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where the _{0}S_{ij} and _{0}\epsilon_{rs} are the components of the second Piola-Kirchhoff stress and Green-Lagrange strain tensors and the _{0}C_{ijrs} are the components of the constant elasticity tensor. Considering three-dimensional stress conditions, we have
{ } _ { 0 } ^ { t } C _ { i j r s } = \lambda \delta _ { i j } \delta _ { r s } + \mu ( \delta _ { i r } \delta _ { j s } + \delta _ { i s } \delta _ { j r } ) \tag {6.185}
where \lambda and \mu are the Lamé constants and \delta_{ij} is the Kronecker delta,
\begin{array}{l} \lambda = \frac {E \nu}{(1 + \nu) (1 - 2 \nu)}; \quad \mu = \frac {E}{2 (1 + \nu)} \\ \delta_ {i j} = \left\{ \begin{array}{l l} 0; & \quad i \neq j \\ 1; & \quad i = j \end{array} \right. \\ \end{array}
The components of the elasticity tensor given in (6.185) are identical to the values given in Table 4.3 (see Exercise 2.10).
Considering this material description we can make a number of important observations. We recognize that in infinitesimal displacement analysis, the relation in (6.184) reduces to the description used in linear elastic analysis because under these conditions the stress and strain variables reduce to the engineering stress and strain measures. However, an important observation is that in large displacement and large rotation but small strain analysis, the relation in (6.184) provides a natural material description because the components of the second Piola-Kirchhoff stress and Green-Lagrange strain tensors do not change under rigid body rotations (see Section 6.2.2 and Examples 6.12 to 6.15). Thus, only the actual straining of material will yield an increase in the components of the stress tensor, and as long as this material straining (accompanied by large rotations and displacements) is small, the use of the relation (6.184) is completely equivalent to using Hooke's law in infinitesimal displacement conditions.
The fundamental observation that “the second Piola-Kirchhoff stress and Green-Lagrange strain components do not change measured in a fixed coordinate system when the material is subjected to rigid body motions” is important not only for elastic analysis. Indeed, this observation implies that any material description which has been developed for infinitesimal displacement analysis using engineering stress and strain measures can directly be employed in large displacement and large rotation but small strain analysis, provided second Piola-Kirchhoff stresses and Green-Lagrange strains are used. Figure 6.7 illustrates this fundamental fact. A practical consequence is, for example, that inelastic material models (see Section 6.6.3) can be directly used in large displacement, large rotation, infinitesimally small strain analysis by simply substituting second Piola-Kirchhoff stresses and Green-Lagrange strains for the engineering stress and strain measures.
The preceding observations are of special importance because, in practice, Hooke's law is applicable only to small strains and because there are many engineering problems in which large displacements, large rotations, but only small elastic strain conditions are encountered. This is, for example, frequently the case in the elastic buckling or collapse analysis of slender (beam or shell) structures.
The stress-strain description given in (6.184) implicitly assumes that a TL formulation is used to analyze the physical problem. Let us now assume that we want to employ a UL formulation but that we are given the constitutive relationship in (6.184). In this case we can write, substituting (6.184) into (6.72),
\int ^ {t} C _ {i j r s} ^ {t} \epsilon_ {r s} \delta_ {0} ^ {t} \epsilon_ {i j} d ^ {0} V = ^ {t} \mathcal {R} \tag {6.186}
