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<!-- source-page: 581 -->
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Note that in (6.141) we have
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$$
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{ } _ { 0 } ^ { t } \overline { { S } } _ { k l } = \frac { 1 } { 2 } \left( \frac { \partial _ { 0 } ^ { t } \overline { { W } } } { \partial _ { 0 } ^ { t } \epsilon _ { k l } } + \frac { \partial _ { 0 } ^ { t } \overline { { W } } } { \partial _ { 0 } ^ { t } \epsilon _ { l k } } \right)
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$$
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$$
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{ } _ { 0 } ^ { t } \overline { { C } } _ { k l r s } = \frac { 1 } { 2 } \left( \frac { \partial _ { 0 } ^ { t } \overline { { S } } _ { k l } } { \partial _ { 0 } ^ { t } \epsilon _ { r s } } + \frac { \partial _ { 0 } ^ { t } \overline { { S } } _ { k l } } { \partial _ { 0 } ^ { t } \epsilon _ { s r } } \right) \tag {6.142}
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$$
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Furthermore, we note that with the interpolations of (6.133) we have
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$$
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\frac {\partial^ {\prime} \tilde {p}}{\partial^ {\prime} \hat {p} _ {i}} = g _ {i} \tag {6.143}
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$$
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and (see Exercise 6.42) $\frac{\partial_0^t\epsilon_{kl}}{\partial^t u_n^L} = \frac{1}{2} (\dot{0} x_{n,k0}h_{L,l} + \dot{0} x_{n,l0}h_{L,k})$ (6.144)
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$$
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\frac {\partial^ {2} {} _ {0} ^ {t} \epsilon_ {k l}}{\partial^ {t} u _ {n} ^ {L} \partial^ {t} u _ {m} ^ {M}} = \frac {1}{2} (_ {0} h h _ {M, l} + _ {0} h h _ {M, k}) \delta_ {n m} \tag {6.145}
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$$
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where a typical nodal point displacement is denoted as $u_n^L$ (with the appropriate indices $n$ and $L$ ). These strain derivatives give the same contributions as do the quantities $_0e_{ij}$ and $_0\eta_{ij}$ used in Table 6.2.
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A study of the above relations shows that if the pressure interpolation is not included, the equations reduce to the total Lagrangian formulation already presented in Section 6.2.3 (see Exercise 6.43).
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The displacement/pressure formulation is effective for the analysis of rubberlike materials in large strains. In this case, the Mooney-Rivlin or Ogden material laws may be used, for which the strain energy density per unit volume $\delta\overline{W}$ is explicitly defined (see Section 6.6.2).
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Let us demonstrate that this formulation, when used in small strain elastic analysis, reduces to the formulation already discussed in Section 5.3.5.
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EXAMPLE 6.19: Show how the displacement/pressure formulation discussed above reduces to the formulation presented in Section 5.3.5 when isotropic linear elasticity with small displacements and small strains is considered.
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Considering the general equations (6.137) to (6.145), we note that in this case:
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The second Piola-Kirchhoff stress $\delta S_{kl}$ reduces to the engineering stress measure $\sigma_{kl}$ .
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The Green-Lagrange strains $_{0}^{t}\epsilon_{kl}$ reduce to the infinitesimally small engineering strains $^{t}e_{kl}$ . The nonlinear strain stiffness matrix in (6.140) is neglected.
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The integration is over the volume V (which is equal to $^{0}V$ ) and the subscript 0 on the constitutive tensors is also not needed.
