EXAMPLE 6.18: Establish the matrices $\mathbf{b}_{L0}, \mathbf{b}_{L1}$ , and $\mathbf{b}_{NL}$ corresponding to the TL formulation for the two-dimensional plane strain element shown in Fig. E6.18. 

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0.5 cm 3 cm 2 1 0x₂, tₓ₂ 3 0x₁, tₓ₁ 4 At time t At time 0 3 cm 1 cm

Figure E6.18 Four-node plane strain element in large displacement/large strain conditions In this case we can directly use the information given in Table 6.5 with $$ ^ \prime u _ {1} ^ {1} = 1; \quad^ {\prime} u _ {2} ^ {1} = 0. 5 $$ $$ \begin{array}{l} ^ {\prime} u _ {1} ^ {2} = 0; \quad^ {\prime} u _ {2} ^ {2} = 0. 5 \\ ^ {\prime} u _ {1} ^ {3} = 0; \quad^ {\prime} u _ {2} ^ {3} = 0 \end{array} \quad^ {0} \mathbf {J} = \left[ \begin{array}{c c} \frac {3}{2} & 0 \\ 0 & \frac {3}{2} \end{array} \right] $$ $$ ^ \prime u _ {1} ^ {4} = 1; \quad^ {\prime} u _ {2} ^ {4} = 0 $$ The interpolation functions of the four-node element are given in Fig. 5.4 (and the required derivatives have been given in Example 5.5), so that we obtain $$ \mathbf {0} \mathbf {B} _ {L 0} = $$ $$ \frac {1}{6} \left[ \begin{array}{c c c c c c c c c} (1 + s) & 0 & - (1 + s) & 0 & - (1 - s) & 0 & (1 - s) & 0 \\ 0 & (1 + r) & 0 & (1 - r) & 0 & - (1 - r) & 0 & - (1 + r) \\ (1 + r) & (1 + s) & (1 - r) & - (1 + s) & - (1 - r) & - (1 - s) & - (1 + r) & (1 - s) \end{array} \right] $$ To evaluate $_{0}^{t}B_{L1}$ we also need the $l_{ij}$ values, where $$ l _ {1 1} = \sum_ {k = 1} ^ {4} _ {0} h _ {k, 1} ^ {\prime} u _ {1} ^ {k} = \frac {2}{3} \left\{h _ {1, r} ^ {\prime} u _ {1} ^ {1} + h _ {4, r} ^ {\prime} u _ {1} ^ {4} \right\} = \frac {1}{3} $$ $$ l _ {1 2} = \sum_ {k = 1} ^ {4} _ {0} h _ {k, 2} ^ {\prime} u _ {1} ^ {k} = \frac {2}{3} \left\{h _ {1, s} ^ {\prime} u _ {1} ^ {1} + h _ {4, s} ^ {\prime} u _ {1} ^ {4} \right\} = 0 $$ $$ l _ {2 1} = \sum_ {k = 1} ^ {4} _ {0} h _ {k, 1} ^ {\prime} u _ {2} ^ {k} = \frac {2}{3} \{h _ {1, r} ^ {\prime} u _ {2} ^ {1} + h _ {2, r} ^ {\prime} u _ {2} ^ {2} \} = 0 $$ $$ l _ {2 2} = \sum_ {k = 1} ^ {4} _ {0} h _ {k, 2} ^ {\prime} u _ {2} ^ {k} = \frac {2}{3} \left\{h _ {1, s} ^ {\prime} u _ {2} ^ {1} + h _ {2, s} ^ {\prime} u _ {2} ^ {2} \right\} = \frac {1}{6} $$ Hence, we have $$ \delta \mathbf {B} _ {L 1} = $$ $$ \frac {1}{3 6} \left[ \begin{array}{c c c c c c c c c} 2 (1 + s) & 0 & - 2 (1 + s) & 0 & - 2 (1 - s) & 0 & 2 (1 - s) & 0 \\ 0 & (1 + r) & 0 & (1 - r) & 0 & - (1 - r) & 0 & - (1 + r) \\ 2 (1 + r) & (1 + s) & 2 (1 - r) & - (1 + s) & - 2 (1 - r) & - (1 - s) & - 2 (1 + r) & (1 - s) \end{array} \right] $$ The nonlinear strain-displacement matrix can also directly be constructed using the derivatives of the interpolation functions and the Jacobian matrix: $$ _ {0} ^ {t} \mathbf {B} _ {N L} = $$ $$ \frac {1}{6} \left[ \begin{array}{c c c c c c c c} (1 + s) & 0 & - (1 + s) & 0 & - (1 - s) & 0 & (1 - s) & 0 \\ (1 + r) & 0 & (1 - r) & 0 & - (1 - r) & 0 & - (1 + r) & 0 \\ 0 & (1 + s) & 0 & - (1 + s) & 0 & - (1 - s) & 0 & (1 - s) \\ 0 & (1 + r) & 0 & (1 - r) & 0 & - (1 - r) & 0 & - (1 + r) \end{array} \right] $$ # 6.3.5 Three-Dimensional Solid Elements The evaluation of the matrices required in three-dimensional isoparametric finite element analysis is accomplished using the same procedures as in two-dimensional analysis. Thus, referring to Section 6.3.4, we simply note that for a typical element we now use the coordinate and displacement interpolations, $$ { } ^ { 0 } x _ { i } = \sum _ { k = 1 } ^ { N } h _ { k } { } ^ { 0 } x _ { i } ^ { k } ; \quad \quad { } ^ { t } x _ { i } = \sum _ { k = 1 } ^ { N } h _ { k } { } ^ { t } x _ { i } ^ { k } ; \quad \quad i = 1 , 2 , 3 \tag {6.127} $$ $$ { } ^ { \prime } u _ { i } = \sum _ { k = 1 } ^ { N } h _ { k } { } ^ { \prime } u _ { i } ^ { k } ; \qquad u _ { i } = \sum _ { k = 1 } ^ { N } h _ { k } u _ { i } ^ { k } ; \qquad i = 1 , 2 , 3 \tag {6.128} $$ where the element interpolation functions $h_{k}$ have been given in Fig. 5.5. Using (6.127) and (6.128) in the same way as in two-dimensional analysis, we can develop the relevant element matrices used in the TL and UL formulations for three-dimensional analysis (see Table 6.6). TABLE 6.6 Matrices used in the three-dimensional element formulation A. Total Lagrangian formulation 1. Incremental strains $$ _ 0 \epsilon_ {i j} = \frac {1}{2} (_ {0} u _ {i, j} + _ {0} u _ {j, i}) + \frac {1}{2} (_ {0} ^ {\prime} u u _ {k, j} + _ {0} u ^ {\prime} u _ {k, j}) + \frac {1}{2} (_ {0} u u _ {k, j}) \quad i = 1, 2, 3; j = 1, 2, 3; k = 1, 2, 3 $$ $$ \text { where } _ {0} u _ {i, j} = \frac {\partial u _ {i}}{\partial^ {0} x _ {j}} $$ 2. Linear strain-displacement transformation matrix Using $_{0}e = _{0}^{t}B_{L}\hat{u}$ where $_0\mathbf{e}^T = [0e_{11} \quad_0e_{22} \quad_0e_{33} \quad 2_0e_{12} \quad 2_0e_{23} \quad 2_0e_{31}]$ ; $$ \hat {\mathbf {u}} ^ {T} = \left[ u _ {1} ^ {1} u _ {2} ^ {1} u _ {3} ^ {1} u _ {1} ^ {2} u _ {2} ^ {2} u _ {3} ^ {2} \dots u _ {1} ^ {N} u _ {2} ^ {N} u _ {3} ^ {N} \right] $$ $$ \mathbf {b} _ {L} = \mathbf {b} _ {L 0} + \mathbf {b} _ {L 1} $$ $$ { } _ { 0 } ^ { \prime } \mathbf { B } _ { L 0 } = \left[ \begin{array} { c c c c c c } _ { 0 } h _ { 1 , 1 } & 0 & 0 & _ { 0 } h _ { 2 , 1 } & \cdots & 0 \\ 0 & _ { 0 } h _ { 1 , 2 } & 0 & 0 & \cdots & 0 \\ 0 & 0 & _ { 0 } h _ { 1 , 3 } & 0 & \cdots & _ { 0 } h _ { N , 3 } \\ _ { 0 } h _ { 1 , 2 } & _ { 0 } h _ { 1 , 1 } & 0 & _ { 0 } h _ { 2 , 2 } & \cdots & 0 \\ 0 & _ { 0 } h _ { 1 , 3 } & _ { 0 } h _ { 1 , 2 } & 0 & \cdots & _ { 0 } h _ { N , 2 } \\ _ { 0 } h _ { 1 , 3 } & 0 & _ { 0 } h _ { 1 , 1 } & _ { 0 } h _ { 2 , 3 } & \cdots & _ { 0 } h _ { N , 1 } \end{array} \right] $$ $$ \text { where } _ {0} h _ {k, j} = \frac {\partial h _ {k}}{\partial^ {0} x _ {j}}; \quad u _ {j} ^ {k} = ^ {t + \Delta t} u _ {j} ^ {k} - ^ {t} u _ {j} ^ {k} $$ TABLE 6.6 (cont.)