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0x2, tx2 t̄τ̄22 t̄x̄2 t̄τ̄12 t̄τ̄11 t̄x̄1 t̄θ̄11 t̄γ̄12 β A Configuration at time t tX1|A tX2|A 0x1, tx1 Original configuration 1.0 1.0 tX1|A 0x1, tx1 tX1|A, tX2|A, β are large t̄ε11, t̄ε22, t̄γ12 << 1 δS11 = t̄τ11 δS22 = t̄τ22 δS12 = t̄τ12
Figure 6.7 Large displacement/large rotation but small strain conditions
Thus, if we define a new constitutive tensor,
{ } ^ { \prime } C _ { m n p q } = \frac { { } ^ { \prime } \rho } { { } ^ { 0 } \rho } { } ^ { \prime } x _ { m , i } { } ^ { \prime } { } ^ { 0 } x _ { n , j } { } ^ { \prime } { } ^ { 0 } C _ { i j r s } { } ^ { \prime } { } ^ { 0 } x _ { p , r } { } ^ { \prime } { } ^ { 0 } x _ { q , s } \tag {6.187}
meaning that t_0 C_{ijrs} = \frac{^0\rho}{^t\rho} t^0 x_{i,m} t^0 x_{j,n} t^t C_{mnpq} t^0 x_{r,p} t^0 x_{s,q} (6.188)
and if we use (see Example 6.10)
\delta_ {t} e _ {m n} = _ {t} ^ {0} x ^ {0} x _ {j, n} \delta_ {0} ^ {t} \epsilon_ {i j} \tag {6.189}
we recognize that (6.186) can be written as
\int_ {t _ {V}} ^ {t} C _ {m n p q} ^ {t} \epsilon_ {p q} ^ {A} \delta_ {t} e _ {m n} d ^ {t} V = ^ {t} \mathcal {R} \tag {6.190}
where \tau_{mn} = \tau_{C_{mnpq}} \epsilon_{pq}^{A} (6.191)
and the \epsilon_{pq}^{A} are the components of the Almansi strain tensor,
{ } _ { t } ^ { t } \epsilon _ { p q } ^ { A } = { } _ { t } ^ { 0 } x _ { r , p } { } _ { t } ^ { 0 } x _ { s , q } { } _ { 0 } ^ { t } \epsilon _ { r s } \tag {6.192}
Like the Green-Lagrange strain tensor, the Almansi strain tensor can also be defined in a number of different but completely equivalent ways, ^{11} namely,
{ } _ { i } \epsilon _ { i j } ^ { A } = \frac { 1 } { 2 } \left( { } _ { i } ^ { \prime } u _ { i , j } + { } _ { i } ^ { \prime } u _ { j , i } - { } _ { i } ^ { \prime } u _ { k , i } { } _ { i } ^ { \prime } u _ { k , j } \right) \tag {6.193}
and we have i\epsilon_{ij}^{A}d^{\prime}x_{i}d^{\prime}x_{j} = \frac{1}{2}\{(d^{\prime}s)^{2} - (d^{0}s)^{2}\} \tag{6.194}
EXAMPLE 6.22: Prove that the definitions of the Almansi strain tensor given in (6.192) to (6.194) are all equivalent.
The relation in (6.192) can be written in matrix form as
{ } _ { t } ^ { t } \boldsymbol { \epsilon } ^ { A } = { } _ { t } ^ { 0 } \mathbf { X } ^ { T } { } _ { 0 } ^ { t } \boldsymbol { \epsilon } { } _ { t } ^ { 0 } \mathbf { X } \tag {a}
But using (6.54) to substitute for \mathfrak{f}\in in (a) and recognizing that
{ } _ { t } ^ { 0 } \mathbf { X } \mathbf { \Phi } _ { 0 } \mathbf { X } = \mathbf { I }
we obtain \epsilon^{A}=\frac{1}{2}(\mathbf{I}-\mathbf{\Sigma}^{0}\mathbf{X}^{T},\mathbf{\Sigma}^{0}\mathbf{X}) (b)
However, we have ^{0}_{i}X = [_{i}\nabla^{0}x^{T}]^{T}
where, in accordance with (6.21),
, \boldsymbol {\nabla} = \left[ \begin{array}{l} \frac {\partial}{\partial^ {t} x _ {1}} \\ \frac {\partial}{\partial^ {t} x _ {2}} \\ \frac {\partial}{\partial^ {t} x _ {3}} \end{array} \right]; \quad^ {0} \mathbf {x} ^ {T} = \left[ \begin{array}{l l l} ^ {0} x _ {1} & ^ {0} x _ {2} & ^ {0} x _ {3} \end{array} \right]
Substituting into (b), we obtain
\epsilon^ {A} = \frac {1}{2} \left\{\mathbf {I} - \left[ _ {t} \nabla \left(^ {t} \mathbf {x} ^ {T} - ^ {t} \mathbf {u} ^ {T}\right) \right] \left[ _ {t} \nabla \left(^ {t} \mathbf {x} ^ {T} - ^ {t} \mathbf {u} ^ {T}\right) \right] ^ {T} \right\}
Since \nabla^{t}x^{T}=I
we thus obtain \mathbf{t}\mathbf{e}^{A} = \frac{1}{2} [\mathbf{I} - (\mathbf{I} - \mathbf{\nabla}^{\prime}\mathbf{u}^{T})(\mathbf{I} - \mathbf{\nabla}^{\prime}\mathbf{u}^{T})^{T}]
or i\epsilon^{A} = \frac{1}{2} [i\nabla^{\prime}\mathbf{u}^{T} + (i\nabla^{\prime}\mathbf{u}^{T})^{T} - (i\nabla^{\prime}\mathbf{u}^{T})(i\nabla^{\prime}\mathbf{u}^{T})^{T}] \tag{c}
and the components of \mathcal{I}\in^{A} in (c) are the relations in (6.193).