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In this case we have
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$$
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\overline {{{C}}} _ {k l r s} = \lambda \delta_ {k l} \delta_ {r s} + \mu (\delta_ {k r} \delta_ {l s} + \delta_ {k s} \delta_ {l r})
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$$
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where $\lambda$ and $\mu$ are the Lamé constants,
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$$
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\lambda = \frac {E \nu}{(1 + \nu) (1 - 2 \nu)}; \quad \mu = \frac {E}{2 (1 + \nu)}
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$$
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<!-- source-page: 582 -->
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with $E$ and $\nu$ the Young's modulus and the Poisson's ratio. The bulk modulus $\kappa$ is
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$$
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\kappa = \frac {E}{3 (1 - 2 \nu)}
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$$
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We have $\overline{p} = -\kappa^{\prime}e_{mm}$
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$$
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\frac {\partial^ {t} \bar {p}}{\partial^ {t} e _ {k l}} = - \kappa \delta_ {k l}
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$$
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$$
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\frac {\partial^ {2} {} ^ {t} \bar {p}}{\partial^ {t} e _ {k l} \partial^ {t} e _ {r s}} = 0
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$$
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so that $^t\sigma_{kl} = ^tS_{kl} - ^t\tilde{p} \delta_{kl}$
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$$
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C U U _ {k l r s} = \overline {{{C}}} _ {k l r s} - \kappa \delta_ {k l} \delta_ {r s}
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$$
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$$
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C U P _ {k l} = - \delta_ {k l}
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$$
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On substituting these quantities into (6.137), we note that the general formulation reduces, in this case, to the formulation already presented in Sections 4.4.3 and 5.3.5.
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# 6.4.2 Updated Lagrangian Formulation
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As we discussed in Section 6.2.3, the updated Lagrangian formulation is conceptually identical to the total Lagrangian formulation but uses the configuration at time t as reference configuration. In this case $\frac{t}{T}S_{ij} = \tau_{ij}$ and $d_{T}^{t}\epsilon_{ij} = d_{t}e_{ij}$ , with the subscript T denoting the configuration $^{8}$ that is fixed and used as reference, and
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$$
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d _ {t} e _ {i j} = \frac {1}{2} \left(\frac {\partial d u _ {i}}{\partial^ {t} x _ {j}} + \frac {\partial d u _ {j}}{\partial^ {t} x _ {i}}\right) \tag {6.146}
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$$
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Following the presentation of the previous section, we thus obtain
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$$
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d _ {T} ^ {i} \overline {{W}} = ^ {i} \overline {{\tau}} _ {i j} d _ {i} e _ {i j} \tag {6.147}
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$$
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and note that $\dot{f}\overline{W}d^{T}V=\dot{f}_{0}\overline{W}d^{0}V$ (6.148)
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If in addition we use
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$$
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\dot {t} Q d ^ {T} V = \dot {0} Q d ^ {0} V; \quad \frac {d ^ {T} V}{d ^ {0} V} = \det \dot {T} \mathbf {X} \tag {6.149}
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$$
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we can write the principle of virtual work (6.134) as
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$$
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\delta \int_ {\tau_ {V}} \left(\frac {t}{t} \overline {{{W}}} + \frac {t}{t} Q\right) d ^ {T} V = ^ {t} \mathcal {R} \tag {6.150}
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$$
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Note that if we use the modification to the total potential $\delta \overline{W}$ in the previous section,
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$$
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_ 0 Q = - \frac {1}{2 \kappa} \left(^ {t} \bar {p} - ^ {t} \tilde {p}\right) ^ {2} \tag {6.151}
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$$
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<!-- source-page: 583 -->
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then
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$$
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\dot {r} Q = - \frac {1}{2 \kappa^ {*}} \left(^ {\prime} \bar {p} - ^ {\prime} \tilde {p}\right) ^ {2} \tag {6.152}
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$$
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with
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$$
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\kappa^ {*} = \kappa \det _ {0} ^ {T} \mathbf {X} \tag {6.153}
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$$
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The governing finite element equations can now be derived by chain differentiation, and provided the same physical material descriptions are used, the same finite element equations are obtained as in the total Lagrangian formulation. The details of the derivation are given by T. Sussman and K. J. Bathe [B].
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# 6.4.3 Exercises
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6.41. Show that using (6.136), the actual solution for the pressure is given by the independently interpolated value $\tilde{p}$ .
