds are real in view of the strong ellipticity conditions in Result 7.16. For non-isotropic models the situation is more complicated.

Surendra Kumar

Numerade Educator

11:06

Problem 16

A viscoelastic material is one which exhibits both viscous and elastic characteristics. A simple viscoelastic constitutive assumption is that there exists a stress response function $\widehat{\boldsymbol{P}}$ : $\mathcal{V}^{2} \times \mathcal{V}^{2} \rightarrow \mathcal{V}^{2}$ such that
$$
\boldsymbol{P}(\boldsymbol{X}, t)=\widehat{\boldsymbol{P}}(\boldsymbol{F}(\boldsymbol{X}, t), \dot{\boldsymbol{F}}(\boldsymbol{X}, t))
$$
Suppose such a body undergoes a motion $\varphi$ in which the body deviates only slightly from its reference configuration in the sense that
$$
\begin{aligned}
&\boldsymbol{\varphi}(\boldsymbol{X}, t)=\boldsymbol{X}+\epsilon \boldsymbol{u}(\boldsymbol{X}, t)+\mathcal{O}\left(\epsilon^{2}\right) \\
&\boldsymbol{F}(\boldsymbol{X}, t)=\boldsymbol{I}+\epsilon \nabla^{x} \boldsymbol{u}(\boldsymbol{X}, t)+\mathcal{O}\left(\epsilon^{2}\right)
\end{aligned}
$$
for some scalar parameter $0 \leq \epsilon \ll 1$. Moreover, suppose the reference configuration is stress-free and that there are no body forces.
(a) Show that, to leading order in $\epsilon$, the field $\boldsymbol{u}$ satisfies
$$
\rho_{0} \ddot{\boldsymbol{u}}=\nabla^{x} \cdot \mathbf{C}_{1}\left(\nabla^{x} \boldsymbol{u}\right)+\nabla^{x} \cdot \mathbf{C}_{2}\left(\nabla^{x} \dot{\boldsymbol{u}}\right)
$$
where $\mathbf{C}_{1}$ and $\mathbf{C}_{2}$ are fourth-order tensors which you should specify.
(b) Show that, if $\mathbf{C}_{1}$ and $\mathbf{C}_{2}$ are isotropic in the sense that
$$
\left.\begin{array}{l}
\mathbf{C}_{1}(\boldsymbol{H})=\lambda_{1} \operatorname{tr}(\boldsymbol{H}) \boldsymbol{I}+2 \mu_{1} \operatorname{sym}(\boldsymbol{H}) \\
\mathbf{C}_{2}(\boldsymbol{H})=\lambda_{2} \operatorname{tr}(\boldsymbol{H}) \boldsymbol{I}+2 \mu_{2} \operatorname{sym}(\boldsymbol{H})
\end{array}\right\} \quad \forall \boldsymbol{H} \in \mathcal{V}^{2}
$$
then $\boldsymbol{u}$ satisfies
$$
\begin{array}{rl}
\rho_{0} \ddot{\boldsymbol{u}}=\mu_{1} \Delta^{x} & \boldsymbol{u}+\left(\lambda_{1}+\mu_{1}\right) \nabla^{x}\left(\nabla^{x} \cdot \boldsymbol{u}\right) \\
& +\mu_{2} \Delta^{x} \dot{\boldsymbol{u}}+\left(\lambda_{2}+\mu_{2}\right) \nabla^{x}\left(\nabla^{x} \cdot \dot{\boldsymbol{u}}\right)
\end{array}
$$

