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EXAMPLE 6.5: Consider the general motion of the body in Fig. 6.2 and establish that the mass density of the body changes as a function of the determinant of the deformation gradient,
^ \prime \rho = \frac {{} ^ {0} \rho}{\det \left(_ {0} ^ {t} \mathbf {X}\right)}
Any infinitesimal volume of material at time 0 can be represented using (see Fig. E6.5)
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Time 0 d⁰Ṽ d⁰x̃₃ d⁰x̃₂ d⁰x̃₁ dᵗṼ dᵗx̃₃ dᵗx̃₂ dᵗx̃₁ Time t x₃ x₂ x₁
Figure E6.5 Infinitesimal volumes at times 0 and t
d ^ {0} \tilde {\mathbf {x}} _ {1} = \left[ \begin{array}{l} 1 \\ 0 \\ 0 \end{array} \right] d s _ {1}; \quad d ^ {0} \tilde {\mathbf {x}} _ {2} = \left[ \begin{array}{l} 0 \\ 1 \\ 0 \end{array} \right] d s _ {2}; \quad d ^ {0} \tilde {\mathbf {x}} _ {3} = \left[ \begin{array}{l} 0 \\ 0 \\ 1 \end{array} \right] d s _ {3}
and d^0\tilde{\mathbf{V}} = ds_1ds_2ds_3
Using (6.22), we have after deformation,
d ^ {t} \tilde {\mathbf {x}} _ {i} = _ {0} ^ {t} \mathbf {X} d ^ {0} \tilde {\mathbf {x}} _ {i} \quad i = 1, 2, 3
where we note that of course the same deformation gradient applies to all material fibers of that infinitesimal volume, and we obtain
\begin{array}{l} d ^ {t} \tilde {V} = \left(d ^ {t} \tilde {\mathbf {x}} _ {1} \times d ^ {t} \tilde {\mathbf {x}} _ {2}\right) \cdot d ^ {t} \tilde {\mathbf {x}} _ {3} \\ = (\det _ {0} ^ {t} \mathbf {X}) d s _ {1} d s _ {2} d s _ {3} \\ = \det _ {0} ^ {t} \mathbf {X} d ^ {0} \tilde {V} \\ \end{array}
But if we assume that no mass is lost during the deformation, we have
^ \prime \rho d ^ {\prime} \tilde {V} = ^ {0} \rho d ^ {0} \tilde {V}
and hence, t\rho = \frac{^0\rho}{\operatorname*{det}_0^t\mathbf{X}}
EXAMPLE 6.6: Consider the element in Fig. E6.6. Evaluate the deformation gradient and the mass density corresponding to the configuration at time t.
The displacement interpolation functions for this element were given in Fig. 5.4. Since the ^0 x_1 , ^0 x_2 axes correspond to the r , s axes, respectively, we have
h _ {1} = \frac {1}{4} (1 + ^ {0} x _ {1}) (1 + ^ {0} x _ {2}); \quad h _ {2} = \frac {1}{4} (1 - ^ {0} x _ {1}) (1 + ^ {0} x _ {2})
h _ {3} = \frac {1}{4} (1 - ^ {0} x _ {1}) (1 - ^ {0} x _ {2}); \quad h _ {4} = \frac {1}{4} (1 + ^ {0} x _ {1}) (1 - ^ {0} x _ {2})
and \frac{\partial h_1}{\partial^0 x_1} = \frac{1}{4} (1 + ^0 x_2); \frac{\partial h_2}{\partial^0 x_1} = -\frac{1}{4} (1 + ^0 x_2)
\frac {\partial h _ {3}}{\partial^ {0} x _ {1}} = - \frac {1}{4} (1 - ^ {0} x _ {2}); \quad \frac {\partial h _ {4}}{\partial^ {0} x _ {1}} = \frac {1}{4} (1 - ^ {0} x _ {2})
\frac {\partial h _ {1}}{\partial^ {0} x _ {2}} = \frac {1}{4} (1 + ^ {0} x _ {1}); \quad \frac {\partial h _ {2}}{\partial^ {0} x _ {2}} = \frac {1}{4} (1 - ^ {0} x _ {1})
\frac {\partial h _ {3}}{\partial^ {0} x _ {2}} = - \frac {1}{4} (1 - ^ {0} x _ {1}); \quad \frac {\partial h _ {4}}{\partial^ {0} x _ {2}} = - \frac {1}{4} (1 + ^ {0} x _ {1})
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Thickness = 1 cm Density at t = 0 is 0p 0x2 2 1 v1 = 0.5 cm 2 cm d0's d's d' s d' s u1 = 1 cm 0x1 3 4 2 cm