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6.42. Let $u_{n} = \Sigma_{L} h_{L} u_{n}^{L}$ and prove that (6.144) and (6.145) hold. Here you may want to recall that
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$$
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\frac {\partial (A _ {M , i} {} ^ {t} u _ {i} ^ {M})}{\partial^ {t} u _ {k} ^ {L}} = A _ {M, i} \delta_ {i k} \delta_ {M L} = A _ {L, k}
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$$
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where $\delta_{ik}$ is the Kronecker delta.
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6.43. Show explicitly that the pressure/displacement mixed formulation reduces to the pure displacement-based formulation if the pressure interpolation is not included.
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6.44. Prove the relations in (6.140) and (6.141).
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6.45. Consider the 4/1 plane strain element shown. Develop in detail all expressions for the calculation of the matrices in (6.137) assuming large strain analysis but do not perform any integrations. (Hint: See Example 4.32.)
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<details>
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<summary>text_image</summary>
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x₂
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2
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x₁
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2
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</details>
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Bulk modulus $\kappa$ Shear modulus $G$
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6.46. Consider the 4/1 element in Exercise 6.45 and develop in detail all expressions for the calculation of the matrices in (6.137) but corresponding to the updated Lagrangian formulation. However, do not perform any integrations.
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6.47. You want to obtain some insight into whether the computer program you use employs the tangent stiffness matrix in plane strain analysis. Consider a single nine-node element in the deformed state shown. Assuming that the calculation of the stresses and the force vector 'F is correct, design a test to determine whether the stiffness matrix calculation for node 1 is probably also correct. For this analysis case the $u / p$ formulation (9/3 element) would be efficient. (Hint: Note that 'K = $\partial'F/\partial'U$ .)
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<!-- source-page: 584 -->
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<details>
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<summary>text_image</summary>
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20 mm
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20 mm
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Time 0
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6 mm
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20 mm
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20 mm
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5 mm
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Time t
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1
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All midnodes are
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halfway between
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corners
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9-node element in plane strain
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Mooney-Rivlin rubber model
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C₁ = 0.6 MPa,
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C₂ = 0.3 MPa,
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κ = 2000 MPa
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</details>
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6.48. Perform the numerical experiment in Exercise 6.47 for the case of the axisymmetric element shown.
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<details>
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<summary>text_image</summary>
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Time 0
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5
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Time t
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10
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10
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10
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10
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Node 1
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All midnodes are halfway
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between corner nodes
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Mooney-Rivlin rubber model
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C₁ = 0.6 MPa,
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C₂ = 0.3 MPa,
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κ = 2000 MPa
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</details>
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6.49. Use a computer program to analyze the thick disk shown. The applied pressure increases uniformly, and the analysis is required up to a maximum displacement of 3 in.
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<details>
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<summary>text_image</summary>
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1 in
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L = 10 in
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E = 200 lb/in²
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v = 0.499
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</details>
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6.50. Use a computer program to analyze the plate with a hole shown on the following page. The plate is stretched by imposing a uniform horizontal displacement at the right end.
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<!-- source-page: 585 -->
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<details>
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<summary>text_image</summary>
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10 in
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6 in
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6 in
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20 in
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3 in
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Prescribed displacement
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to 3 in
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</details>
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$E = 200\mathrm{lb / in}^2$
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$\nu = 0.499$
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Plane strain conditions
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# 6.5 STRUCTURAL ELEMENTS
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A large number of beam, plate, and shell elements have been proposed for nonlinear analysis (see, for example, A. K. Noor [A]). Our objective here is not to survey the various formulations proposed in the literature but to present briefly those elements that we already have discussed for linear analysis in Section 5.4. These beam, plate, and shell elements have evolved from the isoparametric formulation and are particularly attractive because of the consistent formulation, the generality of the elements, and the computational efficiency.
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In the following discussion, we first consider beam and axisymmetric shell elements and then discuss plate and general shell elements.