$_0\mathbf{B}_{L1} = \left[ \begin{array}{cccccc} l_{110}h_{1,1} & l_{210}h_{1,1} & l_{310}h_{1,1} & l_{110}h_{2,1} & \cdots & l_{310}h_{N,1} \\ l_{120}h_{1,2} & l_{220}h_{1,2} & l_{320}h_{1,2} & l_{120}h_{2,2} & \cdots & l_{320}h_{N,2} \\ l_{130}h_{1,3} & l_{230}h_{1,3} & l_{330}h_{1,3} & l_{130}h_{2,3} & \cdots & l_{330}h_{N,3} \\ (l_{110}h_{1,2} + l_{120}h_{1,1}) & (l_{210}h_{1,2} + l_{220}h_{1,1}) & (l_{310}h_{1,2} + l_{320}h_{1,1}) & (l_{110}h_{2,2} + l_{120}h_{2,1}) & \cdots & (l_{310}h_{N,2} + l_{320}h_{N,1}) \\ (l_{120}h_{1,3} + l_{130}h_{1,2}) & (l_{220}h_{1,3} + l_{230}h_{1,2}) & (l_{320}h_{1,3} + l_{330}h_{1,2}) & (l_{120}h_{2,3} + l_{130}h_{2,2}) & \cdots & (l_{320}h_{N,3} + l_{330}h_{N,2}) \\ (l_{110}h_{1,3} + l_{130}h_{1,1}) & (l_{210}h_{1,3} + l_{230}h_{1,1}) & (l_{310}h_{1,3} + l_{330}h_{1,1}) & (l_{110}h_{2,3} + l_{130}h_{2,1}) & \cdots & (l_{310}h_{N,3} + l_{330}h_{N,1}) \end{array} \right]$