Figure E6.6 Four-node element subjected to large deformations
Now we use ^t x_i = \sum_{k=1}^{4} h_k ^t x_i^k
and hence, \frac{\partial^t x_i}{\partial^0 x_j} = \sum_{k=1}^{4} \left( \frac{\partial h_k}{\partial^0 x_j} \right)^t x_i^k
The nodal point coordinates at time t are
^ {\prime} x _ {1} ^ {1} = 2; \quad^ {\prime} x _ {2} ^ {1} = 1. 5; \quad^ {\prime} x _ {1} ^ {2} = - 1; \quad^ {\prime} x _ {2} ^ {2} = 1
^ {\prime} x _ {1} ^ {3} = - 1; \quad^ {\prime} x _ {2} ^ {3} = - 1; \quad^ {\prime} x _ {1} ^ {4} = 1; \quad^ {\prime} x _ {2} ^ {4} = - 1
Hence,
\begin{array}{l} \frac {\partial^ {t} x _ {1}}{\partial^ {0} x _ {1}} = \frac {1}{4} \left[ \left(1 + ^ {0} x _ {2}\right) (2) - \left(1 + ^ {0} x _ {2}\right) (- 1) - \left(1 - ^ {0} x _ {2}\right) (- 1) + \left(1 - ^ {0} x _ {2}\right) (1) \right] \\ = \frac {1}{4} (5 + ^ {0} x _ {2}) \\ \end{array}
and \frac{\partial^t x_1}{\partial^0 x_2} = \frac{1}{4} (1 + ^0 x_1); \quad \frac{\partial^t x_2}{\partial^0 x_1} = \frac{1}{8} (1 + ^0 x_2)
\frac {\partial^ {t} x _ {2}}{\partial^ {0} x _ {2}} = \frac {1}{8} (9 + ^ {0} x _ {1})
so that the deformation gradient is
_ {0} ^ {t} \mathbf {X} = \frac {1}{4} \left[ \begin{array}{l l} (5 + ^ {0} x _ {2}) & (1 + ^ {0} x _ {1}) \\ \frac {1}{2} (1 + ^ {0} x _ {2}) & \frac {1}{2} (9 + ^ {0} x _ {1}) \end{array} \right]
and using (6.26), the mass density in the deformed configuration is
^ t \rho = \frac {3 2 ^ {0} \rho}{(5 + ^ {0} x _ {2}) (9 + ^ {0} x _ {1}) - (1 + ^ {0} x _ {1}) (1 + ^ {0} x _ {2})}
The deformation gradient is also used to measure the stretch of a material fiber and the change in angle between adjacent material fibers due to the deformation. In this calculation we use the right Cauchy-Green deformation tensor,
\boxed { \begin{array}{l} \dot {\mathbf {\Phi}} _ {0} \mathbf {C} = \dot {\mathbf {\Phi}} _ {0} \mathbf {X} ^ {T} \dot {\mathbf {\Phi}} \mathbf {X} \end{array} } \tag {6.27}
We note that \delta^{\prime}\mathbf{C} is, in general, not equal to the left Cauchy-Green deformation tensor,
{ } _ { 0 } ^ { t } \mathbf { B } = { } _ { 0 } ^ { t } \mathbf { X } { } _ { 0 } ^ { t } \mathbf { X } ^ { T } \tag {6.28}
EXAMPLE 6.7: The stretch \lambda of a line element of a general body in motion is defined as \lambda = d's / d^0 s , where d^0 s and d's are the original and current lengths of the line element as shown in Fig. E6.7. Prove that
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0x3, tx3 dts 0n tλ = dt/s d0s d0x3 d0x2 d0x1 0x2, tx2 0x1, tx1 0x3, tx3 0x3, tx3 0n dts 0n d0s 0θ dts 0x2, tx2 0x1, tx1
Figure E6.7 Stretch and rotation of line elements
^ \prime \lambda = (^ {0} \mathbf {n} ^ {T} _ {0} ^ {\prime} \mathbf {C} ^ {0} \mathbf {n}) ^ {1 / 2} \tag {a}
where ^{0}n is a vector of the direction cosines of the line element at time 0. Also, prove that considering two line elements emanating from the same material point, the angle ^{1}\theta between the line elements at time t is given by
\cos^ {\prime} \theta = \frac {{} ^ {0} \mathbf {n} ^ {T} {} _ {0} ^ {\prime} \mathbf {C} {} ^ {0} \hat {\mathbf {n}}}{{} ^ {\prime} \lambda^ {\prime} \hat {\lambda}} \tag {b}
where the hat denotes the second line element (see Fig. E6.7).