Khoobchandra Agrawal

Numerade Educator

02:21

Problem 17

In one spatial dimension, the simplest model of linear viscoelasticity as introduced in Exercise 16 yields an equation of the form
$$
\rho_{0} \frac{\partial^{2}}{\partial t^{2}} u=\eta_{1} \frac{\partial^{2}}{\partial x^{2}} u+\eta_{2} \frac{\partial}{\partial t} \frac{\partial^{2}}{\partial x^{2}} u
$$
where $\rho_{0}, \eta_{1}$ and $\eta_{2}$ are scalar constants, $u(x, t)$ is the scalar displacement field and $(x, t)$ are the space and time coordinates. Here we study solutions of (7.43) of the form
$$
u(x, t)=\exp [i(x-c t)] \quad \text { where } \quad i^{2}=-1
$$
(a) Consider the elastic limit defined by $\eta_{1}>0$ and $\eta_{2}=0$. Show that (7.43) admits wave-like solutions which propagate, but do not damp out in time.
(b) Consider the viscous limit defined by $\eta_{1}=0$ and $\eta_{2}>$
0. Show that (7.43) admits stationary solutions which damp exponentially in time.
(c) Describe mathematically the form of solutions for the case $\eta_{1}>0$ and $\eta_{2}>0$. Distinguish between the cases $4 \rho_{0} \eta_{1}>\eta_{2}^{2}$ and $4 \rho_{0} \eta_{1}<\eta_{2}^{2}$.

Sana Riaz

Numerade Educator

03:10

Problem 18

In small deformation elasticity the infinitesimal strain field $\boldsymbol{E}$ is defined by
$$
\boldsymbol{E}=\frac{1}{2}\left(\nabla^{x} \boldsymbol{u}+\nabla^{x} \boldsymbol{u}^{T}\right) \quad \text { or } \quad E_{i j}=\frac{1}{2}\left(u_{i, j}+u_{j, i}\right)
$$
where $\boldsymbol{u}$ is the displacement field. In this question we address when $(7.44)$ can be solved for $\boldsymbol{u}$ in terms of $\boldsymbol{E}$. That is, given a strain field $\boldsymbol{E}$, when is it possible to solve $(7.44)$ as a differential equation for $\boldsymbol{u} ?$ Notice that, by symmetry, (7.44) yields six independent equations for the three unknown components of $\boldsymbol{u}$. Since there are more equations than unknowns, $\boldsymbol{E}$ must satisfy certain compatibility conditions for $(7.44)$ to be solvable.
(a) Show that, if $(7.44)$ possesses a smooth solution $u$ for a given $\boldsymbol{E}$, then $\boldsymbol{E}$ must necessarily satisfy the following six compatibility equations
$$
\begin{aligned}
&E_{11,22}+E_{22,11}=2 E_{12,12} \\
&E_{22,33}+E_{33,22}=2 E_{23,23} \\
&E_{33,11}+E_{11,33}=2 E_{31,31} \\
&E_{11,23}=\left(-E_{23,1}+E_{31,2}+E_{12,3}\right)_{11} \\
&E_{22,31}=\left(-E_{31,2}+E_{12,3}+E_{23,1}\right)_{22} \\
&E_{33,12}=\left(-E_{12,3}+E_{23,1}+E_{31,2}\right)_{, 3}
\end{aligned}
$$
Hint: Use the fact that the mixed partial derivatives of a smooth solution $\boldsymbol{u}$ must commute.
(b) Show that the six compatibility equations can be written succinctly in index notation as
$$
E_{n j, k m}+E_{k m, j n}-E_{k n, j m}-E_{m j, k n}=0
$$

Sana Riaz

Numerade Educator

09:25

Problem 19

Define two curl operations for a second order tensor $\boldsymbol{S}$ as
$$
\nabla \times \boldsymbol{S}=\epsilon_{p k q} S_{k m, p} \boldsymbol{e}_{q} \otimes \boldsymbol{e}_{m} \quad \text { and } \quad \boldsymbol{S} \times \nabla=\epsilon_{k p q} S_{r k, p} \boldsymbol{e}_{r} \otimes \boldsymbol{e}_{q}
$$
(a) Using the definitions above, show that
$$
-\left[\nabla \times\left(\boldsymbol{S}^{T}\right)\right]^{T}=\boldsymbol{S} \times \nabla \quad \text { and } \quad \nabla \times(\boldsymbol{S} \times \nabla)=(\nabla \times \boldsymbol{S}) \times \nabla
$$
(b) Show that the compatibility conditions from Exercise 18. may be written as
$$
\nabla \times(\boldsymbol{E} \times \nabla)=\boldsymbol{O}
$$

