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are used to evaluate the corresponding element stresses and nodal point forces t+\Delta t\mathbf{F}^{(i-1)} and tangent stiffness matrix t+\Delta t\mathbf{K}^{(i-1)} .
The out-of-balance load vector t+\Delta tR - t+\Delta tF^{(i-1)} corresponds to a load vector that is not yet balanced by element stresses, and hence an increment in the nodal point displacements is required. This updating of the nodal point displacements in the iteration is continued until the out-of-balance loads and incremental displacements are small.
Let us summarize some important considerations regarding the Newton-Raphson iterative solution.
An important point is that the correct calculation of t + \Delta t \mathbf{F}^{(i-1)} from t + \Delta t \mathbf{U}^{(i-1)} is crucial. Any errors in this calculation will, in general, result in an incorrect response prediction.
The correct evaluation of the tangent stiffness matrix r + \Delta r \mathbf{K}^{(i - 1)} is also important. The use of the proper tangent stiffness matrix may be necessary for convergence and, in general, will result in fewer iterations until convergence is reached.
However, because of the expense involved in evaluating and factoring a new tangent stiffness matrix, in practice, it can be more efficient, depending on the nonlinearities present in the analysis, to evaluate a new tangent stiffness matrix only at certain times. Specifically, in the modified Newton-Raphson method a new tangent stiffness matrix is established only at the beginning of each load step, and in quasi-Newton methods secant stiffness matrices are used instead of the tangent stiffness matrix (see Section 8.4). We note that, which scheme to use is only a matter of computational efficiency provided convergence is reached.
The use of the iterative solution requires appropriate convergence criteria. If inappropriate criteria are used, the iteration may be terminated before the necessary solution accuracy is reached or be continued after the required accuracy has been reached.
We discuss these numerical considerations in Section 8.4 but note here that whichever iterative technique is used, the basic requirements are (1) the evaluation of the (tangent) stiffness matrix corresponding to a given state and (2) the evaluation of the nodal force vector corresponding to the stresses in that state (the "state" being given by 'U or '+Δ'U(i-1), i = 1, 2, 3, ...). Hence, our primary focus in this chapter is on explaining how, for a generic state, and we use the state at time t, the tangent stiffness matrices 'K and force vectors 'F for various elements and material stress-strain relations can be evaluated.
Let us now demonstrate these concepts in two examples.
EXAMPLE 6.3: Idealize the simple arch structure shown in Fig. E6.3(a) as an assemblage of two bar elements. Assume that the force in one bar element is given by F_{\text{bar}} = k' \delta , where k is a constant and \delta is the elongation of the bar at time t . (The assumption that k is constant is likely to be valid only for small deformations in the bar, but we use this assumption in order to simplify the analysis.) Establish the equilibrium relation (6.5) for this problem.
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L 15° tR
(a) Bar assemblage subjected to apex load
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1 15° 2 tR/2
(b) Simple model using one bar (truss) element, nodes 1 and 2
Figure E6.3 A simple arch structure
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L 15° tβ tδ tΔ
(c) Geometric variables in typical configuration
line
| tΔ/L (x 10⁻¹) | iR/2kL (x 10⁻³) |
|---|---|
| 0 | 0 |
| 0.1 | 4 |
| 0.05 | 0 |
| 0.0 | -4 |
| 0.05 | 0 |
| 0.1 | 4 |
(d) Load-displacement relation
Figure E6.3 (continued)
This is a large displacement problem, and the response is calculated by focusing attention on the equilibrium of the bar assemblage in the configuration corresponding to a typical time t. Using symmetry as shown in Figs. E6.3(b) and (c), we have
(L - ^ {\prime} \delta) \cos^ {\prime} \beta = L \cos 1 5 ^ {\circ}
(L - ^ {\prime} \delta) \sin^ {\prime} \beta = L \sin 1 5 ^ {\circ} - ^ {\prime} \Delta
hence, \delta = L - \sqrt{L^2 - 2L' \Delta \sin 15^\circ + ' \Delta^2}
\sin^ {\prime} \beta = \frac {L \sin 1 5 ^ {\circ} - ^ {\prime} \Delta}{L - ^ {\prime} \delta}
Equilibrium at time t requires that
2 ^ {\prime} F _ {\text { bar }} \sin^ {\prime} \beta = ^ {\prime} R
hence, the relation in (6.5) is
\frac {^ \prime R}{2 k L} = \left\{- 1 + \frac {1}{\left[ 1 - 2 \frac {^ \prime \Delta}{L} \sin 1 5 ^ {\circ} + \left(\frac {^ \prime \Delta}{L}\right) ^ {2} \right] ^ {1 / 2}} \right\} \left(\sin 1 5 ^ {\circ} - \frac {^ \prime \Delta}{L}\right) \tag {a}
Figure E6.3(d) shows the force-displacement relationship established in (a). It should be noted that between points A and B, for a given load level, we have two possible displacement configurations. If the structure is loaded with 'R monotonically increasing, the displacement path with snap-through from A to B in Fig. E6.3(d) is likely to be followed in an actual physical situation.
