6. Small displacement - statics#

  • Labile - Undetermined.

  • Isostatic - “determined”.

  • Hyperstatic - “overdetermined”.

6.1. Linear isotropic elastic medium#

6.1.1. Constitutive equation#

An isotropic elastic medium has no prefereed orientation. The most general linear relation between stress tensor \(\symbf{\sigma}\) and strain tensor \(\symbf{\varepsilon}\), and temperature difference \(\Delta T\) w.r.t. a reference temperature, \(\Delta T := T - T_0\),

\[\symbf{\sigma} = \mathbf{D} : \symbf{\varepsilon} - \symbf{\beta} \Delta T \ ,\]

for isotropic media involves the rank-\(2\) and rank-\(4\) isotropic tensors, see rank-2-iso, and rank-4-iso. Since stress tensor \(\symbf{\sigma}\) and strain tensor \(\symbf{\varepsilon}\) are symmetric the constitutive law for isotropic elastic media reads

(6.1)#\[\symbf{\sigma} = 2 \mu \symbf{\varepsilon} + \lambda \, \text{tr}(\symbf{\varepsilon}) \mathbb{I} - \beta \Delta T \mathbb{I} \ ,\]

being \(\mu\), \(\lambda\) the Lamé coefficients.

Decomposition of tensor in hydrostatic (proportional to \(\mathbb{I}\)) and deviatoric (traceless) parts reads

(6.2)#\[\symbf{\sigma} = \left( 2 \mu \symbf{\varepsilon} - \frac{2}{3} \mu \, \text{tr}(\varepsilon) \mathbb{I} \right) + \left( \lambda + \frac{2}{3} \mu \right)\text{tr}(\varepsilon) \mathbb{I} - \beta \Delta T \mathbb{I} \]

The expression of strain tensor \(\symbf{\varepsilon}\) as a function of stress tensor and temperature difference

\[\symbf{\varepsilon} = \]

can be easily evaluated, using the relation between the traces of the tensors

Different expression of constitutive laws, and sets of parameters.

Remembering that \(\text{tr}(\mathbb{I}) = 3\) in the 3-dimensional space, evaluating the trace of the relation (6.1) provides the relation between the traces of strain and stress tensors and the temeprature difference

\[\text{tr} (\symbf{\sigma}) = \left( 2 \mu + 3 \lambda \right) \text{tr} (\symbf{\varepsilon}) - 3 \beta \Delta T \ , \]

and thus

\[\text{tr} (\symbf{\varepsilon}) = \frac{1}{ 2 \mu + 3 \lambda} \text{tr} (\symbf{\sigma}) + \frac{3 \beta}{ 2 \mu + 3 \lambda} \Delta T \ . \]

Using the relation between traces, it’s possible to find \(\symbf{\varepsilon}\left(\symbf{\sigma}, \Delta T \right)\),

\[\begin{split}\begin{aligned} \symbf{\varepsilon} & = \frac{1}{2 \mu} \symbf{\sigma} - \frac{\lambda}{2\mu} \text{tr}(\symbf{\varepsilon}) \mathbb{I} + \frac{\beta}{2 \mu} \Delta T \mathbb{I} = \\ & = \frac{1}{2 \mu} \symbf{\sigma} - \frac{\lambda}{2\mu} \left[ \frac{1}{2 \mu + 3 \lambda} \text{tr}(\symbf{\sigma}) + \frac{3 \beta}{2 \mu + 3 \lambda} \Delta T \right] \mathbb{I} + \frac{\beta}{2 \mu} \Delta T \mathbb{I} = \\ & = \frac{1}{2 \mu} \symbf{\sigma} - \frac{\lambda}{2\mu ( 2 \mu + 3 \lambda)} \text{tr}(\symbf{\sigma}) \mathbb{I} + \frac{1}{2 \mu}\left[ 1 - \frac{3 \lambda}{2 \mu + 3 \lambda} \right] \beta \Delta T \mathbb{I} = \\ & = \frac{1}{2 \mu} \symbf{\sigma} - \frac{\lambda}{2\mu ( 2 \mu + 3 \lambda)} \text{tr}(\symbf{\sigma}) \mathbb{I} + \frac{1}{2 \mu + 3 \lambda} \beta \Delta T \mathbb{I} \ . \end{aligned}\end{split}\]