where $l_{ij} = \sum_{k=1}^{N} 0 h_{k,j} u_i^k$ 3. Nonlinear strain-displacement transformation matrix $$ { } _ { 0 } ^ { t } \mathbf { B } _ { N L } = \left[ \begin{array} { c c c } { } _ { 0 } ^ { t } \bar { \mathbf { B } } _ { N L } & \tilde { \mathbf { 0 } } & \tilde { \mathbf { 0 } } \\ \tilde { \mathbf { 0 } } & { } _ { 0 } ^ { t } \tilde { \mathbf { B } } _ { N L } & \tilde { \mathbf { 0 } } \\ \tilde { \mathbf { 0 } } & \tilde { \mathbf { 0 } } & { } _ { 0 } ^ { t } \tilde { \mathbf { B } } _ { N L } \end{array} \right] ; \quad \tilde { \mathbf { 0 } } = \left[ \begin{array} { c } 0 \\ 0 \\ 0 \end{array} \right] $$ where $\mathfrak{h}_{0}\tilde{\mathbf{B}}_{NL} = \begin{bmatrix} 0 & 0 & 0 & h_{2,1} & \cdots & 0 & h_{N,1}\\ 0 & 0 & 0 & h_{2,2} & \cdots & 0 & h_{N,2}\\ 0 & 0 & 0 & h_{2,3} & \cdots & 0 & h_{N,3} \end{bmatrix}$ 4. Second Piola-Kirchhoff stress matrix and vector $$ { } _ { 6 } \mathbf { S } = \left[ \begin{array} { c c c } { } _ { 6 } \tilde { \mathbf { S } } & \overline { { \mathbf { 0 } } } & \overline { { \mathbf { 0 } } } \\ \overline { { \mathbf { 0 } } } & { } _ { 6 } \tilde { \mathbf { S } } & \overline { { \mathbf { 0 } } } \\ \overline { { \mathbf { 0 } } } & \overline { { \mathbf { 0 } } } & { } _ { 6 } \tilde { \mathbf { S } } \end{array} \right] ; \quad \overline { { \mathbf { 0 } } } = \left[ \begin{array} { c c c } 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right] $$ $$ \hat {\mathbf {S}} ^ {T} = \left[ \begin{array}{l l l l l l} \hat {\mathbf {0}} S _ {1 1} & \hat {\mathbf {0}} S _ {2 2} & \hat {\mathbf {0}} S _ {3 3} & \hat {\mathbf {0}} S _ {1 2} & \hat {\mathbf {0}} S _ {2 3} & \hat {\mathbf {0}} S _ {3 1} \end{array} \right] $$ where $\delta\tilde{S}=\left[\begin{matrix}\delta S_{11}&\delta S_{12}&\delta S_{13}\\ \delta S_{21}&\delta S_{22}&\delta S_{23}\\ \delta S_{31}&\delta S_{32}&\delta S_{33}\end{matrix}\right]$ B. Updated Lagrangian formulation 1. Incremental strains $$ _ {i} \epsilon_ {i j} = \frac {1}{2} (_ {i} u _ {i, j} + _ {i} u _ {j, i}) + \frac {1}{2} (_ {i} u u _ {k, j}) \quad i = 1, 2, 3; j = 1, 2, 3; k = 1, 2, 3 $$ where $_{t}u_{i,j} = \frac{\partial u_{i}}{\partial^t x_j}$ 2. Linear strain-displacement transformation matrix Using, $\mathbf{e} = ;\mathbf{B}_L\hat{\mathbf{u}}$ where $\mathbf{e}^T = \left[ \begin{array}{llllll} e_{11} & e_{22} & e_{33} & 2_i e_{12} & 2_i e_{23} & 2_i e_{31} \end{array} \right]$ $$ \hat {\mathbf {u}} ^ {T} = \left[ u _ {1} ^ {1} \quad u _ {2} ^ {1} \quad u _ {3} ^ {1} \quad u _ {1} ^ {2} \quad u _ {2} ^ {2} \quad u _ {3} ^ {2} \quad \dots \quad u _ {1} ^ {N} \quad u _ {2} ^ {N} \quad u _ {3} ^ {N} \right] $$ $$ \mathbf {B} _ {L} = \left[ \begin{array}{c c c c c c} h _ {1, 1} & 0 & 0 & h _ {2, 1} & \dots & 0 \\ 0 & h _ {1, 2} & 0 & 0 & \dots & 0 \\ 0 & 0 & h _ {1, 3} & 0 & \dots & h _ {N, 3} \\ h _ {1, 2} & h _ {1, 1} & 0 & h _ {2, 2} & \dots & 0 \\ 0 & h _ {1, 3} & h _ {1, 2} & 0 & \dots & h _ {N, 2} \\ h _ {1, 3} & 0 & h _ {1, 1} & h _ {2, 3} & \dots & h _ {N, 1} \end{array} \right] $$ where $h_{k,j} = \frac{\partial h_k}{\partial^t x_j}$ ; $u_j^k = ^{t + \Delta t} u_j^k - ^t u_j^k$ ; $N =$ number of nodes 3. Nonlinear strain-displacement transformation matrix $$ { } ^ { \prime } \mathbf { B } _ { N L } = \left[ \begin{array} { c c c } { { } ^ { \prime } \tilde { \mathbf { B } } _ { N L } } & { { } \tilde { \mathbf { 0 } } } & { { } \tilde { \mathbf { 0 } } } \\ { { } \tilde { \mathbf { 0 } } } & { { } ^ { \prime } \tilde { \mathbf { B } } _ { N L } } & { { } \tilde { \mathbf { 0 } } } \\ { { } \tilde { \mathbf { 0 } } } & { { } \tilde { \mathbf { 0 } } } & { { } ^ { \prime } \tilde { \mathbf { B } } _ { N L } } \end{array} \right] ; \qquad \tilde { \mathbf { 0 } } = \left[ \begin{array} { c } 0 \\ 0 \\ 0 \end{array} \right] $$ TABLE 6.6 (cont.)