EXAMPLE 6.4: Calculate the response of the bar assemblage considered in Example 6.1 using the modified Newton-Raphson iteration. Use two equal load steps to reach the maximum load application.
In the modified Newton-Raphson iteration, we use (6.11) and (6.12) but evaluate new tangent stiffness matrices only at the beginning of each step. Hence, the iterative equations are in this analysis
\left(^ {t} K _ {a} + ^ {t} K _ {b}\right) \Delta u ^ {(i)} = ^ {t + \Delta t} R - ^ {t + \Delta t} F _ {a} ^ {(i - 1)} - ^ {t + \Delta t} F _ {b} ^ {(i - 1)} \tag {a}
{ } ^ { t + \Delta t } u ^ { ( i ) } = { } ^ { t + \Delta t } u ^ { ( i - 1 ) } + \Delta u ^ { ( i ) }
with t + \Delta t u^{(0)} = {}^t u
{ } ^ { t + \Delta t } F _ { a } ^ { ( 0 ) } = { } ^ { t } F _ { a } ; \quad { } ^ { t + \Delta t } F _ { b } ^ { ( 0 ) } = { } ^ { t } F _ { b } \tag {b}
^ \prime K _ {a} = \frac {^ \prime C A}{L _ {a}}; \quad^ {\prime} K _ {b} = \frac {^ \prime C A}{L _ {b}}
where {}^t C\left\{ \begin{array}{ll} = E & \text{ if section is elastic } \\ = {E}_{T} & \text{ if section is plastic } \end{array}\right.
For an elastic section,
{ } ^ { t + \Delta t } F ^ { ( i - 1 ) } = E A ^ { t + \Delta t } \epsilon ^ { ( i - 1 ) } \tag {c}
for a plastic section,
{ } ^ { t + \Delta t } F ^ { ( i - 1 ) } = A \left[ E _ { T } \left( { } ^ { t + \Delta t } \epsilon ^ { ( i - 1 ) } - \epsilon _ { y } \right) + \sigma _ { y } \right] \tag {d}
and the strains in the sections are
{ } ^ { t + \Delta t } \epsilon _ { a } ^ { ( i - 1 ) } = \frac { { } ^ { t + \Delta t } u ^ { ( i - 1 ) } } { L _ { a } }
{ } ^ { t + \Delta t } \epsilon _ { b } ^ { ( i - 1 ) } = \frac { { } ^ { t + \Delta t } u ^ { ( i - 1 ) } } { L _ { b } } \tag {e}
In the first load step, we have t = 0 and \Delta t = 1 . Thus, the application of the relations in (a) to (e) gives
t = 1:
\left(^ {0} K _ {a} + ^ {0} K _ {b}\right) \Delta u ^ {(1)} = ^ {1} R - ^ {1} F _ {a} ^ {(0)} - ^ {1} F _ {b} ^ {(0)}
\Delta u ^ {(1)} = \frac {2 \times 1 0 ^ {4}}{1 0 ^ {7} (\frac {1}{1 0} + \frac {1}{5})} = 6. 6 6 6 7 \times 1 0 ^ {- 3} \mathrm{cm}
(i = 1) \quad {} ^ {1} u ^ {(1)} = ^ {1} u ^ {(0)} + \Delta u ^ {(1)} = 6. 6 6 6 7 \times 1 0 ^ {- 3} \mathrm{cm}
{ } ^ { 1 } \epsilon _ { a } ^ { ( 1 ) } = \frac { { } ^ { 1 } u ^ { ( 1 ) } } { L _ { a } } = 6 . 6 6 6 7 \times 1 0 ^ { - 4 } < \epsilon _ { y } \rightarrow \text { s e c t i o n } a \text { i s e l a s t i c }
{ } ^ { 1 } \epsilon _ { b } ^ { ( 1 ) } = \frac { { } ^ { 1 } u ^ { ( 1 ) } } { L _ { b } } = 1 . 3 3 3 3 \times 1 0 ^ { - 3 } < \epsilon _ { y } \rightarrow \text { s e c t i o n } b \text { is elastic }