Modulo elastico. Il modulo elastico \(E\) è definito dalla relazione

\[\begin{split}\begin{aligned} \frac{1}{E} & = \frac{1}{2 \mu} - \frac{\lambda}{2 \mu (2 \mu + 3 \lambda)} = \\ & = \frac{1}{2 \mu} \frac{2 (\mu + \lambda)}{2 \mu + 3 \lambda} \ , \end{aligned}\end{split}\]

e quindi

\[E = \frac{\mu (2 \mu + 3 \lambda)}{\mu + \lambda}\]

Modulo di Poisson. Il modulo di Poisson \(\nu\) è definito dalla relazione

\[\frac{\nu}{E} = \frac{\lambda}{2\mu (2 \mu + 3 \lambda)} \ ,\]

e quindi

\[\nu = \frac{\lambda}{2 (\mu + \lambda)}\]
Thermodynamics

Helmholtz’s free energy is the thermodynamic potential \(F\) defined w.r.t. temperature \(T\) and generalized displacements \(\mathbf{X}_i\) as independet variables. Here unit-volume relation (Explain why! Link to general continuum mechanics, and equations in reference coordinates). Using difference of temperature \(\Delta T = T - T_0\) w.r.t. a reference temperature \(T_0\) instead of temperature \(T\), differential of Helmhtoltz’s free energy per unit volume for a linear elastic medium reads

\[d\mathcal{F}(\varepsilon_{ij}, \Delta T) = - \mathcal{S} d \Delta T + \sigma_{ij} d \varepsilon_{ij} \ ,\]

so that

\[ \sigma_{ij} = \left( \frac{\partial \mathcal{F}}{\partial \varepsilon_{ij}} \right)_{\mathcal{S}} \qquad , \qquad \mathcal{S} = - \left( \frac{\partial \mathcal{F}}{\partial T} \right)_{\varepsilon_{ij}} \ . \]

Using the constitutive law (6.1) for a linear isotropic elastic medium, here written in Cartesian coordinates as

\[\sigma_{ij} = 2 \mu \varepsilon_{ij} + \lambda \varepsilon_{kk} \delta_{ij} - \beta \Delta T \delta_{ij} \ ,\]

integration w.r.t. \(\varepsilon_{ij}\) gives

\[\begin{split}\begin{aligned} \mathcal{F}(\varepsilon_{ij}, \Delta T) & = \frac{1}{2} \left( 2 \mu \varepsilon_{ij} \varepsilon_{ij} + \lambda \varepsilon_{kk} \varepsilon_{ll} \right) - \beta \Delta T \varepsilon_{ll} + \mathcal{F}(0, \Delta T) = \\ & = \frac{1}{2} \left( 2 \mu \symbf{\varepsilon} : \symbf{\varepsilon} + \lambda \left( \text{tr}(\symbf{\varepsilon}) \right)^2 \right) - \beta \Delta T \text{tr}(\symbf{\varepsilon}) + \mathcal{F}(0, \Delta T) \\ \end{aligned}\end{split}\]

being \(\mathcal{F}(0, \Delta T) = F(\Delta T)\) a function of \(\Delta T\) appearing from integration in \(\varepsilon_{ij}\). If you don’t trust this, try do evaluate the partial derivative w.r.t. the components of the strain tensor of the last relation.

Entropy. Entropy is the parital derivative w.r.t. \(T\) of Helmholtz’s free energy and, assuming constant parameters, its expression reads

\[\begin{aligned} \mathcal{S}(\varepsilon_{ij}, \Delta T) = - \left( \frac{\partial \mathcal{F}}{\partial T} \right)_{\varepsilon_{ij}} = \beta \varepsilon_{ll} - F'(\Delta T) \ . \end{aligned}\]

Heat coefficients. Heat coefficient at constant strain per unit-volume reads

\[C_{\varepsilon_{ij}} := T \left( \frac{\partial \mathcal{S}}{\partial T} \right)_{\varepsilon_{ij}} = - (T_0 + \Delta T) F''(\Delta T) \ .\]