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# 6.5.1 Beam and Axisymmetric Shell Elements
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In this section we consider the one-dimensional bending elements that we discussed in Section 5.4.1 for linear analysis; there we considered the plane stress and plane strain planar beam elements, an axisymmetric shell element, and a general three-dimensional beam element. We observed that the planar beam and the axisymmetric shell element formulations are actually cases easily derivable from the general three-dimensional beam element formulation. Hence, we consider here the calculation of the element matrices pertaining to the large displacement--large rotation behavior of a general beam of rectangular cross-sectional area. The relations given can be directly used to also obtain the matrices corresponding to the planar beam elements and axisymmetric shell elements (see Examples 6.20 and 6.21).
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Figure 6.5 shows a typical element in the original configuration and the position at time t. To describe the element behavior we use the same assumptions that we employed in linear analysis (namely, that plane sections initially normal to the neutral axis remain plane and that only the longitudinal stress and two shear stresses are nonzero), but the displacements and rotations of the element can now be arbitrarily large. The element strains are still
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<!-- source-page: 586 -->
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<details>
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<summary>text_image</summary>
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At time t
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At time zero
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b2
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a2
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b1
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r, η
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4
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s, ξ
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t, ξ
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t'V1s
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3
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1
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2
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a1
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b2
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a2
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x3
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x2
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x1
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b1
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3
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r, η
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4
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s, ξ
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t, ξ
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1
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0V1s
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a1
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0V1t
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</details>
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Figure 6.5 Beam element undergoing large displacements and rotations
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assumed to be small, which means that the cross-sectional area does not change. $^{9}$ This is an appropriate assumption for most geometrically nonlinear analyses of beam-type structures.
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Using the general continuum mechanics equations for nonlinear analysis presented in Section 6.2, the beam element matrices for nonlinear analysis are evaluated by a direct extension of the formulation given in Section 5.4.1. The calculations are performed as in the evaluation of the matrices of the finite elements with displacement degrees of freedom only (see Sections 5.4.1 and 6.3).
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With the same notation as in Section 5.4.1, the geometry of the beam element at time $t$ is given by
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$$
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{ } ^ { t } x _ { i } = \sum _ { k = 1 } ^ { q } h _ { k } { } ^ { t } x _ { i } ^ { k } + \frac { t } { 2 } \sum _ { k = 1 } ^ { q } a _ { k } h _ { k } { } ^ { t } V _ { i i } ^ { k } + \frac { s } { 2 } \sum _ { k = 1 } ^ { q } b _ { k } h _ { k } { } ^ { t } V _ { s i } ^ { k } \quad i = 1, 2, 3 \tag {6.154}
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$$
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where the coordinates of a typical point in the beam are $x_{1}$ , $x_{2}$ , $x_{3}$ . Considering the
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<!-- source-page: 587 -->
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configurations at times 0, t, and $t + \Delta t$ , the displacement components are
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$$
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{ } ^ { t } u _ { i } = { } ^ { t } x _ { i } - { } ^ { 0 } x _ { i } \tag {6.155}
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$$
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and $u_{i} = ^{t + \Delta t}x_{i} - ^{t}x_{i}$ (6.156)
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Substituting (6.154) into (6.155) and (6.156), we obtain expressions for the displacement components in terms of the nodal point displacements and changes in the direction cosines of the nodal point director vectors; i.e.,
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$$
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{ } ^ { t } u _ { i } = \sum _ { k = 1 } ^ { q } h _ { k } { } ^ { t } u _ { i } ^ { k } + \frac { t } { 2 } \sum _ { k = 1 } ^ { q } a _ { k } h _ { k } \left( { } ^ { t } V _ { t i } ^ { k } - { } ^ { 0 } V _ { t i } ^ { k } \right) + \frac { s } { 2 } \sum _ { k = 1 } ^ { q } b _ { k } h _ { k } \left( { } ^ { t } V _ { s i } ^ { k } - { } ^ { 0 } V _ { s i } ^ { k } \right) \tag {6.157}
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$$
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and $u_{i} = \sum_{k = 1}^{q}h_{k}u_{i}^{k} + \frac{t}{2}\sum_{k = 1}^{q}a_{k}h_{k}V_{ti}^{k} + \frac{s}{2}\sum_{k = 1}^{q}b_{k}h_{k}V_{si}^{k}$ (6.158)
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where $V_{ti}^{k} = {}^{t + \Delta t}V_{ti}^{k} - {}^{t}V_{ti}^{k}$ (6.159)
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$$
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V _ {s i} ^ {k} = ^ {t + \Delta t} V _ {s i} ^ {k} - ^ {t} V _ {s i} ^ {k} \tag {6.160}
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$$
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The relation in (6.157) is directly employed to evaluate the total displacements and total strains (hence also total stresses) for both the UL and TL formulations and holds for any magnitude of displacement components.