As an example, apply the formulas in (a) and (b) to evaluate the stretches of the specific line elements d^0 s and d^t \hat{s} shown in Fig. E6.6 and evaluate also the angular distortion between them.
To prove (a), we recognize that
(d ^ {t} s) ^ {2} = d ^ {t} \mathbf {x} ^ {T} d ^ {t} \mathbf {x}; \quad d ^ {t} \mathbf {x} = _ {0} ^ {t} \mathbf {X} d ^ {0} \mathbf {x}
so that using (6.27), (d^{\prime}s)^{2} = d^{0}\mathbf{x}^{T} _{0}\mathbf{C} d^{0}\mathbf{x}
Hence, ^{t}\lambda^{2}=\frac{d^{0}\mathbf{x}^{T}}{d^{0}s}\mathbf{0}\mathbf{C}\frac{d^{0}\mathbf{x}}{d^{0}s}
and since ^0\mathbf{n} = \frac{d^0\mathbf{x}}{d^0s}
we have \lambda = (^{0}n^{T}{}_{0}{}^{t}C{}^{0}n)^{1/2}
To prove (b) we use (2.50)
d ^ {\prime} \mathbf {x} ^ {T} d ^ {\prime} \hat {\mathbf {x}} = (d ^ {\prime} s) (d ^ {\prime} \hat {s}) \cos^ {\prime} \theta
Hence, \cos^t\theta = \frac{d^0\mathbf{x}^T_0^\prime\mathbf{X}^T_0^\prime\hat{\mathbf{X}} d^0\hat{\mathbf{x}}}{(d^t s)(d^t\hat{s})} (c)
Since \mathbf{^t}\mathbf{X}\equiv \mathbf{^t}\hat{\mathbf{X}} (it is the deformation gradient at the location of the differential line elements), we obtain from (c),
\cos^ {\prime} \theta = \frac {{} ^ {0} \mathbf {n} ^ {T} {} _ {0} ^ {t} \mathbf {C} {} ^ {0} \hat {\mathbf {n}}}{^ {\prime} \lambda^ {\prime} \hat {\lambda}}
It should be noted that the relations in (a) and (b) show that when \delta \mathbf{C} = \mathbf{I} , the stretches of the line elements are equal to 1 and the angle between line elements has not changed during the motion. Hence, when the Cauchy-Green deformation tensor is equal to the identity matrix, the motion could have been at most a rigid body motion.
If we apply (a) and (b) to the line elements depicted in Fig. E6.6, we obtain at ^0 x_1 = 0 , ^0 x_2 = 0 (see Example 6.6)
\mathbf {\delta} _ {0} ^ {t} \mathbf {C} = \frac {1}{1 6} \left[ \begin{array}{c c} 2 5. 2 5 & 7. 2 5 \\ 7. 2 5 & 2 1. 2 5 \end{array} \right]
{ } ^ { 0 } \mathbf { n } = \left[ \begin{array} { c } 1 \\ 0 \end{array} \right] ; \quad { } ^ { 0 } \hat { \mathbf { n } } = \left[ \begin{array} { c } 0 \\ 1 \end{array} \right]
Hence, using (a), \lambda = 1.256; \hat{\lambda} = 1.152
and using (b), \cos^{\prime}\theta = 0.313; ^\prime \theta = 71.75^{\circ}
Therefore, the angular distortion between the line elements d^{0}s and d^{0}\hat{s} due to the motion from time 0 to time t is 18.25 degrees.
A most important property of the deformation gradient is that it can always be decomposed into a unique product of two matrices, a symmetric stretch matrix \delta \mathbf{U} and an orthogonal matrix \delta \mathbf{R} corresponding to a rotation such that
\delta \mathbf {X} = \delta \mathbf {R} \delta \mathbf {U} \tag {6.29}
We can interpret (6.29), conceptually, to mean that the total deformation is obtained by first applying the stretch and then the rotation. That is, we could write (6.29) also as \delta \mathbf{X} = \tau \mathbf{R} \delta \mathbf{U} , where \tau corresponds to an intermediate (conceptual) time. Then we realize that the decomposition is really an application of the chain rule \delta \mathbf{X} = \tau \mathbf{X} \delta \mathbf{X} , where \tau \mathbf{X} \equiv \tau \mathbf{R} and \delta \mathbf{X} \equiv \delta \mathbf{U} . However, the state corresponding to \tau is only conceptual, and we therefore usually use the notation in (6.29).