Assuming that heat coefficient \(C_{\varepsilon_{ij}}\) is independent from \(T\), integration in \(\Delta T\) gives

\[C_{\varepsilon_{ij}} \ln \left( 1 + \frac{\Delta T}{T_0} \right) = - F'(\Delta T) + F'(0) \ ,\]

and thus an expression for \(F'(\Delta T)\)

\[F'(\Delta T) = - C_{\varepsilon_{ij}} \ln \left( 1 + \frac{\Delta T}{T_0} \right) + F'(0) \ ,\]

to be inserted in the expression of entropy

\[\mathcal{S}(\varepsilon_{ij}, \Delta T) = \beta \varepsilon_{ll} + C_{\varepsilon_{ij}} \ln \left( 1 + \frac{\Delta T}{T_0} \right) - F'(0) \ ,\]

and \(-F'(0) = \mathcal{S}_0\) can be recognized as the entropy per unit volume in the reference condition with \(\varepsilon_{ij} = 0\), \(\Delta T = 0\). The differential of the last expression reads

\[d \mathcal{S} = \beta d \varepsilon_{ll} + \frac{C_{\varepsilon_{ij}}}{T_0} \frac{1}{1 + \frac{\Delta T}{T_0}} \, d T = \beta d \varepsilon_{ll} + \frac{C_{\varepsilon_{ij}}}{T}\, d T \]

or

\[dT = \frac{T}{C_{\varepsilon_{ij}}} d \mathcal{S} - \frac{\beta T}{C_{\varepsilon_{ij}}} \, d \varepsilon_{ll}\]

Furhter integration in \(\Delta T\) of

\[F'(\Delta T) = - C_{\varepsilon_{ij}} \ln \left( 1 + \frac{\Delta T}{T_0} \right) - \mathcal{S}_0 \ ,\]

gives and expression of funtion \(F(\Delta T)\),

\[F(\Delta T) - F(0) = -C_{\varepsilon_{ij}} \left[ (T_0 + \Delta T) \ln \left( 1 + \frac{\Delta T}{T_0} \right) - \Delta T \right] - \mathcal{S}_0 \Delta T \ ,\]

that can be used int he expression of Helmholtz free energy, as an example, as show later.

Internal energy. The relation between the internal energy and Hemlholtz free energy \(\mathcal{F} = \mathcal{E} - T \mathcal{S}\) allows to find the expression of the internal energy per unit volume of an elastic linear isotropic media,

\[\begin{split}\begin{aligned} \mathcal{E} & = \mathcal{F} + T \mathcal{S} = \\ & = \mu \varepsilon_{ij} \varepsilon_{ij} + \frac{\lambda}{2} \left( \varepsilon_{kk} \right)^2 - \beta \Delta T \varepsilon_{kk} + F(0) -C_{\varepsilon_{ij}} \left[ (T_0 + \Delta T) \ln \left( 1 + \frac{\Delta T}{T_0} \right) - \Delta T \right] \\ & \quad - \mathcal{S}_0 \Delta T + (T_0 + \Delta T) \left( \beta \varepsilon_{ll} + C_{\varepsilon_{ij}} \ln \left( 1 + \frac{\Delta T}{T_0} \right) + \mathcal{S}_0 \right) = \\ & = \mu \varepsilon_{ij} \varepsilon_{ij} + \frac{\lambda}{2} \left( \varepsilon_{kk} \right)^2 + \beta T_0 \varepsilon_{kk} + C_{\varepsilon_{ij}} \Delta T + F(0) + T_0 \mathcal{S}_0 \ , \end{aligned}\end{split}\]

so that the reference internal energy and reference Helmholtz free energy can be recognized as

\[\mathcal{F}_0 = F(0) \qquad , \qquad \mathcal{E}_0 = \mathcal{F}_0 + T_0 \mathcal{S}_0 \ ,\]