where

$\tilde{\mathbf{B}}_{NL} = \begin{bmatrix} h_{1,1} & 0 & 0 & h_{2,1} & \cdots & h_{N,1} \\ h_{1,2} & 0 & 0 & h_{2,2} & \cdots & h_{N,2} \\ h_{1,3} & 0 & 0 & h_{2,3} & \cdots & h_{N,3} \end{bmatrix}$

4. Cauchy stress matrix and stress vector $$ ^ {\prime} \boldsymbol {\tau} = \left[ \begin{array}{c c c} ^ {\prime} \widetilde {\boldsymbol {\tau}} & \overline {{\boldsymbol {0}}} & \overline {{\boldsymbol {0}}} \\ \overline {{\boldsymbol {0}}} & ^ {\prime} \widetilde {\boldsymbol {\tau}} & \overline {{\boldsymbol {0}}} \\ \overline {{\boldsymbol {0}}} & \overline {{\boldsymbol {0}}} & ^ {\prime} \widetilde {\boldsymbol {\tau}} \end{array} \right]; \quad^ {\prime} \hat {\boldsymbol {\tau}} = \left[ \begin{array}{c} ^ {\prime} \boldsymbol {\tau} _ {1 1} \\ ^ {\prime} \boldsymbol {\tau} _ {2 2} \\ ^ {\prime} \boldsymbol {\tau} _ {3 3} \\ ^ {\prime} \boldsymbol {\tau} _ {1 2} \\ ^ {\prime} \boldsymbol {\tau} _ {2 3} \\ ^ {\prime} \boldsymbol {\tau} _ {3 1} \end{array} \right] \quad \overline {{\boldsymbol {0}}} = \left[ \begin{array}{c c c} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right] $$ where $\tilde{\tau}=\begin{bmatrix}^{\prime}\tau_{11}&^{\prime}\tau_{12}&^{\prime}\tau_{13}\\^{\prime}\tau_{21}&^{\prime}\tau_{22}&^{\prime}\tau_{23}\\^{\prime}\tau_{31}&^{\prime}\tau_{32}&^{\prime}\tau_{33}\end{bmatrix}$ # 6.3.6 Exercises 6.31. Consider the problem shown and evaluate the following quantities in terms of the given data: $_{0}e_{ij}, _{0}\eta_{ij}, _{0}u_{k,i}, _{0}u_{k,j}, _{0}x_{i,k}$ . 

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Time 0 Time t 0L = 2 tU = 0.8 x Displacement = u 0.2 0.2 No change in cross- sectional area during deformations Node 2 Node 1 No displacement One 2-node element idealization is used

6.32. Consider the truss element shown. The truss has a cross-sectional area $A$ and a Young's modulus $E$ . We assume small strain conditions, i.e., $\Delta / ^0 L \ll 1$ . 

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0L Δ↑

(a) Evaluate the total stiffness matrix as a function of $\Delta$ and plot the linear strain stiffness matrix element $\mathbf{\delta}_{0}\mathbf{K}_{L}$ and nonlinear strain stiffness matrix element $\mathbf{\delta}_{0}\mathbf{K}_{NL}$ as a function of $\Delta$ . (b) Let R be the external load applied to obtain the displacement $\Delta$ . Plot the force R as a function of $\Delta$ . 6.33. Consider the snap-action toggle shown in its initial configuration. Assume small strain conditions and that each element has a cross-sectional area $A$ and Young's modulus $E$ . (a) For each element, calculate the linear and nonlinear strain stiffness matrices $\mathbf{t}_{0}\mathbf{K}_{L}$ and $\mathbf{t}_{0}\mathbf{K}_{NL}$ and the force vector $\mathbf{t}_{0}\mathbf{F}$ . (b) Calculate the linear and nonlinear strain stiffness matrices $\mathbf{^t}\mathbf{K}_L$ and $\mathbf{^t}\mathbf{K}_{NL}$ and the force vector $\mathbf{^t}\mathbf{F}$ of the complete toggle. Eliminate prescribed degrees of freedom. (c) Using your results from part (b), establish the force-deflection curve P versus $\Delta$ . 