To show that (6.194) also holds, we use the relation in (b) to obtain
d ^ {\prime} \mathbf {x} ^ {T} \mathbf {\Phi} _ {i} \in^ {A} d ^ {\prime} \mathbf {x} = \frac {1}{2} \left(d ^ {\prime} \mathbf {x} ^ {T} d ^ {\prime} \mathbf {x} - d ^ {0} \mathbf {x} ^ {T} d ^ {0} \mathbf {x}\right) \tag {d}
because d^0\mathbf{x} = {}^0\mathbf{X}d'\mathbf{x}
But (d) can also be written as
d ^ {t} \mathbf {x} ^ {T} \mathbf {\Phi} _ {t} ^ {t} \in^ {A} d ^ {t} \mathbf {x} = \frac {1}{2} \left[ \left(d ^ {t} s\right) ^ {2} - \left(d ^ {0} s\right) ^ {2} \right] \tag {e}
because d^t\mathbf{x}^T d^t\mathbf{x} = (d^t s)^2; \quad d^0\mathbf{x}^T d^0\mathbf{x} = (d^0 s)^2
and (e) is equivalent to (6.194).
Of course, using (6.190) with the Almansi strain and the constitutive tensor \dot{t} C_{mnpq} is quite equivalent to transforming the second Piola-Kirchhoff stress \dot{t} S_{ij} (obtained using \dot{t} S_{ij} = \dot{t} C_{ijrs} \dot{t} \epsilon_{rs} ) to the Cauchy stress and then using (6.13) to evaluate \dot{\mathcal{R}} . Indeed, if \dot{t} C_{ijrs} is known, this procedure is computationally more efficient, and the definition and use of the Almansi strain with (6.190) may be regarded as only of theoretical interest.
However, in the following example we prove an important result, which can be stated in summary as follows.
Consider the TL and UL formulations in Tables 6.2 and 6.3,
\int C _ {i j r s 0} e _ {r s} \delta_ {0} e _ {i j} d ^ {0} V + \int ^ {t} S _ {i j} \delta_ {0} \eta_ {i j} d ^ {0} V = ^ {t + \Delta r} \mathcal {R} - \int ^ {t} S _ {i j} \delta_ {0} e _ {i j} d ^ {0} V \tag {6.195}
\int_ {t _ {V}} ^ {t} C _ {i j r s t} e _ {r s} \delta_ {t} e _ {i j} d ^ {t} V + \int_ {t _ {V}} ^ {t} \tau_ {i j} \delta_ {t} \eta_ {i j} d ^ {t} V = ^ {t + \Delta t} \mathcal {R} - \int_ {t _ {V}} ^ {t} \tau_ {i j} \delta_ {t} e _ {i j} d ^ {t} V \tag {6.196}
The corresponding integral terms in the formulations are identical provided the transformations for the stresses given in (6.69) and for the constitutive tensors given in (6.187) are used. Hence, whether we choose the TL or the UL continuum formulation is decided merely by considerations of numerical efficiency.
EXAMPLE 6.23: Consider the total and updated Lagrangian formulations in incremental form (see Tables 6.2 and 6.3).
(a) Derive the relationship that should be satisfied between the tensors _{0}C_{ijrs} and _{t}C_{ijrs} so that the incremental relations
_ 0 S _ {i j} = _ {0} C \epsilon_ {r s} \tag {a}
and _{t}S_{ij} = _{t}C_{ijrs} \epsilon_{rs} (b)
refer to the same physical material response.
(b) Show that when the relationship derived in part (a) is satisfied, each integral term in the linearized TL formulation is identical to its corresponding term in the UL formulation.
A constitutive law relates a stress measure to a strain measure. Since there are different stress and corresponding strain measures, the constitutive law for a given material may take different forms, but these forms are related by the fact that they all describe the same given material. Hence, if equations (a) and (b) describe the same material, _{0}C_{ijrs} and _{1}C_{ijrs} must be related by purely kinematic transformations.