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We use the relation in (6.158) in the linearization of the principle of virtual work and need to express the components $V_{ti}^{k}$ and $V_{si}^{k}$ of the vectors $V_{t}^{k}$ , $V_{s}^{k}$ in terms of nodal rotational degrees of freedom. Depending on the size of the incremental step, the actual rotation corresponding to the vectors $V_{t}^{k}$ and $V_{s}^{k}$ may be a large rotation, and therefore cannot be represented by vector component rotations about the Cartesian axes. However, we recall that our objective is to express the continuum linear and nonlinear strain increments in Tables 6.2 and 6.3 by finite element degrees of freedom and corresponding interpolations so as to achieve a full linearization of the principle of virtual work (see Section 6.3.1). For this purpose we define the vector of nodal rotational degrees of freedom $\theta_{k}$ with components measured about the Cartesian axes and use the second-order approximations (see Exercise 6.56)
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$$
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\mathbf {V} _ {t} ^ {k} = \boldsymbol {\theta} _ {k} \times {} ^ {t} \mathbf {V} _ {t} ^ {k} + \frac {1}{2} \boldsymbol {\theta} _ {k} \times \left(\boldsymbol {\theta} _ {k} \times {} ^ {t} \mathbf {V} _ {t} ^ {k}\right) \tag {6.161}
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$$
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$$
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\mathbf {V} _ {s} ^ {k} = \boldsymbol {\theta} _ {k} \times {} ^ {t} \mathbf {V} _ {s} ^ {k} + \frac {1}{2} \boldsymbol {\theta} _ {k} \times (\boldsymbol {\theta} _ {k} \times {} ^ {t} \mathbf {V} _ {s} ^ {k}) \tag {6.162}
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$$
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The only purpose of using $\pmb{\theta}_k$ is to evaluate (approximations to) the new director vectors, and $\pmb{\theta}_k$ is discarded thereafter.
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Substituting from (6.161) and (6.162) into (6.158) we obtain the expression for $u_{i}$ to evaluate the continuum linear and nonlinear incremental strain tensors in Tables 6.2 and 6.3. Since the relations in (6.161) and (6.162) involve quadratic expressions, we neglect all higher-order terms in the solution variables to obtain the fully linearized form of the principle of virtual work equation—linearized about the state at time t with respect to the solution variables (the nodal point displacements and rotations). With this process, the exact tangent stiffness matrix is arrived at and employed in the incremental finite element
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<!-- source-page: 588 -->
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solution. However, we should note that the continuum linear strain increments in Tables 6.2 and 6.3 now include quadratic terms in rotations, and hence the right-hand-side terms
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$$
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\int_ {0 _ {V}} ^ {t} S _ {i j} \delta_ {0} e _ {i j} d ^ {0} V \quad \text { and } \quad \int_ {t _ {V}} ^ {t} \tau_ {i j} \delta_ {t} e _ {i j} d ^ {t} V
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$$
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in (6.74) and (6.75) contribute, in this case, to the tangent stiffness matrices of the TL and UL formulations. The same incremental equations are of course also obtained if we use the procedure in Section 6.3.1 to develop these equations.
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A kinematic assumption in this interpolation is that “plane sections remain plane,” and hence warping is not included. However, warping displacement behavior can be added to the assumed deformations as discussed in Section 5.4.1.