The relation in (6.29) is referred to as the polar decomposition of the deformation gradient, and we prove and demonstrate this property in the following examples.
To simplify the notation in the following discussion of continuum mechanics relations, we shall frequently not show the superscripts and subscripts t and 0 but always imply them, and when there is doubt, we shall also actually show them. For example, (6.29) is written as X = RU.
EXAMPLE 6.8: Show that the deformation gradient X can always be decomposed as follows:
\mathbf {X} = \mathbf {R U} \tag {a}
where \mathbf{R} is an orthogonal (rotation) matrix and \mathbf{U} is a stretch (symmetric) matrix.
To prove the relationship in (a), we consider the Cauchy-Green deformation tensor C and represent this tensor in its principal coordinate axes. For this purpose we solve the eigenproblem
\mathbf {C p} = \lambda \mathbf {p} \tag {b}
The complete solution of (b) can be written as (see Section 2.5)
\mathbf {C P} = \mathbf {P C} ^ {\prime}
where the columns of \mathbf{P} are the eigenvectors of \mathbf{C} , and \mathbf{C}' is a diagonal matrix storing the corresponding eigenvalues. We also have
\mathbf {P} ^ {T} \mathbf {C P} = \mathbf {C} ^ {\prime} \tag {c}
and \mathbf{C}' is the representation of the Cauchy-Green deformation tensor in its principal coordinate axes. The representation of the deformation gradient in this coordinate system, denoted as \mathbf{X}' , is similarly obtained
\mathbf {X} ^ {\prime} = \mathbf {P} ^ {T} \mathbf {X} \mathbf {P} \tag {d}
where we note that (c) and (d) are really tensor transformations from the original to a new coordinate system (see Section 2.4).
Using these relations and \mathbf{C} = \mathbf{X}^T\mathbf{X} , we have
\mathbf {C} ^ {\prime} = \mathbf {X} ^ {\prime T} \mathbf {X} ^ {\prime}
and we note that the matrix
\mathbf {R} ^ {\prime} = \mathbf {X} ^ {\prime} (\mathbf {C} ^ {\prime}) ^ {- 1 / 2}
is an orthogonal matrix; i.e., \mathbf{R}^{\prime T}\mathbf{R}^{\prime} = \mathbf{I} . Hence, we can write
\mathbf {X} ^ {\prime} = \mathbf {R} ^ {\prime} \mathbf {U} ^ {\prime} \tag {e}
where \mathbf{U}^{\prime} = (\mathbf{C}^{\prime})^{1 / 2}
and to evaluate U' we use the positive values of the square roots of the diagonal elements of C' . The positive values must be used because the diagonal values in U' represent the stretches in the new coordinate system.
The relation in (e) is the decomposition of the deformation gradient X' into the product of the orthogonal matrix R' and the stretch matrix U' . This decomposition has been accomplished in the principal axes of C but is also valid in any other (admissible) coordinate system because the deformation gradient is a tensor (see Section 2.4). Indeed, we can now obtain R and U directly corresponding to the decomposition in (a); i.e.,
\mathbf {R} = \mathbf {P} \mathbf {R} ^ {\prime} \mathbf {P} ^ {T}
\mathbf {U} = \mathbf {P} \mathbf {U} ^ {\prime} \mathbf {P} ^ {T}
where we used the inverse of the transformation employed in (d).
EXAMPLE 6.9: Consider the four-node element and its deformation shown in Fig. E6.9. (a) Evaluate the deformation gradient and its polar decomposition at time t. (b) Assume that the motion from time t to time t + \Delta t consists only of a counterclockwise rigid body rotation of 45 degrees. Evaluate the new deformation gradient.