To derive the kinematic transformations we express _tS_{ij} in terms of _0S_{ij} , and _t\epsilon_{rs} in terms of _0\epsilon_{rs} .
We have _{i}S_{ij} = ^{i+\Delta_{i}^{\prime}}S_{ij} - ^{i}\tau_{ij} (c)
Using ^{'}\tau_{ij} = \frac{^{'}\rho}{^0\rho} _{0}x_{i,r} _{0}S_{rs} _{0}x_{j,s} (d)
and ^{t + \Delta t}{}_{t}S_{ij} = \frac{t\rho}{0}\rho _{0}^{t}x_{i,r}^{t + \Delta t}{}_{0}S_{rs} _{0}^{t}x_{j,s}
and (c), we obtain
{ } _ { t } S _ { i j } = \frac { { } ^ { t } \rho } { { } _ { 0 } \rho } { } _ { 0 } ^ { t } x _ { i , r } { } _ { 0 } ^ { t } x _ { j , s } { } _ { 0 } S _ { r s } \tag {e}
We also have for the strain terms
_ 0 \epsilon_ {i j} = ^ {1 + \Delta_ {0} ^ {I}} \epsilon_ {i j} - _ {0} ^ {I} \epsilon_ {i j}
and \epsilon_{ij} = t + \Delta t \epsilon_{ij}
Hence, 0\epsilon_{ij} = \frac{1}{2}\left(t^{+}\Delta t_{0}x_{k,i}^{t + \Delta t_{0}}x_{k,j} - t_{0}x_{k,i}t_{0}x_{k,j}\right) (f)
and \epsilon_{ij}=\frac{1}{2}\left(^{t+\Delta t}_{t}x_{k,i}{}^{t+\Delta t}_{t}x_{k,j}-\delta_{ij}\right) (g)
We should note here that \epsilon_{ij} is the Green-Lagrange strain based on the displacements from time t to time t + \Delta t , with the reference configuration at time t.^{12}
Using (f) and (g), we obtain
\begin{array}{l} _ 0 \epsilon_ {i j} = \frac {1}{2} _ {0} ^ {t} x ^ {t} x _ {q, j} \left(^ {t + \Delta_ {t}} _ {t} x _ {k, p} ^ {t + \Delta_ {t}} _ {t} x _ {k, q} - \delta_ {p q}\right) \\ = \delta x _ {p, i} \delta x _ {q, j}, \epsilon_ {p q} \tag {h} \\ \end{array}
We may now use (e) and (h) in the material law (b), which gives
\frac {^ {t} \rho}{^ {0} \rho} ^ {t} x _ {i, a} ^ {t} x _ {j, b} ^ {0} S _ {a b} = _ {t} C _ {i j r s} ^ {0} x _ {p, r} ^ {0} x _ {q, s} ^ {0} \epsilon_ {p q}
or _0S_{ij} = \left(\frac{^0\rho}{^t\rho}{}^0 x_{i,m}{}^0{}^0 x_{j,n}{}^0{}^t C_{mnpq}{}^0{}^0{}^r x_{r,p}{}^0{}^t x_{s,q}\right)_0 \epsilon_{rs}
Hence, for the same material to be described, the relation between the constitutive tensors is
{ } _ { 0 } C _ { i j r s } = \frac { { } ^ { 0 } \rho } { { } ^ { i } \rho } { } ^ { 0 } x _ { i , m } { } ^ { 0 } x _ { j , n } { } ^ { i } C _ { m n p q } { } ^ { 0 } x _ { r , p } { } ^ { 0 } x _ { s , q } \tag {i}
We note that the same material law transformation as earlier stated in (6.188) must be used if C_{ijrs} is known and the TL formulation with (a) is to be employed. Of course, the transformation in (6.187) would be applicable if _{0}C_{ijrs} were known and the UL formulation were to be used.
Next, we want to show that each term in the TL formulation is identical to its corresponding term in the UL formulation. Considering the right-hand sides, r^{+\Delta t}\mathcal{R} is of course the same in both formulations, and
\int_ {0 _ {V}} \delta_ {0} S _ {i j} \delta_ {0} e _ {i j} d ^ {0} V = \int_ {t _ {V}} ^ {t} \tau_ {i j} \delta_ {t} e _ {i j} d ^ {t} V
because \delta_0e_{ij} = \delta_0^t\epsilon_{ij} .