{ } ^ { 1 } F _ { a } ^ { ( 1 ) } = 6 . 6 6 6 7 \times 1 0 ^ { 3 } \mathrm{~N}
{ } ^ { 1 } F _ { b } ^ { ( 1 ) } = 1 . 3 3 3 3 \times 1 0 ^ { 4 } \mathrm{~N}
\left(^ {0} K _ {a} + ^ {0} K _ {b}\right) \Delta u ^ {(2)} = ^ {1} R - ^ {1} F _ {a} ^ {(1)} - ^ {1} F _ {b} ^ {(1)}
∴ Convergence is achieved in one iteration
{ } ^ { 1 } u = 6 . 6 6 6 7 \times 1 0 ^ { - 3 } \mathrm{cm}
t = 2:
{ } ^ { 1 } K _ { a } = \frac { E A } { L _ { a } } ; \quad { } ^ { 1 } K _ { b } = \frac { E A } { L _ { b } }
{ } ^ { 2 } F _ { a } ^ { ( 0 ) } = { } ^ { 1 } F _ { a } ; \quad { } ^ { 2 } F _ { b } ^ { ( 0 ) } = { } ^ { 1 } F _ { b }
\begin{array}{l} \left(^ {1} K _ {a} + ^ {1} K _ {b}\right) \Delta u ^ {(1)} = ^ {2} R - ^ {2} F _ {a} ^ {(0)} - ^ {2} F _ {b} ^ {(0)} \\ \Delta u ^ {(1)} = \frac {(4 \times 1 0 ^ {4}) - (6 . 6 6 6 7 \times 1 0 ^ {3}) - (1 . 3 3 3 3 \times 1 0 ^ {4})}{1 0 ^ {7} \left(\frac {1}{1 0} + \frac {1}{5}\right)} \\ = 6. 6 6 6 7 \times 1 0 ^ {- 3} \mathrm{cm} \\ (i = 1) ^ {2} u ^ {(1)} = ^ {2} u ^ {(0)} + \Delta u ^ {(1)} = 1. 3 3 3 3 \times 1 0 ^ {- 2} \mathrm{cm} \\ ^ 2 \epsilon_ {a} ^ {(1)} = 1. 3 3 3 3 \times 1 0 ^ {- 3} < \epsilon_ {y} \rightarrow \text { section } a \text { is elastic } \\ ^ 2 \epsilon_ {b} ^ {(1)} = 2. 6 6 6 7 \times 1 0 ^ {- 3} > \epsilon_ {y} \rightarrow \text { section } b \text { is plastic } \\ { } ^ { 2 } F _ { \mathrm{a} } ^ { ( 1 ) } = 1 . 3 3 3 3 \times 1 0 ^ { 4 } \mathrm{~N~} \\ { } ^ { 2 } F _ { b } ^ { ( 1 ) } = \left[ E _ { T } \left( ^ { 2 } \epsilon _ { b } ^ { ( 1 ) } - \epsilon _ { y } \right) + \sigma _ { y } \right] A = 2 . 0 0 6 7 \times 1 0 ^ { 4 } \mathrm{~N} \\ \left(^ {1} K _ {a} + ^ {1} K _ {b}\right) \Delta u ^ {(2)} = ^ {2} R - ^ {2} F _ {a} ^ {(1)} - ^ {2} F _ {b} ^ {(1)} \\ \Delta u ^ {(2)} = 2. 2 \times 1 0 ^ {- 3} \mathrm{cm} \\ (i = 2) \quad {} ^ {2} u ^ {(2)} = ^ {2} u ^ {(1)} + \Delta u ^ {(2)} = 1. 5 5 3 3 \times 1 0 ^ {- 2} \mathrm{cm} \\ { } ^ { 2 } \epsilon _ { a } ^ { ( 2 ) } = 1 . 5 5 3 3 \times 1 0 ^ { - 3 } < \epsilon _ { y } \\ { } ^ { 2 } \epsilon _ { b } ^ { ( 2 ) } = 3 . 1 0 6 6 \times 1 0 ^ { - 3 } > \epsilon _ { y } \\ \therefore {} ^ {2} F _ {a} ^ {(2)} = 1. 5 5 3 3 \times 1 0 ^ {4} \mathrm{N} \\ { } ^ { 2 } F _ { b } ^ { ( 2 ) } = 2 . 0 1 1 1 \times 1 0 ^ { 4 } \mathrm{~N~} \\ \left(^ {1} K _ {a} + ^ {1} K _ {b}\right) \Delta u ^ {(3)} = ^ {2} R - ^ {2} F _ {a} ^ {(2)} - ^ {2} F _ {b} ^ {(2)} \\ \Delta u ^ {(3)} = 1. 4 5 2 1 \times 1 0 ^ {- 3} \mathrm{cm} \\ \end{array}
The procedure is repeated, and the results of successive iterations are tabulated in the accompanying table.