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Time t Time 0 x₂ x₁ 30° 2 3 3 4
Figure E6.9 Four-node element subjected to stretching and rotation
To evaluate the deformation gradient at time t, we can here conveniently use \delta X = r R_{0} \tau U , where the hypothetical (or conceptual) configuration \tau corresponds to the stretching of the fibers only. Hence,
\mathbf {r} \mathbf {R} = \left[ \begin{array}{c c} \frac {\sqrt {3}}{2} & \frac {- 1}{2} \\ \frac {1}{2} & \frac {\sqrt {3}}{2} \end{array} \right]; \quad \mathbf {s} \mathbf {U} = \left[ \begin{array}{c c c} \frac {4}{3} & & 0 \\ & & \frac {3}{2} \end{array} \right]
and
_ {6} \mathbf {X} = \left[ \begin{array}{c c} \frac {2}{\sqrt {3}} & - \frac {3}{4} \\ \frac {2}{3} & \frac {3 \sqrt {3}}{4} \end{array} \right]
Of course, the same result is also obtained by writing x_{i} in terms of ^0 x_j , i = 1, 2; j = 1, 2 , and using the definition of \delta \mathbf{X} given in (6.19).
Let us next subject the element to the counterclockwise rotation of 45 degrees. The deformation gradient is then
\begin{array}{l} { } ^ { t + \Delta t } \mathbf { X } = \left[ \begin{array} { c c } \cos 4 5 ^ { \circ } & - \sin 4 5 ^ { \circ } \\ \sin 4 5 ^ { \circ } & \cos 4 5 ^ { \circ } \end{array} \right] \left[ \begin{array} { c c } \frac { 2 } { \sqrt { 3 } } & - \frac { 3 } { 4 } \\ \frac { 2 } { 3 } & \frac { 3 \sqrt { 3 } } { 4 } \end{array} \right] \\ = \frac {1}{\sqrt {2}} \left[ \begin{array}{l l} \frac {2 \sqrt {3} - 2}{3} & - \frac {3 + 3 \sqrt {3}}{4} \\ \frac {2 \sqrt {3} + 2}{3} & \frac {- 3 + 3 \sqrt {3}}{4} \end{array} \right] \\ \end{array}
The proof in Example 6.8 also indicates how any deformation gradient can be decomposed into the product in (6.29). Assume that \mathbf{X} is given and we want to find \mathbf{R} and \mathbf{U} ; then we may calculate \mathbf{C} = \mathbf{X}^T\mathbf{X} = \mathbf{U}^2 and, using (2.109), we have (for n = 2 or 3), \mathbf{U} = \sum_{i=1}^{n}\sqrt{\lambda_i}\mathbf{p}_i\mathbf{p}_i^T with \mathbf{C}\mathbf{p}_i = \lambda_i\mathbf{p}_i . With \mathbf{U} given, we obtain \mathbf{R} from \mathbf{R} = \mathbf{X}\mathbf{U}^{-1} .
The preceding relations can now be used to evaluate additional kinematic relations that describe the motion of the body. That is, it can be proven (see Exercise 6.7) that we also have
\mathbf {X} = \mathbf {V R} \tag {6.30}
where \mathbf{V} is also a symmetric matrix
\mathbf {V} = \mathbf {R U R} ^ {T} \tag {6.31}
We refer to U as the right stretch matrix and to V as the left stretch matrix.
Example 6.8 shows that we have the spectral decomposition of U,
\mathbf {U} = \mathbf {R} _ {L} \boldsymbol {\Lambda} \mathbf {R} _ {L} ^ {T} \tag {6.32}
Physically, \mathbf{A} corresponds to the principal stretches and \mathbf{R}_L stores the directions of these stretches, with the rigid body rotation removed since this rotation appears in \mathbf{R} . (In Example 6.8, the matrix \mathbf{P} is equal to \mathbf{R}_L .) We also have
\mathbf {V} = \mathbf {R} _ {E} \boldsymbol {\Lambda} \mathbf {R} _ {E} ^ {T} \tag {6.33}
where \mathbf{R}_E = \mathbf{R}\mathbf{R}_L (6.34)
We note that R_{E} stores the base vectors of the principal stretches in the stationary coordinate system x_{i} .