In order to write the differential of the internal energy w.r.t. its natural independent variables \(\mathcal{E}(\varepsilon_{ij}, \mathcal{S})\), the temperature difference must be written as a function of strain and entropy. Using relation todo, and performing derivatives of the composite funciton \(\mathcal{E}(\varepsilon_{ij}, \Delta T(\varepsilon_{ij}, \mathcal{S}))\)

\[\begin{split}\begin{aligned} d \mathcal{E} & = \left( \frac{\partial \mathcal{E}}{\partial \varepsilon_{ij}} \right)_{\Delta T} d \varepsilon_{ij} + \left( \frac{\partial \mathcal{E}}{\partial \Delta T} \right)_{\varepsilon_{ij}} \left[ \left( \frac{\partial \Delta T}{\partial \varepsilon_{ij}} \right)_{\mathcal{S}} d \varepsilon_{ij} + \left( \frac{\partial \Delta T}{\partial \mathcal{S}} \right)_{\varepsilon_{ij}} d \mathcal{S} \right] = \\ & = \left( 2 \mu \varepsilon_{ij} + \lambda \varepsilon_{kk} \delta_{ij} + \beta T_0 \delta_{ij} \right) d \varepsilon_{ij} + C_{\varepsilon_{ij}} \left[ -\frac{\beta T}{C_{\varepsilon_{ij}}} \delta_{ij} \, d \varepsilon_{ij} + \frac{T}{C_{\varepsilon_{ij}}} \, d \mathcal{S} \right] = \\ & = \left( 2 \mu \varepsilon_{ij} + \lambda \varepsilon_{kk} \delta_{ij} - \beta \Delta T \delta_{ij} \right) d \varepsilon_{ij} + T \, d \mathcal{S} \ , \end{aligned}\end{split}\]

i.e. we found what we should have expected, i.e.

\[d \mathcal{E} = \sigma_{ij} \, d \varepsilon_{ij} + T \, d\mathcal{S} \ , \]

and thus

\[\begin{split}\begin{aligned} \sigma_{ij} & = \left( \frac{\partial \mathcal{F}}{\partial \varepsilon_{ij}} \right)_{T} = \left( \frac{\partial \mathcal{E}}{\partial \varepsilon_{ij}} \right)_{\mathcal{S}} \\ T & = \left( \frac{\partial \mathcal{E}}{\partial \mathcal{S}} \right)_{\varepsilon_{ij}}\\ \mathcal{S} & = - \left( \frac{\partial \mathcal{F}}{\partial T} \right)_{\varepsilon_{ij}}\\ \end{aligned}\end{split}\]
Isothermal and isentropic elastic coefficients

Assuming small enough \(\Delta T = T - T_0\) so that linear approximation of the relation between entropy, temperature and strain holds,

\[\Delta \mathcal{S} = \beta \, \varepsilon_{ll} + \frac{C_{\varepsilon_{ij}}}{T_0} \Delta T \ ,\]

it’s possible to write the stress tensor as a function of strain and entropy

\[\begin{split}\begin{aligned} \sigma_{ij} & = 2 \mu \varepsilon_{ij} + \lambda \varepsilon_{ll} \delta_{ij} - \beta \Delta T \delta_{ij} = \\ & = 2 \mu \varepsilon_{ij} + \lambda \varepsilon_{ll} \delta_{ij} - \beta \left[ \frac{T_0}{C_{\varepsilon_{ij}}} \Delta \mathcal{S} - \frac{T_0}{C_{\varepsilon_{ij}}} \beta \, \varepsilon_{ll} \right] \delta_{ij} = \\ & = 2 \mu \varepsilon_{ij} + \left( \lambda + \frac{\beta^2 T_0}{C_{\varepsilon_{ij}}} \right) \varepsilon_{ll} \delta_{ij} - \beta \frac{T_0}{C_{\varepsilon_{ij}}} \Delta \mathcal{S} = \\ & = 2 \mu_s \varepsilon_{ij} + \lambda_s \varepsilon_{ll} \delta_{ij} - \beta \frac{T_0}{C_{\varepsilon_{ij}}} \Delta \mathcal{S} \ . \end{aligned}\end{split}\]

having defined Lamé coefficients in isentropic conditions as functions of the cofficients in isothermal conditions,

\[\begin{split}\begin{aligned} \lambda_s & = \lambda + \frac{\beta^2 T_0}{C_{\varepsilon_{ij}}} \\ \mu_s & = \mu \end{aligned}\end{split}\]