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Element ① ② θ P 0.5 Δ 5 5

Initial stress-free configuration 6.34. Consider the three-element truss structure shown. Derive the tangent stiffness matrix $\delta \mathbf{K}$ and force vector $\delta \mathbf{F}$ corresponding to the configuration at time $t$ allowing for large displacements, large rotations, and large strains. Assume that the constitutive relationship is $\delta S_{11} = C\delta \epsilon_{11}$ , with $C$ given as some function of the strain. 

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Node 1 t u₁¹ t u₂¹ Time t ⁰A = 1 ⁰A = 1 Time 0 x₂ x₁ 5 5 5

6.35. Consider the evaluation of the stiffness matrix of the four-node element shown when calculated with the information given in Table 6.5. Let the two-dimensional element be loaded with a deformation-dependent pressure between nodes 1 and 2. Establish the terms that should be added to the stiffness matrix if the effect of the pressure is included in the linearization to obtain the exact tangent stiffness matrix. Consider plane stress, plane strain, and axisymmetric conditions. 

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p|time 0 p|time t x2 x1 Time 0 1 2 1 2 Time t p|time t = f(txi)

6.36. The initial configuration and configuration at time t of a four-node plane strain element are as shown. The material law is linear, $\delta S_{ij} = \delta C_{ijrs} \delta \epsilon_{rs}$ , with $E = 20,000 \, N/m^{2}$ and $\nu = 0.3$ . (a) Calculate the nodal point forces required to hold the element in equilibrium at time t. Use an appropriate finite element formulation. (b) If the element now rotates rigidly from time $t$ to time $t + \Delta t$ by an angle of 90 degrees counterclockwise, calculate the new nodal point forces corresponding to the configuration at time $t + \Delta t$ . 

scatter

| Point | x1 | x2 | Label | |-------|----|----|-------| | 1 | 4 | 2 | 1 | | 2 | 4 | 2 | 2 | | 3 | 4 | 2 | 3 | | 4 | 4 | 2 | 4 | | 5 | 6 | 2 | 1 | | 6 | 6 | 2 | 2 | | 7 | 6 | 2 | 3 | | 8 | 6 | 2 | 4 | | 9 | 6 | 2 | 3 | | 10 | 6 | 2 | 2 | | 11 | 6 | 2 | 1 | | 12 | 6 | 2 | 0 | | 13 | 6 | 2 | 1 | | 14 | 6 | 2 | 2 | | 15 | 6 | 2 | 3 | | 16 | 6 | 2 | 4 | | 17 | 6 | 2 | 5 | | 18 | 6 | 2 | 6 | | 19 | 6 | 2 | 7 | | 20 | 6 | 2 | 8 | | 21 | 6 | 2 | 9 | | 22 | 6 | 2 | 10 | | 23 | 6 | 2 | 11 | | 24 | 6 | 2 | 12 | | 25 | 6 | 2 | 13 | | 26 | 6 | 2 | 14 | | 27 | 6 | 2 | 15 | | 28 | 6 | 2 | 16 | | 29 | 6 | 2 | 17 | | 30 | 6 | 2 | 18 | | 31 | 6 | 2 | 19 | | 32 | 6 | 2 | 20 | | 33 | 6 | 2 | 21 | | 34 | 6 | 2 | 22 | | 35 | 6 | 2 | 23 | | 36 | 6 | 2 | 24 | | 37 | 6 | 2 | 25 | | 38 | 6 | 2 | 26 | | 39 | 6 | 2 | 27 | | 40 | 6 | 2 | 28 | | 41 | 6 | 2 | 29 | | 42 | 6 | 2 | 30 | | 43 | 6 | 2 | 31 | | 44 | 6 | 2 | 32 | | 45 | 6 | 2 | 33 | | 46 | 6 | 2 | 34 | | 47 | 6 | 2 | 35 | | 48 | 6 | 2 | 36 | | 49 | 6 | 2 | 37 | | 50 | 6 | 2 | 38 | | 51 | 6 | 2 | 39 | | 52 | 6 | 2 | 40 | | 53 | 6 | 2 | 41 | | 54 | 6 | 2 | 42 | | 55 | 6 | 2 | 43 | | 56 | 6 | 2 | 44 | | 57 | 6 | 2 | 45 | | 58 | 6 | 2 | 46 | | 59 | 6 | 2 | 47 | | 60 | 6 | 2 | 48 | | 61 | 6 | 2 | 49 | | 62 | 6 | 2 | 50 | | 63 | 6 | 2 | 51 | | 64 | 6 | 2 | 52 | | 65 | 6 | 2 | 53 | | 66 | 6 | 2 | 54 | | 67 | 6 | 2 | 55 | | 68 | 6 | 2 | 56 | | 69 | 6 | 2 | 57 | | 70 | 6 | 2 | 58 | | 71 | 6 | 2 | 59 | | 72 | 6 | 2 | 60 | | 73 | 6 | 2 | 61 | | 74 | 6 | 2 | 62 | | 75 | 6 | 2 | 63 | | 76 | 6 | 2 | 64 | | 77 | 6 | 2 | 65 | | 78 | 6 | 2 | 66 | | 79 | 6 | 2 | 67 | | 80 | 6 | 2 | 68 | | 81 | 6 | 2 | 69 | | 82 | 6 | 2 | 70 | | 83 | 6 | 2 | 71 | | 84 | 6 | 2 | 72 | | 85 | 6 | 2 | 73 | | 86 | 6 | 2 | 74 | | 87 | 6 | 2 | 75 | | 88 | 6 | 2 | 76 | | 89 | 6 | 2 | 77 | | 90 | 6 | 2 | 78 | | 91 | 6 | 2 | 79 | | 92 | 6 | 2 | 80 | | 93 | 6 | 2 | 81 | | 94 | 6 | 2 | 82 | | 95 | 6 | 2 | 83 | | 96 | 6 | 2 | 84 | | 97 | 6 | 2 | 85 | | 98 | 6 | 2 | 86 | | 99 | 6 | 2 | 87 | | 100 | 6 | 2 | 88 |