The fact that \delta_0e_{ij} = \delta_0^t\epsilon_{ij} needs some explanation. In this evaluation of \delta_0^t\epsilon_{ij} we calculate the variation in the Green-Lagrange strain corresponding to the configuration at time t , and the equation says that this value is equal to the linear strain increment corresponding to the virtual displacement \delta u_i . Let us recall that when taking the variation about the configuration at time t + \Delta t , we used \delta^{+ \Delta t}_{0}\epsilon_{ij} = \delta_{0}e_{ij} + \delta_{0}\eta_{ij} (see Table 6.2). If the incremental displacements are zero, i.e., u_i = 0 , then the configuration at time t + \Delta t is identical to the configuration at time t . Hence,
\delta^ {t + \Delta t} _ {0} \epsilon_ {i j} | _ {u _ {i} = 0} = \delta_ {0} ^ {t} \epsilon_ {i j}
It follows that, considering \delta u_{i} as a variation on u_{i} ,
\begin{array}{l} \delta^ {t + \Delta t} _ {0} \epsilon_ {i j} | _ {u _ {i} = 0} = \delta_ {0} ^ {t} \epsilon_ {i j} + \delta_ {0} \epsilon_ {i j} | _ {u _ {i} = 0} \\ = \delta_ {0} \epsilon_ {i j} | _ {u _ {i} = 0} \quad \text { here } \delta_ {0} ^ {\prime} \epsilon_ {i j} = 0 \text { because } _ {0} ^ {\prime} \epsilon_ {i j} \text { is independent of } u _ {i}. \\ = \delta_ {0} e _ {i j} | _ {u _ {i} = 0} + \delta_ {0} \eta_ {i j} | _ {u _ {i} = 0} \\ = \delta_ {0} e _ {i j} \\ \end{array}
because \delta_0\eta_{ij} is a linear function in u_{i} and therefore \delta_0\eta_{ij}|_{u_i = 0} = 0 .
Next, we prove that
\int_ {0 _ {V}} \delta S _ {i j} \delta_ {0} \eta_ {i j} d ^ {0} V = \int_ {t _ {V}} ^ {t} \tau_ {i j} \delta_ {t} \eta_ {i j} d ^ {t} V \tag {j}
However, from (h) it also follows that, grouping the terms nonlinear in the incremental displacements u_{i} ,
{ } _ { 0 } \eta _ { i j } = { } _ { 0 } ^ { \prime } x _ { p , i } { } _ { 0 } ^ { \prime } x _ { q , j } { } _ { t } \eta _ { p q }
and hence,
\delta_ {0} \eta_ {i j} = _ {0} ^ {t} x ^ {t} x _ {q, j} \delta_ {t} \eta_ {p q} \tag {k}
Substituting from (d) and (k) into (j), with appropriate changes on indices, directly verifies (j).
Finally, we prove that
\int C _ {i j r s 0} e _ {r s} \delta_ {0} e _ {i j} d ^ {0} V = \int C _ {i j r s t} e _ {r s} \delta_ {t} e _ {i j} d ^ {t} V \tag {1}
Here we again use (h), which also gives
{ } _ { 0 } e _ { i j } = { } _ { 0 } ^ { \prime } x _ { p , i } { } _ { 0 } ^ { \prime } x _ { q , j } { } _ { t } e _ { i j } \tag {m}
and hence, \delta_0e_{ij} = \delta x_{p,i} \delta x_{q,j} \delta_t e_{ij} (n)
Substituting from (i), (m), and (n) with appropriate changes on indices, into (l) also directly verifies (l).
In summary, we note that if _{0}C_{ijrs} and _{t}C_{ijrs} in the incremental stress-strain relations (a) and (b) are for any material related in such a way as to represent the same physical material response, then the TL and UL incremental continuum mechanics formulations are identical. This observation pertains not only to elastic materials but is general and holds for any material.