Elastic modulus and Poisson ratio. Starting from the relations

\[\begin{split}\begin{aligned} E & = \frac{\mu (2 \mu + 3 \lambda)}{\mu + \lambda} \\ \frac{E}{\nu} & = \frac{2 \mu (2 \mu + 3 \lambda)}{\lambda} \\ \end{aligned}\end{split}\]
\[G E + \lambda E = 2 G^2 + 3 G \lambda\]
\[\lambda = \frac{G (E - 2 G)}{ 3G - E }\]
\[\begin{split}\begin{aligned} \nu & = \frac{E \lambda}{2G(2G+3 \lambda)} = \\ & = \frac{\lambda}{2G(2G+3 \lambda)} \frac{G (2G + 3\lambda)}{G + \lambda} = \\ & = \frac{\lambda}{2(G + \lambda)} = \\ & = \frac{G(E-2G)}{3G - E} \frac{1}{2 \left( G + \frac{G(E-2G)}{3G-E} \right)} = \\ & = \frac{G}{2G} \frac{E-2G}{3G - E} \frac{3G - E}{3G - E + E - 2G } = \\ & = \frac{1}{2} \frac{E - 2G}{G} \ . \end{aligned}\end{split}\]

so that

\[\nu = \frac{E - 2G}{2G} \qquad , \qquad G = \frac{E}{2(1+\nu)}\]
Heat capacity, thermal expansion coefficients, compressibility coefficients

Thermal expansion coefficient reads

\[\alpha_x := \frac{1}{V} \left( \frac{\partial V}{\partial T}\right)_x\]

with \(\dfrac{dV}{V} = d \text{tr}(\symbf{\varepsilon})\) for small displacement and strain regime. At constant strain, \(d \sigma_{ij} = 0\),

\[ \alpha_{\sigma} = \left( \dfrac{\partial \varepsilon_{ll}}{\partial T} \right)_{\sigma} \]
Some algebra
\[\begin{split}\begin{aligned} & = \frac{3 \beta}{2 \mu + 3 \lambda} = \\ & = 3 \beta \frac{2 G \nu }{E \lambda} = \\ & = 3 \beta \frac{\nu }{(1+ \nu) \lambda} = \\ & = 3 \beta \frac{\nu }{(1+ \nu)} \frac{3G-E}{G(E-2G)} = \\ & = 3 \beta \frac{\nu }{(1+ \nu)} \frac{3-2(1+\nu)}{G\left(2(1+\nu) -2 \right)} = \\ & = 3 \frac{\beta}{2G} \frac{\nu }{(1+ \nu)} \frac{1-2\nu}{\nu} = \\ & = 3 \frac{\beta}{E} (1 - 2 \nu) \ . \end{aligned}\end{split}\]

so that

\[\beta = K \alpha_{\sigma} = \left( \frac{2}{3} \mu + \lambda \right) \, \alpha_{\sigma} = \frac{E}{3(1-2\nu)} \, \alpha_{\sigma} \]

Heat capacity. The heat capacity at constant strain \(C_{\varepsilon_{ij}}\) has been assumed to be constant (or just independent from temperature). Entropy can be written as functions of strain and temperature or - exploiting the relation between traces of tensors - stress and temperature

\[\begin{split}\begin{aligned} \mathcal{S} - \mathcal{S}_0 & = \beta \varepsilon_{ll} + C_{\varepsilon_{ij}} \ln \left( 1 + \frac{\Delta T}{T_0} \right) \\ & = \beta \left( \frac{1}{2\mu+3\lambda} \sigma_{ll} + \alpha_{\sigma} \Delta T \right) + C_{\varepsilon_{ij}} \ln \left( 1 + \frac{\Delta T}{T_0} \right) \\ & = \beta \left( \frac{1}{3 \, K} \sigma_{ll} + \alpha_{\sigma} \Delta T \right) + C_{\varepsilon_{ij}} \ln \left( 1 + \frac{\Delta T}{T_0} \right) \\ \end{aligned}\end{split}\]