Jacob Fry

Numerade Educator

01:20

Problem 20

Consider a linear elastic body with mass density $\rho_{0}$, first PiolaKirchhoff stress field $\boldsymbol{P}$ and infinitesimal strain field $\boldsymbol{E} .$ Suppose the body is isotropic with Lamé constants $\lambda$ and $\mu$, and suppose the body is subject to a body force per unit mass $\boldsymbol{b}_{m}$. (Here and throughout the remaining exercises we shall omit the subscript $m$ on $\boldsymbol{b}_{m}$ for clarity.)
(a) Show that the isotropic, linear elastic stress-strain relation $\boldsymbol{P}=\lambda(\operatorname{tr} \boldsymbol{E}) \boldsymbol{I}+2 \mu \boldsymbol{E}$ can be inverted as
$$
\boldsymbol{E}=-\frac{\nu}{\gamma}(\operatorname{tr} \boldsymbol{P}) \boldsymbol{I}+\left(\frac{1+\nu}{\gamma}\right) \boldsymbol{P}
$$
where the Young's modulus $\gamma$ and Poisson's ratio $\nu$ are related to the Lamé constants $\lambda$ and $\mu$ by
$$
\mu=\frac{\gamma}{2(1+\nu)}, \quad \lambda=\frac{\nu \gamma}{(1+\nu)(1-2 \nu)}
$$
(b) Show that, if the body is in equilibrium, then
$$
P_{i k, j k}+\left(\rho_{0} b_{i}\right)_{, j}=0
$$

Hast Aggarwal

Numerade Educator

01:02

Problem 21

Consider a linear elastic, isotropic body as in Exercise 20. As before, we suppose the body is in equilibrium.
(a) Show that, when expressed in terms of the stress tensor $P_{i j}$ the compatibility equations of Exercise 18 can be reduced to
$$
P_{i j, k k}+P_{k k, i j}-P_{i k, j k}-P_{j k, i k}=\left(\frac{\nu}{1+\nu}\right)\left[\delta_{i j} P_{p p, k k}+P_{p p, i j}\right]
$$
(b) Use the results from part (a) and Exercise $20(\mathrm{~b})$ to show that $P_{i j}$ satisfies the equation
$$
\Delta^{x} P_{i j}+\left(\frac{1}{1+\nu}\right) \psi_{, i j}-\left(\frac{\nu}{1+\nu}\right) \delta_{i j} \Delta^{x} \psi=-\left(\rho_{0} b_{i}\right)_{, j}-\left(\rho_{0} b_{j}\right)_{, i}
$$
where $\psi=\operatorname{tr}(\boldsymbol{P})$
(c) Again use the results from part (a) and Exercise $20(\mathrm{~b})$ to show that $P_{i j}$ satisfies the equation
$$
P_{i j, i j}=\left(\frac{1-\nu}{1+\nu}\right) \Delta^{x} \psi \quad \text { and hence } \quad \Delta^{x} \psi=-\left(\frac{1+\nu}{1-\nu}\right) \beta
$$
where $\beta=\nabla^{X} \cdot\left(\rho_{0} \boldsymbol{b}\right)$(d) Combine the results from parts (b) and (c) to show that
$$
\Delta^{x} P_{i j}+\left(\frac{1}{1+\nu}\right) \psi_{, i j}=-\left(\frac{\nu}{1-\nu}\right) \delta_{i j} \beta-\left(\rho_{0} b_{i}\right)_{, j}-\left(\rho_{0} b_{j}\right)_{i i}
$$