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The linear and nonlinear strain displacement matrices of the beam element corresponding to the UL formulation can now be evaluated using the approach employed in linear analysis. That is, using (6.158), the strain components are calculated corresponding to the global axes and are then transformed to obtain the strain components corresponding to the local beam axes, $\eta$ , $\xi$ , $\zeta$ . Since the element stiffness matrix is evaluated using numerical integration, the transformation from global to local strain components must be performed during the numerical integration at each integration point.
|
||||
|
||||
Considering the TL formulation, we recognize that, first, derivatives analogous to those used in the UL formulation are required, but the derivatives are taken with respect to the coordinates at time 0. In addition, however, in order to include the initial displacement effect, the derivatives of the displacements at time t with respect to the original coordinates are needed. These derivatives are evaluated using $(6.157)$ .
|
||||
|
||||
The above interpolations lead to the displacement-based finite element formulation which, as discussed in Section 5.4.1, yields a very slowly converging discretization. In order to obtain an effective scheme a mixed interpolation should be used which, for the beam formulation, is equivalent to employing an appropriate Gauss integration order for the r-direction integration: namely one-point integration for the two-node element, two-point integration for the three-node element, and three-point integration for the four-node element.
|
||||
|
||||
The finite element equations thus arrived at are
|
||||
|
||||
$$
|
||||
{ } ^ { t } \mathbf { K } \left[ \begin{array} { l } \vdots \\ \mathbf { u } _ { k } \\ \boldsymbol { \theta } _ { k } \\ \vdots \end{array} \right] = { } ^ { t + \Delta t } \mathbf { R } - { } ^ { t } \mathbf { F } \tag {6.163}
|
||||
$$
|
||||
|
||||
Having solved (6.163) for $u_{k}$ and $\theta_{k}$ , we obtain approximations for the nodal point displacements and director vectors at time $t + \Delta t$ using
|
||||
|
||||
$$
|
||||
{ } ^ { t + \Delta t } \mathbf { u } _ { k } = { } ^ { t } \mathbf { u } _ { k } + \mathbf { u } _ { k } \tag {6.164}
|
||||
$$
|
||||
|
||||
and $t + \Delta t\mathbf{V}_t^k = {}^t\mathbf{V}_t^k +\int_{\theta_k}d\theta_k\times {}^\tau \mathbf{V}_t^k$ (6.165)
|
||||
|
||||
$$
|
||||
{ } ^ { t + \Delta t } \mathbf { V } _ { s } ^ { k } = { } ^ { t } \mathbf { V } _ { s } ^ { k } + \int _ { \theta _ { k } } d \theta _ { k } \times { } ^ { t } \mathbf { V } _ { s } ^ { k } \tag {6.166}
|
||||
$$
|
||||
|
||||
<!-- source-page: 589 -->
|
||||
|
||||
The integrations in (6.165) and (6.166) can be performed in one step using an orthogonal matrix for finite rotations (see, for example, J. H. Argyris [B] and Exercise 6.55) or in a number of steps using a simple Euler forward method (see Section 9.6). Of course, $\theta_{k}$ (and $u_{k}$ ) are approximations to the actual required increments (because of the linearization of the principle of virtual work), but with the integrations in (6.165) and (6.166) we intend to arrive at a more accurate evaluation of the new director vectors than by simply substituting into (6.161) and (6.162).
|
||||
|
||||
The above presentation corresponds of course to the first iteration of the usual Newton-Raphson iterative solution process or to a typical iteration when the last calculated values of coordinates and director vectors are used.
|
||||
|
||||
It should be noted that this beam element formulation admits very large displacements and rotations and has an important advantage when compared with the formulation of a straight beam element based on Hermitian displacement interpolations: all individual displacement components are expressed using the same functions because the displacement expressions are derived from the geometry interpolation. Thus there is no directionality in the displacement interpolations, and the change in the geometry of the beam structure with increasing deformations is modeled more accurately than by using straight beam elements based on Hermitian functions, as for example presented by K. J. Bathe and S. Bolourchi [A].
|
||||
|
||||
We mentioned earlier that this general beam formulation can be used to derive the matrices pertaining to the formulations of planar beam elements for plane stress or plane strain conditions or axisymmetric shell elements. We demonstrate such derivations in the following examples.