The preceding discussion shows that the UL formulation may be used if the constitutive relationship corresponding to the TL formulation is known (and this observation holds for any material law that can be written in the form used in the TL formulation), and vice versa. This equivalence between the UL and TL formulations of course holds for any level of strain, but in most practical analyses, the linear elastic material behavior (Hooke's law) is valid only for small strain conditions. In that case, for an isotropic elastic material, the results using either (6.184) and (6.185), or directly
^ \prime \tau_ {i j} = ^ {\prime} C _ {i j r s} ^ {\prime} \epsilon_ {r s} ^ {A} \tag {6.197}
{ } _ { i } ^ { \prime } C _ { i j r s } = \lambda \delta _ { i j } \delta _ { r s } + \mu \left( \delta _ { i r } \delta _ { j s } + \delta _ { i s } \delta _ { j r } \right) \tag {6.198}
\lambda = \frac {E \nu}{(1 + \nu) (1 - 2 \nu)}; \quad \mu = \frac {E}{2 (1 + \nu)}
where \lambda and \mu are the same constants as in (6.185), are practically the same. Hence, the same constants can be used to define the material law for the total and updated Lagrangian formulations, and only small differences would be observed in the solution results for large displacements and rotations as long as the strains are small. The reason is that considering large displacements and rotations but small strains, the transformations on the constitutive tensors given in (6.187) and (6.188) reduce to mere rotations. Therefore, since the material is assumed to be isotropic, the transformations do not change the components of the constitutive tensors and the use of either (6.184) and (6.185) or (6.197) and (6.198) to characterize the material response is quite equivalent.
However, when large strains are modeled using (6.184) and (6.197) with the same elastic material constants, completely different response predictions must be expected. Figure 6.8 shows the results obtained in a simple analysis of this kind. We note that the force-displacement response is totally different when using the two descriptions and that an instability is observed in the TL formulation at the displacement (-2 + 2/\sqrt{3}) .
line
| tU (cm) | tP/A̅ (10⁸ N/cm²) |
|---|---|
| -2.0 | 0.0 |
| -1.5 | 0.0 |
| -1.0 | 0.0 |
| -0.5 | 0.0 |
| 0.5 | 0.2 |
| 1.0 | 0.6 |
| 1.5 | 1.0 |
| 2.0 | 1.4 |
| 2.0 | -0.2 |
(a) P - \Delta response of 8-node element under uniform loading (E = 10^{7} \mathrm{~N/cm}^{2}, \nu = 0.30) , \overline{A} is restrained to be constant.
\tilde {E} = \frac {E (1 - \nu)}{(1 + \nu) (1 - 2 \nu)}
(i) Using \delta S_{11} = \tilde{E}_0^\prime \epsilon_{11}
^ {\prime} P = \frac {\tilde {E} \bar {A}}{2} \left(1 + \frac {^ {\prime} u}{^ {0} L}\right) \left(\left(1 + \frac {^ {\prime} u}{^ {0} L}\right) ^ {2} - 1\right)
(ii) Using ^t\tau_{11} = \tilde{E}^t\epsilon_{11}^A
^ \prime P = \frac {\tilde {E} \overline {{A}}}{2} \left(1 - \left(\frac {{} ^ {0} L}{^ {0} L + ^ {\prime} u}\right) ^ {2}\right)
(b) Basic relations
Figure 6.8 One-dimensional response analysis; here the TL formulation is unstable in large compressive strain
Of course, if the transformations of (6.187) or (6.188) , whichever are applicable, were performed, the same force-displacement curves would be obtained using either the total or the updated Lagrangian formulations (see Exercise 6.62).
Let us further demonstrate in the following example the differences in response that are observed when using different stress-strain measures with the same material constants.
EXAMPLE 6.24: Consider the four-node element shown in Fig. E6.24. The displacements of the element are given as a function of time.
Calculate the Cauchy stresses using the following two stress measures:
(i) Use the total formulation of the second Piola-Kirchhoff stress and Green-Lagrange strain tensors,
\delta S _ {i j} = \delta C _ {i j r s} \delta \epsilon_ {r s} \tag {a}
(ii) Use the rate formulation of the Jaumann stress rate and the velocity strain tensors (see L. E. Malvern [A]),
{ } ^ { t } \bar { \tau } _ { i j } = { } _ { t } C _ { i j r s } { } ^ { t } D _ { r s } \tag {b}
and let the constitutive tensors \delta C_{ijrs} and tC_{ijrs} be given by the same matrix in Table 4.2.
We note that the components of the Jaumann stress rate tensor are given by (see L. E. Malvern [A])
{ } ^ { t } \dot { \tau } _ { i j } = { } ^ { t } \dot { \tau } _ { i j } + { } ^ { t } \tau _ { i p } { } ^ { t } W _ { p j } + { } ^ { t } \tau _ { j p } { } ^ { t } W _ { p i } \tag {c}
where the ^{t}W_{ij} are the components of the spin tensor [see (6.43)]. The relation (c) expresses that the rate of change of the Cauchy stress, ^{t}\tau_{ij} , is equal to the Jaumann stress rate (which gives the rate of change in Cauchy stress due to material straining) plus the effect of rate of rigid body rotation of the material (and hence, rate of rotation of stress). The Jaumann stress rate is used in practice, although the rate formulation results in numerical integration errors and a nonphysical behavior (see Section 6.6 and M. Kojić and K. J. Bathe [A]).