and thus, constaint strain heat capacity per unit volume reads

\[C_{\varepsilon_{ij}} := T \left( \frac{\partial \mathcal{S}}{\partial T} \right)_{\varepsilon} = C_{\varepsilon_{ij}} \ ,\]

as defined above, whilte constant stress heat capacity per unit volume reads

\[\begin{split}\begin{aligned} C_{\sigma_{ij}} & := T \left( \frac{\partial \mathcal{S}}{\partial T} \right)_{\sigma} = \\ & = T \left( \beta \alpha_{\sigma} + C_{\varepsilon_{ij}} \frac{1}{T_0} \frac{1}{1 + \frac{\Delta T}{T_0}} \right) \\ & = T K \alpha^2_{\sigma} + C_{\varepsilon_{ij}} \\ \end{aligned}\end{split}\]

Compressibility coefficients. todo check: Below, \(T\) or \(T_0\)? It should be a minor change, since we’re assuming small \(\Delta T\) so that linearized constitutive equation holds?

\[\begin{split}\begin{aligned} K & = \frac{2}{3} \mu + \lambda \\ K_s & = \frac{2}{3} \mu_s + \lambda_s = \frac{2}{3} \mu + \lambda + \frac{\beta^2 T}{C_{\varepsilon_{ij}}} = K + \frac{\beta^2 T}{C_{\varepsilon_{ij}}} = K + \frac{K^2 \alpha^2_{\sigma} T}{C_{\varepsilon}} \end{aligned}\end{split}\]
\[\begin{split}\begin{aligned} K_s & = K \left( 1 + \frac{\alpha^2 K T}{C_{\varepsilon}} \right) \\ \rightarrow \quad & \frac{1}{K_s} = \frac{1}{K} \frac{C_{\varepsilon}}{C_{\sigma}} \\ \rightarrow \quad & \frac{1}{K_s} = \frac{1}{K} \frac{C_{\sigma} - \alpha_{\sigma}^2 K T}{C_{\sigma}} \\ \rightarrow \quad & \frac{1}{K_s} = \frac{1}{K} - \frac{\alpha_{\sigma}^2 T}{C_{\sigma}} \\ \end{aligned}\end{split}\]

Elastic modulus and Poisson ratio.

\[ E_s = \frac{E}{1 - E \, \dfrac{\alpha^2_\sigma T}{9 C_\sigma}} \qquad , \qquad \nu_s = \frac{\nu + E \, \dfrac{\alpha^2_\sigma T}{9 C_\sigma}}{1 - E \, \dfrac{\alpha^2_\sigma T}{9 C_\sigma}} \]
Some algebra
\[\begin{split}\begin{aligned} K & = \lambda + \frac{2}{3} G = \\ & = \frac{GE - 2G^2}{3G-E} + \frac{2}{3} G = \\ & = G \frac{3 E - 6 G + 6 G - 2 E}{3(3 G - E)} = \frac{E G}{3 (3 G - E)} \end{aligned}\end{split}\]
\[E = \frac{9 G K}{ 3 K + G } = \frac{1}{ \frac{1}{3 G} + \frac{1}{9 K}} \]
\[\begin{split}\begin{aligned} E_s & = \frac{1}{ \dfrac{1}{3 G_s} + \dfrac{1}{9 K_s}} = \\ & = \frac{1}{\dfrac{1}{3 G} + \dfrac{1}{9 K} - \dfrac{\alpha^2_\sigma T}{9 C_\sigma}} = \\ & = \frac{\left(\dfrac{1}{3 G} + \dfrac{1}{9 K}\right)^{-1}}{\left(\dfrac{1}{3 G} + \dfrac{1}{9 K} - \dfrac{\alpha^2_\sigma T}{9 C_\sigma}\right)\left(\dfrac{1}{3 G} + \dfrac{1}{9 K}\right)^{-1}} = \\ & = \frac{E}{1 - E \, \dfrac{\alpha^2_\sigma T}{9 C_\sigma}} = \\ \end{aligned}\end{split}\]

Fix admonition reference

6.2. Beam models#

6.3. Beam structures#