| i | $\Delta u^{(i)}$ (cm) | $^2 u^{(i)}$ (cm) |
| 3 | $1.4521 \times 10^{-3}$ | $1.6985 \times 10^{-2}$ |
| 4 | $9.5832 \times 10^{-4}$ | $1.7944 \times 10^{-2}$ |
| 5 | $6.3249 \times 10^{-4}$ | $1.8576 \times 10^{-2}$ |
| 6 | $4.1744 \times 10^{-4}$ | $1.8994 \times 10^{-2}$ |
| 7 | $2.7551 \times 10^{-4}$ | $1.9269 \times 10^{-2}$ |
After seven iterations, we have
{ } ^ { 2 } u \doteq { } ^ { 2 } u ^ { ( 7 ) } = 1 . 9 2 6 9 \times 1 0 ^ { - 2 } \mathrm{~cm}
6.2 FORMULATION OF THE CONTINUUM MECHANICS INCREMENTAL EQUATIONS OF MOTION
The objective in the introductory discussion of nonlinear analysis in Section 6.1 was to describe various nonlinearities and the form of the basic finite element equations that are used to analyze the nonlinear response of a structural system. To show the procedure of analysis, we simply stated the finite element equations, discussed their solution, and gave a
physical argument why the nonlinear response is appropriately predicted using these equations. We demonstrated the applicability of the approach in the solution of two very simple problems merely to give some insight into the steps of analysis used. In each of these analyses the applicable finite element matrices and vectors were developed using physical arguments.
The physical approach of analysis used in Examples 6.3 and 6.4 is very instructive and yields insight into the analysis; however, when considering a more complex solution, a consistent continuum-mechanics-based approach should be employed to develop the governing finite element equations. The objective in this section is to present the governing continuum mechanics equations for a displacement-based finite element solution. As in Section 4.2.1, we use the principle of virtual work but now include the possibility that the body considered undergoes large displacements and rotations and large strains and that the stress-strain relationship is nonlinear. The governing continuum mechanics equations to be presented can therefore be regarded as an extension of the basic equation given in (4.7). In the linear analysis of a general body, the equation in (4.7) was used as the basis for the development of the governing linear finite element equations [given in (4.17) to (4.27)]. Considering the nonlinear analysis of a general body, after having developed suitable continuum mechanics equations, we will proceed in a completely analogous manner to establish the nonlinear finite element equations that govern the nonlinear response of the body (see Section 6.3).