|
||||
|
||||
EXAMPLE 6.20: Consider the two-node beam element shown in Fig. E6.20. Evaluate the coordinate and displacement interpolations and derivatives that are required for the calculation of the strain-displacement matrices of the UL and TL formulations.
|
||||
|
||||
|
||||
|
||||
<details>
|
||||
<summary>text_image</summary>
|
||||
|
||||
tV_s^2 = [-sin tθ_2 / cos tθ_2]
|
||||
tθ_2
|
||||
2
|
||||
h
|
||||
At time t
|
||||
At time 0
|
||||
0V_s^1
|
||||
s
|
||||
0V_s^2 = [0/1]
|
||||
h
|
||||
1
|
||||
r
|
||||
2
|
||||
x_1, tU_1
|
||||
0L
|
||||
x_2, tU_2
|
||||
tV_s^1
|
||||
s
|
||||
r
|
||||
h
|
||||
α
|
||||
1
|
||||
At time t
|
||||
</details>
|
||||
|
||||
Figure E6.20 Two-node beam element in large displacements and rotations
|
||||
|
||||
<!-- source-page: 590 -->
|
||||
|
||||
Using the variables in Fig. E6.20, we have corresponding to (6.154),
|
||||
|
||||
$$
|
||||
^ {\prime} x _ {1} = \left(\frac {1 - r}{2}\right) ^ {\prime} x _ {1} ^ {1} + \left(\frac {1 + r}{2}\right) ^ {\prime} x _ {1} ^ {2} - \frac {s h}{2} \left(\frac {1 - r}{2}\right) \sin^ {\prime} \theta_ {1} - \frac {s h}{2} \left(\frac {1 + r}{2}\right) \sin^ {\prime} \theta_ {2}
|
||||
$$
|
||||
|
||||
$$
|
||||
^ {\prime} x _ {2} = \left(\frac {1 - r}{2}\right) ^ {\prime} x _ {2} ^ {1} + \left(\frac {1 + r}{2}\right) ^ {\prime} x _ {2} ^ {2} + \frac {s h}{2} \left(\frac {1 - r}{2}\right) \cos^ {\prime} \theta_ {1} + \frac {s h}{2} \left(\frac {1 + r}{2}\right) \cos^ {\prime} \theta_ {2}
|
||||
$$
|
||||
|
||||
$$
|
||||
{ } ^ { 0 } x _ { 1 } = \left( \frac { 1 + r } { 2 } \right) { } ^ { 0 } L
|
||||
$$
|
||||
|
||||
$$
|
||||
{ } ^ { 0 } x _ { 2 } = \frac { s h } { 2 }
|
||||
$$
|
||||
|
||||
Hence, the displacement components are at any point at time t,
|
||||
|
||||
$$
|
||||
{ } ^ { \prime } u _ { 1 } = \left( \frac { { } ^ { \prime } x _ { 1 } ^ { 1 } + { } ^ { \prime } x _ { 1 } ^ { 2 } - { } ^ { 0 } L } { 2 } \right) + \left( \frac { { } ^ { \prime } x _ { 1 } ^ { 2 } - { } ^ { \prime } x _ { 1 } ^ { 1 } - { } ^ { 0 } L } { 2 } \right) r - \frac { s h } { 2 } \left[ \left( \frac { 1 - r } { 2 } \right) \sin { { } ^ { \prime } \theta _ { 1 } } + \left( \frac { 1 + r } { 2 } \right) \sin { { } ^ { \prime } \theta _ { 2 } } \right]
|
||||
$$
|
||||
|
||||
$$
|
||||
{ } ^ { \prime } u _ { 2 } = \left( \frac { { } ^ { \prime } x _ { 2 } ^ { 1 } + { } ^ { \prime } x _ { 2 } ^ { 2 } } { 2 } \right) + \left( \frac { { } ^ { \prime } x _ { 2 } ^ { 2 } - { } ^ { \prime } x _ { 2 } ^ { 1 } } { 2 } \right) r + \frac { s h } { 2 } \left[ \left( \frac { 1 - r } { 2 } \right) \cos { { } ^ { \prime } \theta _ { 1 } } + \left( \frac { 1 + r } { 2 } \right) \cos { { } ^ { \prime } \theta _ { 2 } } - 1 \right]