E = 5 0 0 0
v = 0. 3 0
Figure E6.24 Four-node element subjected to motion
Consider case (i). The deformation gradient is
_ {0} ^ {t} \mathbf {X} = \left[ \begin{array}{c c} 1 & 2 t \\ 0 & 1 \end{array} \right]
and hence,
_ {0} ^ {t} \epsilon = \left[ \begin{array}{c c} 0 & t \\ t & 2 t ^ {2} \end{array} \right]
Using Table 4.2 with the given values of E and \nu , we obtain as nonzero values C_{1111} = 6731 , C_{2211} = C_{1122} = 2885 , C_{2222} = 6731 , C_{1212} = 1923 .
Hence, using the total Lagrangian description, we have
_ {0} ^ {t} S _ {1 1} = 5 7 7 0 t ^ {2}; \quad_ {0} ^ {t} S _ {2 2} = 1 3, 4 6 2 t ^ {2}; \quad_ {0} ^ {t} S _ {1 2} = 3 8 4 6 t
and using the standard transformations between the second Piola-Kirchhoff and Cauchy stresses (to two significant figures),
\left[ \begin{array}{l} ^ {t} \tau_ {1 1} \\ ^ {t} \tau_ {2 2} \\ ^ {t} \tau_ {1 2} \end{array} \right] = \left[ \begin{array}{l} 2 1, 0 0 0 t ^ {2} + 5 4, 0 0 0 t ^ {4} \\ 1 3, 0 0 0 t ^ {2} \\ 3 8 0 0 t + 2 7, 0 0 0 t ^ {3} \end{array} \right] \tag {d}
Next consider case (ii). The velocity strain tensor \mathbf{D} is computed as given in (6.42). Hence,
^ {\prime} \mathbf {L} = \left[ \begin{array}{c c} 0 & 2 \\ 0 & 0 \end{array} \right]; \qquad^ {\prime} \mathbf {D} = \left[ \begin{array}{c c} 0 & 1 \\ 1 & 0 \end{array} \right]; \qquad^ {\prime} \mathbf {W} = \left[ \begin{array}{c c} 0 & + 1 \\ - 1 & 0 \end{array} \right]
Now we use the same constitutive matrix C to obtain
\left[ \begin{array}{l} ^ {t} \overline {{T}} _ {1 1} \\ ^ {t} \overline {{T}} _ {2 2} \\ ^ {t} \overline {{T}} _ {1 2} \end{array} \right] = \left[ \begin{array}{l} 0 \\ 0 \\ 3 8 4 6 \end{array} \right]
We note that the Jaumann stress rate is independent of time. However, the material also rotates as expressed in 'W' and the time rates of the Cauchy stress components are given by
\left[ \begin{array}{l} ^ {\prime} \dot {\tau} _ {1 1} \\ ^ {\prime} \dot {\tau} _ {2 2} \\ ^ {\prime} \dot {\tau} _ {1 2} \end{array} \right] = \left[ \begin{array}{c} 2 ^ {\prime} \tau_ {1 2} \\ - 2 ^ {\prime} \tau_ {1 2} \\ 3 8 4 6 + ^ {\prime} \tau_ {2 2} - ^ {\prime} \tau_ {1 1} \end{array} \right]
These differential equations can be solved to obtain (again to two significant figures and hence using G = \frac{E}{2(1 + \nu)} \doteq 1900 )
\left[ \begin{array}{l} ^ {t} \tau_ {1 1} \\ ^ {t} \tau_ {2 2} \\ ^ {t} \tau_ {1 2} \end{array} \right] = \left[ \begin{array}{c} 1 9 0 0 (1 - \cos 2 t) \\ - 1 9 0 0 (1 - \cos 2 t) \\ 1 9 0 0 \sin 2 t \end{array} \right] \tag {e}
We note that the results given in (d) and (e) are quite different when t is larger than about 0.1 and that in each material description normal stresses are generated (that are zero when infinitesimally small strains are assumed). Also, the oscillatory behavior of the Cauchy stresses in (e) with period \pi is peculiar.