6.2.1 The Basic Problem
In Section 6.1 we underlined the fact that in a nonlinear analysis the equilibrium of the body considered must be established in the current configuration. We also pointed out that in general it is necessary to employ an incremental formulation and that a time variable is used to conveniently describe the loading and the motion of the body.
In the development to follow, we consider the motion of a general body in a stationary Cartesian coordinate system, as shown in Fig. 6.2, and assume that the body can experience large displacements, large strains, and a nonlinear constitutive response. The aim is to evaluate the equilibrium positions of the complete body at the discrete time points 0, \Delta t , 2 \Delta t , 3 \Delta t , \ldots , where \Delta t is an increment in time. To develop the solution strategy, assume that the solutions for the static and kinematic variables for all time steps from time 0 to time t, inclusive, have been obtained. Then the solution process for the next required equilibrium position corresponding to time t + \Delta t is typical and is applied repetitively until the complete solution path has been solved for. Hence, in the analysis we follow all particles of the body in their motion, from the original to the final configuration of the body, which means that we adopt a Lagrangian (or material) formulation of the problem. This approach stands in contrast to an Eulerian formulation which is usually used in the analysis of fluid mechanics problems, in which attention is focused on the motion of the material through a stationary control volume. Considering the analysis of solids and structures, a Lagrangian formulation usually represents a more natural and effective analysis approach than an Eulerian formulation. For example, using an Eulerian formulation of a structural problem with large displacements, new control volumes have to be created (because the boundaries of the solid change continuously), and the nonlinearities in the convective acceleration terms are difficult to deal with (see Section 7.4).
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Configuration corresponding to variation in displacement δu on t+Δt u δu = [δu1] [δu2] [δu3] P(t+Δt x1, t+Δt x2, t+Δt x3) Configuration at time t+Δt Surface area t+Δt S Volume t+Δt V Configuration at time t Surface area t S Volume t V P(0 x1, 0 x2, 0 x3) Configuration at time 0 Surface area 0 S Volume 0 V x1(or 0 x1, t x1, t+Δt x1) t+Δt x_i = 0 x_i + t+Δt u_i t_x_i = 0 x_i + t u_i t+Δt x_i = t x_i + u_i i = 1, 2, 3
Figure 6.2 Motion of body in Cartesian coordinate frame
In our Lagrangian incremental analysis approach we express the equilibrium of the body at time t + \Delta t using the principle of virtual displacements. Using tensor notation (see Section 2.4), this principle requires that
\boxed {\int_ {t + \Delta t V} ^ {t + \Delta t} \tau_ {i j} \delta_ {t + \Delta t} e _ {i j} d ^ {t + \Delta t} V = ^ {t + \Delta t} \mathcal {R}} \tag {6.13}
where
^{t+\Delta t}\tau_{ij}= Cartesian components of the Cauchy stress tensor (forces per unit areas in the deformed geometry)
\delta_{t + \Delta t}e_{ij} = \frac{1}{2}\left(\frac{\partial\delta u_i}{\partial^{t + \Delta t}x_j} +\frac{\partial\delta u_j}{\partial^{t + \Delta t}x_i}\right) = strain tensor corresponding to virtual displacements
\delta u_{i} = components of virtual displacement vector imposed on configuration at time t + \Delta t , a function of t + \Delta t x_{j}, j = 1, 2, 3
t + \Delta t x_{i} = Cartesian coordinates of material point at time t + \Delta t
t + \Delta t V = \text{volume at time } t + \Delta t
and
\boxed {^ {t + \Delta t} \mathcal {R} = \int_ {t + \Delta t V} ^ {t + \Delta t} f _ {i} ^ {B} \delta u _ {i} d ^ {t + \Delta t} V + \int_ {t + \Delta t S _ {f}} ^ {t + \Delta t} f _ {i} ^ {S} \delta u _ {i} ^ {S} d ^ {t + \Delta t} S} \tag {6.14}
where
^ {t + \Delta t} f _ {i} ^ {B} = \text { components of externally applied forces per unit volume at time } t + \Delta t
^ {t + \Delta t} f _ {i} ^ {S} = \text { components of externally applied surface tractions per unit surface area at time } t + \Delta t
^ {t + \Delta t} S _ {f} = \text { surface at time } t + \Delta t \text { on which external tractions are applied }
\begin{array}{l} \delta u _ {i} ^ {S} = \delta u _ {i} \text { evaluated on the surface } ^ {t + \Delta t} S _ {f} \text {(the} \delta u _ {i} \text { components are zero at and corresponding to the } \\ \text { prescribed displacements on the surface } ^ {t + \Delta t} S _ {u}) \end{array}
In (6.13), the left-hand side is the internal virtual work and the right-hand side is the external virtual work. The relation is derived as in linear infinitesimal displacement analysis (see Example 4.2), but the current configuration at time t + \Delta t (with the stresses and forces at that time) is used. Hence, the derivation of (6.13) is based on the following equilibrium equations.