To proceed further with our description of the motion of the material particles in the body, we consider next the time rates of change of the quantities defined above. For this development we define
\dot {\mathbf {R}} = \boldsymbol {\Omega} _ {R} \mathbf {R} \tag {6.35}
\dot {\mathbf {R}} _ {L} = \mathbf {R} _ {L} \boldsymbol {\Omega} _ {L} \tag {6.36}
\dot {\mathbf {R}} _ {E} = \mathbf {R} _ {E} \boldsymbol {\Omega} _ {E} \tag {6.37}
where \Omega_{R} , \Omega_{L} , and \Omega_{E} are skew-symmetric spin tensors, and clearly, using (6.34),
\boldsymbol {\Omega} _ {R} = \mathbf {R} _ {E} \left(\boldsymbol {\Omega} _ {E} - \boldsymbol {\Omega} _ {L}\right) \mathbf {R} _ {E} ^ {T} \tag {6.38}
The velocity gradient L is defined as the gradient of the velocity field with respect to the current position x_{j} of the material particles,
\mathbf {L} = \left[ \frac {\partial^ {t} \dot {u} _ {i}}{\partial^ {t} x _ {j}} \right] \tag {6.39}
or
\mathbf {L} = \dot {\mathbf {X}} \mathbf {X} ^ {- 1} \tag {6.40}
The symmetric part of L is the velocity strain tensor D (also called the rate-of-deformation tensor or stretching tensor), and the skew-symmetric part is the spin tensor W (also called the vorticity tensor). Hence,
\mathbf {L} = \mathbf {D} + \mathbf {W} \tag {6.41}
Using the polar decomposition of X we obtain from (6.40),
\mathbf {D} = \frac {1}{2} \mathbf {R} \left(\dot {\mathbf {U}} \mathbf {U} ^ {- 1} + \mathbf {U} ^ {- 1} \dot {\mathbf {U}}\right) \mathbf {R} ^ {T} \tag {6.42}
\mathbf {W} = \boldsymbol {\Omega} _ {R} + \frac {1}{2} \mathbf {R} (\dot {\mathbf {U}} \mathbf {U} ^ {- 1} - \mathbf {U} ^ {- 1} \dot {\mathbf {U}}) \mathbf {R} ^ {T} \tag {6.43}
Substituting for U from (6.32), we can write
\mathbf {D} = \mathbf {R} _ {E} \mathbf {D} _ {E} \mathbf {R} _ {E} ^ {T} \tag {6.44}
\mathbf {W} = \mathbf {R} _ {E} \mathbf {W} _ {E} \mathbf {R} _ {E} ^ {T} \tag {6.45}
where \mathbf{D}_E = \dot{\mathbf{\Lambda}}\mathbf{\Lambda}^{-1} + \frac{1}{2} (\mathbf{\Lambda}^{-1}\mathbf{\Omega}_L\mathbf{\Lambda} - \mathbf{\Lambda}\mathbf{\Omega}_L\mathbf{\Lambda}^{-1}) (6.46)
\mathbf {W} _ {E} = \boldsymbol {\Omega} _ {E} - \frac {1}{2} \left(\boldsymbol {\Lambda} ^ {- 1} \boldsymbol {\Omega} _ {L} \boldsymbol {\Lambda} + \boldsymbol {\Lambda} \boldsymbol {\Omega} _ {L} \boldsymbol {\Lambda} ^ {- 1}\right) \tag {6.47}
Hence, we obtain for the elements of \dot{\Lambda} ,
[ \dot {\mathbf {A}} ] _ {\alpha \alpha} = \lambda_ {\alpha} [ \mathbf {D} _ {E} ] _ {\alpha \alpha} \quad \text { no sum on } \alpha \tag {6.48}
where the \lambda_{\alpha} are the stretches, and for the elements of \Omega_L and \Omega_E , assuming that \lambda_{\alpha} \neq \lambda_{\beta} ,
[ \boldsymbol {\Omega} _ {L} ] _ {\alpha \beta} = \frac {2 \lambda_ {\beta} \lambda_ {\alpha}}{\lambda_ {\beta} ^ {2} - \lambda_ {\alpha} ^ {2}} [ \mathbf {D} _ {E} ] _ {\alpha \beta} \tag {6.49}
[ \boldsymbol {\Omega} _ {E} ] _ {\alpha \beta} = [ \mathbf {W} _ {E} ] _ {\alpha \beta} + \frac {\lambda_ {\beta} ^ {2} + \lambda_ {\alpha} ^ {2}}{\lambda_ {\beta} ^ {2} - \lambda_ {\alpha} ^ {2}} [ \mathbf {D} _ {E} ] _ {\alpha \beta} \tag {6.50}
We note that D_{E} and W_{E} are the velocity strain and spin tensors referred to the principal axes of the deformation at time t. Hence, by representing the velocity strain and spin tensors in the basis given by R_{E} , we obtain relationships that we can use directly to evaluate the components of \dot{\Lambda} , \Omega_{L} , and \Omega_{E} .