6.6.2 Rubberlike Material Behavior
We introduced in Section 6.4 the displacement/pressure formulations that are much suited for the analysis of rubberlike materials because such materials exhibit an almost incompressible response. The basic ingredient in these formulations is the strain energy density \delta \overline{W} , which is defined by the specific material model used.
Various definitions of \delta \overline{W} are available, but two commonly used models are the Mooney-Rivlin and Ogden models (see R. S. Rivlin [A] and R. W. Ogden [A]).
The conventional Mooney-Rivlin material model is described by the strain energy density per unit original volume
\dot {0} \tilde {W} = C _ {1} (\dot {0} I _ {1} - 3) + C _ {2} (\dot {0} I _ {2} - 3); \quad \dot {0} I _ {3} = 1 \tag {6.199}
where C_{1} and C_{2} are material constants and the invariants _{0}^{t}I_{i} are given in terms of the components of the Cauchy-Green deformation tensor (see 6.27)
_ {0} ^ {t} I _ {1} = _ {0} ^ {t} C _ {k k}
\delta I _ {2} = \frac {1}{2} \left[ (\delta I _ {1}) ^ {2} - \delta C _ {i j} \delta C _ {i j} \right] \tag {6.200}
_ {0} ^ {t} I _ {3} = \det _ {0} ^ {t} \mathbf {C}
Note that the value \delta \widetilde{W} in (6.199) is not yet a strain energy density \delta \overline{W} that we use in our formulation, as we discuss below.
We note that a so-called neo-Hookean material description is obtained with C_{2}=0 , and if small strains are considered 2(C_{1}+C_{2}) is the shear modulus and 6(C_{1}+C_{2}) is the Young's modulus.
EXAMPLE 6.25: Consider the one-dimensional response of the bar shown in Fig. E6.25. Plot the force-displacement relationship for the following two cases: (i) C_1 = 100 , C_2 = 0 and (ii) C_1 = 75 , C_2 = 25 .
line
| Displacement Δ/⁰L | F/⁰A (C₁ = 100) | F/⁰A (C₂ = 0) | F/⁰A (C₁ = 75) | F/⁰A (C₂ = 25) |
|---|---|---|---|---|
| -0.5 | -6.0 | -6.5 | -7.0 | -7.5 |
| 0.0 | 0.0 | 0.0 | 0.0 | 0.0 |
| 0.5 | 2.0 | 2.5 | 3.0 | 3.5 |
| 1.0 | 4.0 | 4.5 | 5.0 | 5.5 |
| 1.5 | 5.5 | 6.0 | 6.5 | 7.0 |
| 2.0 | 6.5 | 7.0 | 7.5 | 8.0 |
| 2.5 | 7.5 | 8.0 | 8.5 | 9.0 |
| 3.0 | 8.0 | 8.5 | 9.0 | 9.5 |
Figure E6.25 One-dimensional response of rubber bar
Using \mathfrak{t}_0 S_{ij} = \partial_0^t \tilde{W} / \partial_0^t \epsilon_{ij} (see 6.129) with the Mooney-Rivlin material model in (6.199) specialized to the case considered, we obtain (using the stretch \lambda as the only variable to evaluate \mathfrak{t}_0^t \tilde{W} ).
F = 2 ^ {0} A \left[ C _ {1} \left(\lambda - \lambda^ {- 2}\right) + C _ {2} \left(1 - \lambda^ {- 3}\right) \right]
\lambda = 1 + \frac {\Delta}{^ 0 L}
Substituting the values for C_1 and C_2 , we obtain the curves shown in the figure.
The description in (6.199) assumes that the material is totally incompressible (since \delta I_{3} = 1 ). A better assumption is that the bulk modulus is several thousand times as large as the shear modulus, which then means that the material is almost incompressible. This assumption is incorporated by dropping the restriction that \delta I_{3} = 1 and including a hydrostatic work term in the strain energy function to obtain
\dot {0} \tilde {\tilde {W}} = C _ {1} \left(_ {0} ^ {\prime} I _ {1} - 3\right) + C _ {2} \left(_ {0} ^ {\prime} I _ {2} - 3\right) + W _ {H} \left(_ {0} ^ {\prime} I _ {3}\right) \tag {6.201}
However, this expression cannot be used directly in the displacement/pressure formulations because all three terms contribute to the pressure. To obtain an appropriate expression, we