6.37. During a TL analysis, we find that a plane strain element is deformed as shown. 

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y 0.1 0.2 2 1 Time t 0.2 3 4 0.2 x 0.3 Time 0 All dimensions are in meters

The stress state, not including $\tau_{zz}$ , is $$ ^ \prime \tau = \left[ \begin{array}{l l} 5. 8 4 9 \times 1 0 ^ {7} & 6. 9 7 1 \times 1 0 ^ {7} \\ 6. 9 7 1 \times 1 0 ^ {7} & 1. 5 1 4 \times 1 0 ^ {8} \end{array} \right] \mathrm{Pa} $$ The Poisson's ratio $\nu = 0.3$ and the tangent Young's modulus is $E$ . Compute $\delta K_{11}$ . 6.38. The two-dimensional four-node isoparametric finite element shown is used in an axisymmetric analysis. Evaluate the last row in the $\mathbf{\delta_{0}B_{L}}$ , $\mathbf{\delta_{0}B_{NL}}$ and $\mathbf{\delta_{1}B_{L}}$ , $\mathbf{\delta_{1}B_{NL}}$ matrices at the material particle P corresponding to the TL and UL formulations. The last row in the strain-displacement matrices corresponds to the circumferential strain. 

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x2 7.5 1 (13, 13) Time 0 s r 2 (5, 7) 3 (5, 2) 4 (11, 2) x1


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x2 2 (6, 14) P 1 (18, 16) Time t s r 3 (9, 6) 4 (20, 5) x1

6.39. Consider the four-node plane stress element shown. Using the total Lagrangian formulation calculate the following. (a) The element of the tangent stiffness matrix corresponding to the incremental displacement $u_{1}^{1}$ ; i.e., evaluate element (1, 1) of the matrix $(_{0}^{\prime}\mathbf{K}_{L} + _{0}^{\prime}\mathbf{K}_{NL})$ . (b) The element of the force vector $\delta\mathbf{F}$ corresponding to $u_{1}^{1}$ ; i.e., evaluate element (1) of $\delta\mathbf{F}$ , where $\delta\mathbf{F}$ is the force vector corresponding to the current element stresses. Assume that Young's modulus $E$ and Poisson's ratio $\nu = 0.3$ relate the incremental second Piola-Kirchhoff stresses to the incremental Green-Lagrange strains and assume thickness $h$ at time 0. 

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0.5 x2 4 x1 t0S12 = 0 t0S22 = 60 t0S11 = 100 6 1.5 u1 At time t At time 0

Constant stresses $_{0}^{t}S_{11}$ and $_{0}^{t}S_{22}$ and all other stresses are zero 6.40. A two-node finite element for modeling large strain torsion problems is to be constructed. The element has a circular cross section and is straight, and all cross sections are parallel to the $x_{1}$ , $x_{2}$ plane as shown. The kinematic assumption to be employed in the element is that each cross section rotates rigidly about its center. This is illustrated in the figure. Notice that the total rotation of fiber AA is fully described by $\theta_{3}$ and that the fiber rotation can be large. Also, note that the fiber AA does not stretch or shrink and that the center of the fiber (point C) remains fixed. 

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x2 1 2 x3 L x2 D x1 x1 x2 A typical material fiber A' A' C A tθ3 2 P(tx1, tx2, tx3) P(0x1, 0x2, 0x3) Fiber AA rotates to A'A' Enlarged view