Within ^{t+\Delta t}V for i = 1, 2, 3,
\frac {\partial^ {t + \Delta t} \tau_ {i j}}{\partial^ {t + \Delta t} x _ {j}} + ^ {t + \Delta t} f _ {i} ^ {B} = 0 \quad \text { sum over } j = 1, 2, 3 \tag {6.15a}
and on the surface t+\Delta t S_{f} , for i = 1, 2, 3,
{ } ^ { t + \Delta t } \tau _ { i j } { } ^ { t + \Delta t } n _ { j } = { } ^ { t + \Delta t } f _ { i } ^ { S } \quad \text { sum over } j = 1, 2, 3 \tag {6.15b}
where the t+\Delta t n_j are the components of the unit normal to the surface t+\Delta t S_f at time t + \Delta t .
As shown in Example 4.2, the equation (6.15a) is multiplied by arbitrary continuous virtual displacements \delta u_{i} that are zero at and corresponding to the prescribed displacements. The integration of the expression obtained from (6.15a) over the volume at time t+\Delta t and the use of the divergence theorem and (6.15b) then directly yield the relation in (6.13).
We note that the strain tensor components \delta_{t+\Delta t}e_{ij} corresponding to the imposed virtual displacements are like the components of the infinitesimal strain tensor, but the derivatives are with respect to the current coordinates at time t + \Delta t . The use of the strain tensor \delta_{t+\Delta t}e_{ij} in (6.13) is the direct result of the transformation by the divergence theorem used in the derivation of (6.13), and this strain tensor is obtained irrespective of the magnitude of the virtual displacements.
However, we now recognize that the virtual displacements \delta u_{i} may be thought of as a variation in the real displacements t^{+\Delta t}u_{i} (subject to the constraint that these variations must be zero at and corresponding to the prescribed displacements). These displacement variations result in variations in the current strains of the body, and we shall later, in particular, use the variation in the Green-Lagrange strain components corresponding to \delta u_{i} (see Example 6.10).
It is most important to recognize that the virtual work principle stated in (6.13) is simply an application of the equation in (4.7) (used in linear analysis) to the body considered in the configuration at time t + \Delta t . Therefore, all previous discussions and results
pertaining to the use of the virtual work principle in linear analysis are now directly applicable, with the current configuration at time t + \Delta t being considered. ^{1}
A fundamental difficulty in the general application of (6.13) is that the configuration of the body at time t + \Delta t is unknown. This is an important difference compared with linear analysis in which it is assumed that the displacements are infinitesimally small so that in (6.13) to (6.15) the original configuration is used. The continuous change in the configuration of the body entails some important consequences for the development of an incremental analysis procedure. For example, an important consideration must be that the Cauchy stresses at time t + \Delta t cannot be obtained by simply adding to the Cauchy stresses at time t a stress increment that is due only to the straining of the material. Namely, the calculation of the Cauchy stresses at time t + \Delta t must also take into account the rigid body rotation of the material because the components of the Cauchy stress tensor also change when the material is subjected to only a rigid body rotation.
The fact that the configuration of the body changes continuously in a large deformation analysis is dealt with in an elegant manner by using appropriate stress and strain measures and constitutive relations, as discussed in detail in the next sections.