|
||||
$$
|
||||
|
||||
The incremental displacements are given by (6.158); hence,
|
||||
|
||||
$$
|
||||
\begin{array}{l} u _ {1} = \frac {1 - r}{2} u _ {1} ^ {1} + \frac {1 + r}{2} u _ {1} ^ {2} + \frac {s h}{2} \left(\frac {1 - r}{2}\right) \left[ (- \cos^ {\prime} \theta_ {1}) \theta_ {1} + \frac {1}{2} \sin^ {\prime} \theta_ {1} (\theta_ {1}) ^ {2} \right] \\ + \frac {s h}{2} \left(\frac {1 + r}{2}\right) \left[ \left(- \cos^ {\prime} \theta_ {2}\right) \theta_ {2} + \frac {1}{2} \sin^ {\prime} \theta_ {2} \left(\theta_ {2}\right) ^ {2} \right] \tag {a} \\ \end{array}
|
||||
$$
|
||||
|
||||
$$
|
||||
u _ {2} = \frac {1 - r}{2} u _ {2} ^ {1} + \frac {1 + r}{2} u _ {2} ^ {2} + \frac {s h}{2} \left(\frac {1 - r}{2}\right) \left[ (- \sin^ {\prime} \theta_ {1}) \theta_ {1} - \frac {1}{2} \cos^ {\prime} \theta_ {1} (\theta_ {1}) ^ {2} \right]
|
||||
$$
|
||||
|
||||
$$
|
||||
+ \frac {s h}{2} \left(\frac {1 + r}{2}\right) \left[ (- \sin^ {\prime} \theta_ {2}) \theta_ {2} - \frac {1}{2} \cos^ {\prime} \theta_ {2} \left(\theta_ {2}\right) ^ {2} \right] \tag {b}
|
||||
$$
|
||||
|
||||
We note the quadratic terms in nodal rotations, which are underlined with a dashed line. Using (a) and (b) to evaluate the continuum incremental strain terms $_0e_{ij}$ , $_t e_{ij}$ , $_0\eta_{ij}$ , and $_t\eta_{ij}$ in Tables 6.2 and 6.3, we recognize that the fully linearized finite element equations are obtained by including the underlined terms in the evaluation of $\int_{0_V} {}^t S_{ij} \delta_0 e_{ij} d^0 V$ and $\int_{t_V} {}^t \tau_{ij} \delta_t e_{ij} d^t V$ . These terms add for the structural elements a contribution to the nonlinear strain stiffness matrices. However, these quadratic terms in rotation do not contribute in the linearized form of the other integrals because they result in those integrals in higher-order terms that are neglected in the linearization.
|
||||
|
||||
In considering the UL formulation, the required derivatives for the Jacobian are
|
||||
|
||||
$$
|
||||
\frac {\partial^ {\prime} x _ {1}}{\partial r} = \frac {L \cos \alpha}{2} - \frac {s h}{4} \left(\sin^ {\prime} \theta_ {2} - \sin^ {\prime} \theta_ {1}\right)
|
||||
$$
|
||||
|
||||
$$
|
||||
\frac {\partial^ {\prime} x _ {1}}{\partial s} = \left(- \frac {h}{2}\right) \left[ \left(\frac {1 - r}{2}\right) \sin^ {\prime} \theta_ {1} + \left(\frac {1 + r}{2}\right) \sin^ {\prime} \theta_ {2} \right]
|
||||
$$
|
||||
|
||||
$$
|
||||
\frac {\partial^ {\prime} x _ {2}}{\partial r} = \frac {L \sin \alpha}{2} + \frac {s h}{4} \left(\cos^ {\prime} \theta_ {2} - \cos^ {\prime} \theta_ {1}\right)
|
||||
$$
|
||||
Reference in New Issue
Block a user