Raj Bala

Numerade Educator

02:16

Problem 22

Consider a linear elastic, isotropic body in equilibrium as in Exercise 20 , and suppose the body experiences a planar deformation in the sense that
$$
\boldsymbol{u}(x, y, z)=u_{x}(x, y) \boldsymbol{e}_{1}+u_{y}(x, y) \boldsymbol{e}_{2}
$$
where for convenience we use $(x, y, z)$ to denote the material coordinates $\left(X_{1}, X_{2}, X_{3}\right)$, and we use subscripts $x, y$ and $z$ to denote components, not partial derivatives.
(a) Show that the infinitesimal strain tensor $\boldsymbol{E}$ for a planar deformation has a matrix representation of the form
$$
[\boldsymbol{E}]=\left(\begin{array}{ccc}
\epsilon_{x} & \epsilon_{x y} & 0 \\
\epsilon_{x y} & \epsilon_{y} & 0 \\
0 & 0 & 0
\end{array}\right)
$$
for some functions $\epsilon_{x}, \epsilon_{y}$ and $\epsilon_{x y}$.
(b) Under the assumption of isotropy, show that the first PiolaKirchhoff stress tensor $\boldsymbol{P}$ for a planar deformation has a matrix representation of the form
$$
[\boldsymbol{P}]=\left(\begin{array}{ccc}
\sigma_{x} & \tau_{x y} & 0 \\
\tau_{x y} & \sigma_{y} & 0 \\
0 & 0 & \sigma_{z}
\end{array}\right)
$$
where
$$
\begin{aligned}
\sigma_{x} &=\lambda e+2 \mu \epsilon_{x}, & \tau_{x y} &=2 \mu \epsilon_{x y} \\
\sigma_{y} &=\lambda e+2 \mu \epsilon_{y}, & \sigma_{z} &=\lambda e
\end{aligned}
$$
and $e=\epsilon_{x}+\epsilon_{y}$. Moreover, show that $\sigma_{z}=\nu\left(\sigma_{x}+\sigma_{y}\right)$, where Poisson's ratio $\nu$ is defined as
$$
\nu=\frac{\lambda}{2(\lambda+\mu)}
$$
(c) Assuming the body is subject to a planar body force $b=$ $b_{x}(x, y) e_{1}+b_{y}(x, y) e_{2}$, the equilibrium equations and the compatibility equation in Exercise 21(c) yield three independent equations for the three independent stress components $\sigma_{x}, \sigma_{y}$ and $\tau_{x y}$, namely
$$
\begin{aligned}
&\frac{\partial \sigma_{x}}{\partial x}+\frac{\partial \tau_{x y}}{\partial y}+\rho_{0} b_{x}=0 \\
&\frac{\partial \tau_{x y}}{\partial x}+\frac{\partial \sigma_{y}}{\partial y}+\rho_{0} b_{y}=0 \\
&\Delta\left(\sigma_{x}+\sigma_{y}\right)=-\left(\frac{1}{1-\nu}\right)\left(\frac{\partial\left(\rho_{0} b_{x}\right)}{\partial x}+\frac{\partial\left(\rho_{0} b_{y}\right)}{\partial y}\right)
\end{aligned}
$$
Assuming zero body force show that the above equations are satisfied by
$$
\sigma_{x}=\frac{\partial^{2} \phi}{\partial y^{2}}, \quad \sigma_{y}=\frac{\partial^{2} \phi}{\partial x^{2}}, \quad \tau_{x y}=-\frac{\partial^{2} \phi}{\partial x \partial y}
$$
provided that $\phi$ satisfies the biharmonic equation
$$
\Delta^{2} \phi=\frac{\partial^{4} \phi}{\partial x^{4}}+2 \frac{\partial^{4} \phi}{\partial x^{2} \partial y^{2}}+\frac{\partial^{4} \phi}{\partial y^{4}}=0
$$
The field $\phi$ is typically called an Airy stress function.

Chai Santi

Numerade Educator