Considering the discussions to follow, we recognize that a difficult point in the presentation of continuum mechanics relations for general large deformation analysis is the use of an effective notation because there are many different quantities that need to be dealt with. The symbols used should display all necessary information but should do so in a compact manner in order that the equations can be read with relative ease. For effective use of a notation, an understanding of the convention employed is most helpful, and for this purpose we summarize here briefly some basic facts and conventions used in our notation.
In our analysis we consider the motion of the body in a fixed (stationary) Cartesian coordinate system as displayed in Fig. 6.2. All kinematic and static variables are measured in this coordinate system, and throughout our description we use tensor notation.
The coordinates of a generic point P in the body at time 0 are ^0 x_1, ^0 x_2, ^0 x_3 ; at time t they are ^t x_1, ^t x_2, ^t x_3 ; and at time t + \Delta t they are ^{t + \Delta t} x_1, ^{t + \Delta t} x_2, ^{t + \Delta t} x_3 , where the left superscripts refer to the configuration of the body and the subscripts to the coordinate axes.
The notation for the displacements of the body is similar to the notation for the coordinates: at time t the displacements are ^{t}u_{i}, i = 1, 2, 3 , and at time t + \Delta t the displacements are ^{t+\Delta t}u_{i}, i = 1, 2, 3 . Therefore, we have
\left. \begin{array}{l} ^ {t} x _ {i} = ^ {0} x _ {i} + ^ {t} u _ {i} \\ ^ {t + \Delta t} x _ {i} = ^ {0} x _ {i} + ^ {t + \Delta t} u _ {i} \end{array} \right\} \quad i = 1, 2, 3 \tag {6.16}
The increments in the displacements from time t to time t + \Delta t are denoted as
u _ {i} = ^ {t + \Delta t} u _ {i} - ^ {t} u _ {i}; \quad i = 1, 2, 3 \tag {6.17}
During motion of the body, its volume, surface area, mass density, stresses, and strains are changing continuously. We denote the specific mass, area, and volume of the body at times 0, t, and t + \Delta t as ^{0}\rho , ^{t}\rho , ^{t+\Delta t}\rho ; ^{0}A , ^{t}A , ^{t+\Delta t}A ; and ^{0}V , ^{t}V , ^{t+\Delta t}V , respectively.
Since the configuration of the body at time t + \Delta t is not known, we will refer applied forces, stresses, and strains to a known equilibrium configuration. In analogy to the notation
used for coordinates and displacements, a left superscript indicates in which configuration the quantity (body force, surface traction, stress, etc.) occurs; in addition, a left subscript indicates the configuration with respect to which the quantity is measured. For example, the surface and body force components at time t + \Delta t , but measured in configuration 0, are ^{t+\Delta t}_{0}f_{i}^{S} , ^{t+\Delta t}_{0}f_{i}^{B} , i = 1, 2, 3. Here we have the exception that if the quantity under consideration occurs in the same configuration in which it is also measured, the left subscript may not be used; e.g., for the Cauchy stresses we have
{ } ^ { t + \Delta t } \tau _ { i j } \equiv { } _ { t + \Delta t } ^ { t + \Delta t } \tau _ { i j }
In the formulation of the governing equilibrium equations we also need to consider derivatives of displacements and coordinates. In our notation a comma denotes differentiation with respect to the coordinate following, and the left subscript denoting time indicates the configuration in which this coordinate is measured; thus we have, for example,
{ } _ { 0 } ^ { t + \Delta t } u _ { i , j } = \frac { \partial ^ { t + \Delta t } u _ { i } } { \partial ^ { 0 } x _ { j } }
and t + \Delta t^0 x_{m,n} = \frac{\partial^0x_m}{\partial^{t + \Delta t}x_n} (6.18)
Using these conventions, we shall define new symbols when they are first encountered.