(a) Calculate the deformation gradient $\delta X$ in terms of the initial coordinates and $\theta_{3}$ . (b) Calculate the Green-Lagrange strain tensor $\delta \epsilon$ . Clearly identify any terms that are associated with large strain effects. (c) Calculate the mass density ratio $\rho/^{0}\rho$ in terms of the initial coordinates and $\theta_{3}$ . (d) Establish the strain-displacement matrix of the element. # 6.4 DISPLACEMENT/PRESSURE FORMULATIONS FOR LARGE DEFORMATIONS As discussed in Section 4.4.3, for (almost) incompressible analysis, a pure displacement-based procedure is, in general, not effective and instead, a displacement/pressure formulation is attractive. Materials in large deformations frequently behave as almost incompressible, and it is therefore important to extend the total and updated Lagrangian formulations of the previous sections to incompressible analysis. Typical applications are in the large strain analysis of rubberlike materials and in the large strain inelastic analysis of metals. The formulations we present here are a direct and natural extension of the pure displacement-based large deformation formulations given in the previous section and of the pressure/displacement formulations that we discussed for linear analysis in Sections 4.4.3 and 5.3.5. # 6.4.1 Total Lagrangian Formulation We make the fundamental assumption that the material description used has an incremental potential $d_0^t\overline{W}$ such that $$ d _ {0} ^ {t} \overline {{{W}}} = _ {0} ^ {t} \overline {{{S}}} _ {i j} d _ {0} ^ {t} \epsilon_ {i j} \tag {6.129} $$ and hence $\delta_{0}\overline{S}_{ij} = \frac{\partial_{0}^{t}\overline{W}}{\partial_{0}^{t}\epsilon_{ij}}$ (6.130) where the overbar in $d_{0}^{i}\overline{W}$ and on the second Piola-Kirchhoff stress (and other quantities in the following discussion) denotes that the quantity is computed only from the displacement fields. Since we shall interpolate the displacements and the pressure independently, the actual stress $_{0}^{i}S_{ij}$ will also contain the interpolated pressure. We note that such incremental potential is given for elastic materials and also for inelastic materials provided the normality rule holds. A consequence of $(6.129)$ is that the tensor $$ _ 0 \overline {{C}} _ {i j r s} = \frac {\partial_ {0} ^ {t} \overline {{S}} _ {i j}}{\partial_ {0} ^ {t} \epsilon_ {r s}} = \frac {\partial^ {2} {} _ {0} ^ {t} \overline {{W}}}{\partial_ {0} ^ {t} \epsilon_ {r s} \partial_ {0} ^ {t} \epsilon_ {i j}} \tag {6.131} $$ has the symmetry property $$ _ {0} \overline {{C}} _ {i j r s} = _ {0} \overline {{C}} _ {r s i j} $$ and the pure displacement and displacement/pressure formulations produce symmetric coefficient matrices. Using (6.130), the principle of virtual displacements at time $t$ in total Lagrangian form with displacements as the only variables can be written as $$ \begin{array}{l} \int_ {0 _ {V}} \frac {\partial_ {0} ^ {t} \overline {{{W}}}}{\partial_ {0} ^ {t} \epsilon_ {i j}} \delta_ {0} ^ {t} \epsilon_ {i j} d ^ {0} V = \int_ {0 _ {V}} \delta_ {0} ^ {t} \overline {{{W}}} d ^ {0} V \tag {6.132} \\ = \delta \left(\int_ {0 _ {V}} \overline {{{{W}}}} d ^ {0} V\right) = ^ {\prime} \mathcal {R} \\ \end{array} $$ The linearization and finite element discretization of $(6.132)$ was presented in Section 6.3. We now use $(6.132)$ as the starting equation to develop the displacement/pressure formulation for large deformations. The basic element interpolations that we shall use are $$ { } ^ { \prime } u _ { i } = \sum _ { k = 1 } ^ { N } h _ { k } { } ^ { \prime } u _ { i } ^ { k } ; \quad { } ^ { \prime } \tilde { p } = \sum _ { i = 1 } ^ { q } g _ { i } { } ^ { \prime } \hat { p } _ { i } \tag {6.133} $$ where the $h_{k}$ are the displacement interpolation functions and the $g_{i}$ are the pressure interpolation functions, with $\tilde{p}$ as the total element pressure at time t. Note that the interpolation of the pressure may correspond to the u/p or to the u/p-c formulation (see Section 5.3.5). The key step in the construction of the displacement/pressure formulation is to properly modify the potential to include the effect of the interpolated pressure. For this purpose we add to the potential $\delta \overline{W}$ a properly chosen potential $\delta Q$ , which is a function of the displacements and the separately interpolated pressure $t\tilde{p}$ (see T. Sussman and K. J. Bathe [B]). The principle of virtual work is then given by $$ \delta \left(\int_ {0 _ {V}} \delta W d ^ {0} V\right) = ^ {t} \mathcal {R} \tag {6.134} $$ where $$ { } _ { 0 } ^ { \prime } W = { } _ { 0 } ^ { \prime } \overline { { W } } + { } _ { 0 } ^ { \prime } Q \tag {6.135} $$ and we now consider the variation with respect to the interpolated displacements and the interpolated pressure. The modified potential $\delta W$ must fulfill the requirements that use of (6.134) gives $\tilde{p}$ as the actual solution for the pressure and also yields a physically reasonable constraint between the interpolated pressure and the pressure computed only from the displacements. A potential that fulfills these requirements for the isotropic materials considered later is given by $$ { } _ { 0 } ^ { \prime } W = { } _ { 0 } ^ { \prime } \overline { { W } } - \frac { 1 } { 2 \kappa } ( { } ^ { \prime } \tilde { p } - { } ^ { \prime } \tilde { p } ) ^ { 2 } \tag {6.136} $$ where $\kappa$ is the constant bulk modulus of the material. Using (6.136), the governing finite element equations can be derived with the approach for linearization presented in Section 6.3.1. Hence, we obtain for a typical element, $$ \left( \begin{array}{c c} ^ {\prime} \mathbf {K U U} & ^ {\prime} \mathbf {K U P} \\ ^ {\prime} \mathbf {K P U} & ^ {\prime} \mathbf {K P P} \end{array} \right) \binom {\hat {\mathbf {u}}} {\hat {\mathbf {p}}} = \binom {^ {\prime + \Delta t} \mathbf {R}} {\mathbf {0}} - \binom {^ {\prime} \mathbf {F U}} {^ {\prime} \mathbf {F P}} \tag {6.137} $$ where $\hat{u}$ and $\hat{p}$ are vectors of the increments in the nodal point displacements $\hat{u}_{i}$ and nodal point or element internal pressure variables $\hat{p}_{i}$ [note that here $\hat{u}_{i}$ is any one of the components $\hat{u}_{i}^{k}$ in (6.122), (6.128), and (6.133)]. The vectors $\hat{F}U$ and $\hat{F}P$ contain the entries $$ { } ^ { t } F U _ { i } = \frac { \partial } { \partial { } ^ { t } \hat { u } _ { i } } \left( \int _ { 0 _ { V } } { } ^ { t } W d ^ { 0 } V \right) \tag {6.138} $$ $$ { } ^ { \prime } F P _ { i } = \frac { \partial } { \partial ^ { \prime } \hat { p } _ { i } } \left( \int _ { 0 _ { V } } { } ^ { \prime } W d ^ { 0 } V \right) $$ and the matrices 'KUU, 'KUP, 'KPU, and 'KPP contain the elements $$ ^ {\prime} K U U _ {i j} = \frac {\partial^ {\prime} F U _ {i}}{\partial^ {\prime} \hat {u} _ {j}} $$ $$ ^ {t} K U P _ {i j} = \frac {\partial^ {t} F U _ {i}}{\partial^ {t} \hat {p} _ {j}} = \frac {\partial^ {t} F P _ {j}}{\partial^ {t} \hat {u} _ {i}} = ^ {t} K P U _ {j i} \tag {6.139} $$ $$ { } ^ { t } K P P _ { i j } = \frac { \partial ^ { t } F P _ { i } } { \partial ^ { t } \hat { p } _ { j } } $$ Using chain differentiation, we obtain $$ ^ {t} F U _ {i} = \int_ {0 _ {V}} ^ {t} S _ {k l} \frac {\partial_ {0} ^ {t} \epsilon_ {k l}}{\partial^ {t} \hat {u} _ {i}} d ^ {0} V $$ $$ ^ {\prime} F P _ {i} = \int_ {0 _ {V}} \frac {1}{\kappa} \left(^ {\prime} \bar {p} - ^ {\prime} \tilde {p}\right) \frac {\partial^ {\prime} \tilde {p}}{\partial^ {\prime} \hat {p} _ {i}} d ^ {0} V $$ $$ { } ^ { t } K U U _ { i j } = \int _ { 0 _ { V } } { } _ { 0 } C U U _ { k l r s } \frac { \partial _ { 0 } ^ { t } \epsilon _ { k l } } { \partial ^ { t } \hat { u } _ { i } } \frac { \partial _ { 0 } ^ { t } \epsilon _ { r s } } { \partial ^ { t } \hat { u } _ { j } } d ^ { 0 } V + \int _ { 0 _ { V } } { } _ { 0 } S _ { k l } \frac { \partial ^ { 2 } { } _ { 0 } \epsilon _ { k l } } { \partial ^ { t } \hat { u } _ { i } \partial ^ { t } \hat { u } _ { j } } d ^ { 0 } V \tag {6.140} $$ $$ { } ^ { t } K U P _ { i j } = \int _ { 0 _ { V } } { } ^ { 0 } C U P _ { k l } \frac { \partial _ { 0 } ^ { t } \epsilon _ { k l } } { \partial ^ { t } \hat { u } _ { i } } \frac { \partial ^ { t } \tilde { p } } { \partial ^ { t } \hat { p } _ { j } } d ^ { 0 } V $$ $$ ^ {\prime} K P P _ {i j} = \int_ {0 _ {V}} - \frac {1}{\kappa} \frac {\partial^ {t} \tilde {p}}{\partial^ {t} \hat {p} _ {i}} \frac {\partial^ {t} \tilde {p}}{\partial^ {t} \hat {p} _ {j}} d ^ {0} V $$ where $t_0 S_{kl} = t_0 \bar{S}_{kl} - \frac{1}{\kappa} (t \bar{p} - t \tilde{p}) \frac{\partial^t \bar{p}}{\partial_0^t \epsilon_{kl}}$ $$ { } _ { 0 } C U U _ { k l r s } = { } _ { 0 } \overline { { C } } _ { k l r s } - \frac { 1 } { \kappa } \frac { \partial ^ { t } \bar { p } } { \partial _ { 0 } ^ { t } \epsilon _ { k l } } \frac { \partial ^ { t } \bar { p } } { \partial _ { 0 } ^ { t } \epsilon _ { r s } } - \frac { 1 } { \kappa } ( { } ^ { t } \bar { p } - { } ^ { t } \tilde { p } ) \frac { \partial ^ { 2 } { } ^ { t } \bar { p } } { \partial _ { 0 } ^ { t } \epsilon _ { k l } \partial _ { 0 } ^ { t } \epsilon _ { r s } } \tag {6.141} $$ $$ _ 0 C U P _ {k l} = \frac {1}{\kappa} \frac {\partial^ {t} \bar {p}}{\partial_ {0} ^ {t} \epsilon_ {k l}} $$