We now want to define strain tensors that are valuable in finite element analysis. The Green-Lagrange strain tensor \delta\epsilon is defined as
{ } _ { 0 } ^ { \prime } \boldsymbol { \epsilon } = { } _ { 0 } ^ { \prime } \mathbf { R } _ { L } [ \frac { 1 } { 2 } ( { } ^ { \prime } \mathbf { \Lambda } ^ { 2 } - \mathbf { I } ) ] { } _ { 0 } ^ { \prime } \mathbf { R } _ { L } ^ { T } \tag {6.51}
The Hencky (or logarithmic) strain tensor is defined as
{ } _ { 0 } ^ { t } \mathbf { E } ^ { H } = { } _ { 0 } ^ { t } \mathbf { R } _ { L } ( \ln { } ^ { t } \boldsymbol { \Lambda } ) { } _ { 0 } ^ { t } \mathbf { R } _ { L } ^ { T } \tag {6.52}
We note that since \delta \mathbf{R} does not enter the definitions in (6.51) and (6.52), both strain tensors are independent of the rigid body motions of the particles.
The Green-Lagrange strain tensor is frequently written in terms of the right stretch tensor \delta \mathbf{U} ; that is, using (6.51), we obtain
\begin{array}{r l} \mathcal {I} _ {0} \boldsymbol {\epsilon} & = \frac {1}{2} \left[ \left(\mathcal {I} _ {0} \mathbf {R} _ {L} ^ {\prime} \boldsymbol {\Lambda} \mathcal {I} _ {0} \mathbf {R} _ {L} ^ {T}\right) \left(\mathcal {I} _ {0} \mathbf {R} _ {L} ^ {\prime} \boldsymbol {\Lambda} \mathcal {I} _ {0} \mathbf {R} _ {L} ^ {T}\right) - \mathbf {I} \right] \\ & = \frac {1}{2} \left(\mathcal {I} _ {0} \mathbf {U} \mathcal {I} _ {0} \mathbf {U} - \mathbf {I}\right) \end{array} \tag {6.53}
Also, we can write the Green-Lagrange strain tensor in terms of the Cauchy-Green deformation tensor,
\begin{array}{l} _ {0} ^ {t} \boldsymbol {\epsilon} = \frac {1}{2} \left(_ {0} ^ {t} \mathbf {U} _ {0} ^ {t} \mathbf {R} ^ {T} _ {0} ^ {t} \mathbf {R} _ {0} ^ {t} \mathbf {U} - \mathbf {I}\right) \\ = \frac {1}{2} \left(_ {0} ^ {i} \mathbf {X} ^ {T} _ {0} ^ {j} \mathbf {X} - \mathbf {I}\right) \tag {6.54} \\ = \frac {1}{2} (_ {0} ^ {\prime} \mathbf {C} - \mathbf {I}) \\ \end{array}
Furthermore, evaluating the components in terms of displacements [i.e., using (6.16) and (6.19) in (6.54)], we have,
\delta \epsilon_ {i j} = \frac {1}{2} \left(\delta u _ {i, j} + \delta u _ {j, i} + \delta u _ {k, i} \delta u _ {k, j}\right) \tag {6.55}
We should note that in the definition of the Green-Lagrange strain tensor, all derivatives are with respect to the initial coordinates of the material particles. For this reason, we say that the strain tensor is defined with respect to the initial coordinates of the body. Also note that, although only up to quadratic terms of displacement derivatives appear in (6.55), this is the complete strain tensor; i.e., we have not neglected any higher-order terms.