6.2.2 The Deformation Gradient, Strain, and Stress Tensors
We mentioned in the previous section that in a large deformation analysis special attention must be given to the fact that the configuration of the body is changing continuously. This change in configuration can be dealt with in an elegant manner by defining auxiliary stress and strain measures. The objective in defining them is to express the internal virtual work in (6.13) in terms of an integral over a volume that is known and to be able to incrementally decompose the stresses and strains in an effective manner. There are various different stress and strain tensors that, in principle, could be used (see L. E. Malvern [A], Y. C. Fung [A], A. E. Green and W. Zerna [A], and R. Hill [A]). However, if the objective is to obtain an effective overall finite element solution procedure, only a few stress and strain measures need to be considered, see also E. N. Dvorkin and M. B. Goldschmit [A]. In the following we first consider the motion of a general body and define kinematic measures of this motion. We then introduce appropriate strain and the corresponding stress tensors. These are used later in the chapter to develop the incremental general finite element equations.
Consider the body in Fig. 6.2 at a generic time t. A fundamental measure of the deformation of the body is given by the deformation gradient, defined as ^{2}
\boxed { \begin{array}{c} \mathbf {\partial} _ {0} ^ {\prime} \mathbf {X} = \left[ \begin{array}{c c c} \frac {\partial^ {\prime} x _ {1}}{\partial^ {0} x _ {1}} & \frac {\partial^ {\prime} x _ {1}}{\partial^ {0} x _ {2}} & \frac {\partial^ {\prime} x _ {1}}{\partial^ {0} x _ {3}} \\ \frac {\partial^ {\prime} x _ {2}}{\partial^ {0} x _ {1}} & \frac {\partial^ {\prime} x _ {2}}{\partial^ {0} x _ {2}} & \frac {\partial^ {\prime} x _ {2}}{\partial^ {0} x _ {3}} \\ \frac {\partial^ {\prime} x _ {3}}{\partial^ {0} x _ {1}} & \frac {\partial^ {\prime} x _ {3}}{\partial^ {0} x _ {2}} & \frac {\partial^ {\prime} x _ {3}}{\partial^ {0} x _ {3}} \end{array} \right] \end{array} }
or _{0}^{t}\mathbf{X} = (_{0}\nabla^{t}\mathbf{x}^{T})^{T} (6.20)
where _{0}\nabla is the gradient operator
{ } _ { 0 } \boldsymbol { \nabla } = \left[ \begin{array} { c } \frac { \partial } { \partial ^ { 0 } x _ { 1 } } \\ \frac { \partial } { \partial ^ { 0 } x _ { 2 } } \\ \frac { \partial } { \partial ^ { 0 } x _ { 3 } } \end{array} \right] ; \qquad { } ^ { t } \mathbf { x } ^ { T } = \left[ \begin{array} { c c c } { } ^ { t } x _ { 1 } & { } ^ { t } x _ { 2 } & { } ^ { t } x _ { 3 } \end{array} \right] \tag {6.21}
The deformation gradient describes the stretches and rotations that the material fibers have undergone from time 0 to time t. Namely, let d^{0}x be a differential material fiber at time 0; then, by the chain rule of differentiation, this material fiber at time t is given by
d ^ {t} \mathbf {x} = _ {0} ^ {t} \mathbf {X} d ^ {0} \mathbf {x} \tag {6.22}
Using chain differentiation, it also follows that
d ^ {0} \mathbf {x} = ^ {0} _ {t} \mathbf {X} d ^ {t} \mathbf {x} \tag {6.23}
where ^{0}X is the inverse deformation gradient. From (6.22) and (6.23) we obtain
d ^ {0} \mathbf {x} = (_ {i} ^ {0} \mathbf {X}) (_ {0} ^ {i} \mathbf {X}) d ^ {0} \mathbf {x} \tag {6.24}
and hence [because (6.24) must hold for any differential length d^{0}x ], we have
{ } _ { t } ^ { 0 } \mathbf { X } = ( { } _ { 0 } ^ { t } \mathbf { X } ) ^ { - 1 } \tag {6.25}
Therefore, the inverse deformation gradient {}^0\mathbf{X} is actually the inverse of the deformation gradient {}^0\mathbf{X} .
An application of (6.18) is given by the evaluation of the mass density t\rho of the body at time t , namely,
\rho = \frac {^ 0 \rho}{\det \left(_ {0} ^ {\prime} \mathbf {X}\right)} \tag {6.26}
We prove and illustrate this relationship in the following examples.