The Green-Lagrange and Hencky strain tensors are clearly of the general form
\mathbf {E} _ {g} = \mathbf {R} _ {L} g (\mathbf {\Lambda}) \mathbf {R} _ {L} ^ {T} \tag {6.56}
where g(\Lambda) = \mathrm{diag}[g(\lambda_i)] . Hence, the rate of change of the strain tensors can be written as
\dot {\mathbf {E}} _ {g} = \mathbf {R} _ {L} \dot {\mathbf {E}} _ {L} \mathbf {R} _ {L} ^ {T} \tag {6.57}
where we have
\dot {\mathbf {E}} _ {L} = \dot {\boldsymbol {\Lambda}} g ^ {\prime} (\boldsymbol {\Lambda}) + \boldsymbol {\Omega} _ {L} g (\boldsymbol {\Lambda}) - g (\boldsymbol {\Lambda}) \boldsymbol {\Omega} _ {L} \tag {6.58}
Expanding this equation, we can identify the components of \dot{E}_{L} as
[ \dot {\mathbf {E}} _ {L} ] _ {\alpha \beta} = \gamma_ {\alpha \beta} [ \mathbf {D} _ {E} ] _ {\alpha \beta} \tag {6.59}
where for the Green-Lagrange strain tensor,
\gamma_ {\alpha \beta} = \lambda_ {\alpha} \lambda_ {\beta} \tag {6.60}
and for the Hencky strain tensor,
\gamma_ {\alpha \beta} = \left\{ \begin{array}{c l} 1 & \text { if } \lambda_ {\alpha} = \lambda_ {\beta} \\ \frac {2 \lambda_ {\alpha} \lambda_ {\beta}}{\lambda_ {\beta} ^ {2} - \lambda_ {\alpha} ^ {2}} \ln \frac {\lambda_ {\beta}}{\lambda_ {\alpha}} & \text { otherwise } \end{array} \right. \tag {6.61}
Using (6.57) and (6.59), we can now establish an important relationship between the time rate of change of the Green-Lagrange strain tensor \dot{o} and the velocity strain tensor 'D. Using (6.57), (6.59), (6.60), and (6.44), we obtain
{ } _ { 0 } ^ { \prime } \mathbf { R } _ { L } ^ { T } { } _ { 0 } ^ { \prime } \dot { \boldsymbol { \epsilon } } { } _ { 0 } ^ { \prime } \mathbf { R } _ { L } = { } ^ { \prime } \boldsymbol { \Lambda } { } _ { 0 } ^ { \prime } \mathbf { R } _ { E } ^ { T } { } ^ { \prime } \mathbf { D } { } _ { 0 } ^ { \prime } \mathbf { R } _ { E } { } ^ { \prime } \boldsymbol { \Lambda } \tag {6.62}
and hence, using (6.32) and (6.34), we obtain the operations
\boxed {0 ^ {\prime} \dot {\boldsymbol {\epsilon}} = _ {0} ^ {\prime} \mathbf {X} ^ {T} {} ^ {\prime} \mathbf {D} _ {0} ^ {\prime} \mathbf {X}} \quad (\text { as a "pull - back" }) \tag {6.63}
\boxed {^ {\prime} \mathbf {D} = ^ {0} _ {t} \mathbf {X} ^ {T} _ {0} ^ {\prime} \dot {\epsilon} ^ {0} _ {t} \mathbf {X}} \quad (\text { as a ``push - forward'' })
or in component form (with super- and subscripts)
\begin{array}{r l} & \dot {0} \dot {\epsilon} _ {i j} = \dot {0} x _ {m, i} \dot {0} x _ {n, j} ^ {\prime} D _ {m n} \\ & \dot {D} _ {m n} = \dot {0} x _ {i, m} \dot {0} x _ {j, n} \dot {0} \dot {\epsilon} _ {i j} \end{array} \tag {6.64}
Of course, we can obtain the same result, but with less insight, by simply differentiating the Green-Lagrange strain tensor with respect to time,
{ } _ { 0 } ^ { \prime } \dot { \boldsymbol { \epsilon } } = \frac { 1 } { 2 } ( { } _ { 0 } ^ { \prime } \dot { \mathbf { X } } ^ { T } { } _ { 0 } ^ { \prime } \mathbf { X } + { } _ { 0 } ^ { \prime } \mathbf { X } ^ { T } { } _ { 0 } ^ { \prime } \dot { \mathbf { X } } ) \tag {6.65}
Using (6.40) and (6.41) to substitute into (6.65), we directly obtain (6.63). We demonstrate this derivation for virtual displacement increments, or variations in the current displacements, in the following example.
EXAMPLE 6.10: Consider a body in its deformed configuration at time t (see Fig. E6.10). The current coordinates of the material particles of the body are 'x_i , i = 1, 2, 3 , and the current displacements are 'u_i = 'x_i - ^0 x_i .
Assume that a virtual displacement field is applied, which we denote as \delta u_{i} (see Fig. E6.10). This virtual displacement field can be thought of as a variation on the current displacements; hence, we may write \delta u_{i} \equiv \delta^{\prime}u_{i} . However, the variation on the current displacements must correspond to a variation on the current Green-Lagrange strain components, \delta_0^{\prime}\epsilon_{ij} , and also to a small strain tensor \delta_{t}e_{mn} referred to the current configuration. Evaluate the components \delta_0^{\prime}\epsilon_{ij} and show that
\delta_ {0} ^ {t} \epsilon_ {i j} = \delta_ {0} ^ {t} x _ {m, i} \delta_ {0} ^ {t} x _ {n, j} \delta_ {t} e _ {m n} \tag {a}