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김경종
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<details>
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<summary>line</summary>
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| Basic variable φ | Applied force, f | Slope H(φ⁰) | Slope H(φ¹) |
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| ---------------- | ---------------- | ----------- | ----------- |
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| φ⁰ | ψ⁰ | - | - |
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| φ¹ | ψ¹ | - | - |
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| φ² | ψ² | - | - |
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</details>
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Fig. 2.5 Tangential stiffness solution algorithm for a single variable situation.
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respectively. If in (2.13) the tangential stiffness matrix is replaced, at all steps of the computation, by the stiffness corresponding to the initial trial value of $\varphi$ a complete factorisation, or reduction, of the assembled equations can be avoided. $^{(3)}$ In this case a complete equation solution need only be performed for the first iteration and subsequent approximations to the nonlinear solution performed, via the expression
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$$
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\Delta \varphi^ {r} = - [ H (\varphi^ {0}) ] ^ {- 1} \psi (\varphi^ {r}). \tag {2.14}
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$$
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Since the same stiffness matrix $H(\varphi^{0})$ is employed at each stage, the reduced equations can be stored in their reduced or factored form and a second or subsequent solution merely necessitates the reduction of the right-hand side $(\psi(\varphi^{r}))$ terms, together with a backsubstitution. This has the immediate advantage of significantly reducing the computing cost per iteration but reduces the convergence rate as can be seen from Fig. 2.6 where the scheme is schematically illustrated. The iterative algorithm is identical to that described in the preceding section. This method can be shown to be unconditionally convergent $^{(4)}$ and can even be employed in situations where the material exhibits negative stiffness. The relative economies of the initial stiffness and tangential stiffness methods depend to a large extent on the degree of nonlinearity inherent in the problem under consideration. The optimum algorithm is generally provided by an amalgamation of both processes, in which the stiffnesses are changed at selected iterative intervals only.
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<details>
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<summary>line</summary>
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| Basic variable, φ | Applied force, f |
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| ----------------- | ---------------- |
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| φ⁰ | ψ⁰ |
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| φ¹ | ψ¹ |
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| φ² | ψ² |
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| φ³ | ψ³ |
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</details>
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Fig. 2.6 Initial stiffness solution algorithm for a single variable situation.
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# 2.3 Systems governed by a quasi-harmonic equation
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Many physical situations in engineering science are governed by a quasi-harmonic equation containing coefficients which are dependent on the unknown variable or its derivatives according to some prescribed law. The most common problem of this type occurs in heat conduction under steady-state conditions when the material conductivity is itself a function of temperature. This phenomenon also arises in diffusion problems where the diffusivity of the medium often varies with the concentration of the diffusing matter. Further physical examples are provided in Ref. (5).
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For a one-dimensional situation the governing equation to be considered is
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$$
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\frac {d}{d x} \left(K \frac {d \phi}{d x}\right) + Q = 0, \tag {2.15}
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$$
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in which $\phi$ is the unknown function and the terms K and Q may be functions of the position coordinate, x. The problem becomes nonlinear if K and/or Q are also functions of the unknown $\phi$ or its derivatives, according to some prescribed function.
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Two types of boundary condition will be considered:
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(a) The value of the unknown specified on the boundary
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$$
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\phi = \phi_ {B}. \tag {2.16}
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$$
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(b) The gradient of the unknown at the boundary specified to be zero
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<!-- source-page: 33 -->
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$$
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\frac {d \phi}{d n} = \frac {d \phi}{d x} = 0. \tag {2.17}
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$$
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(A more general form of this latter boundary condition is considered in Ref. 6.)
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Equation (2.15) can be transformed to finite element form by suitable discretisation and use of the Galerkin weighted residual process. $^{(5,6)}$ The scalar product of equation (2.15) with any arbitrary weighting function, W, must be zero if $\phi$ satisfies (2.15) throughout any region $\Gamma$ , so that
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$$
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\int_ {\Gamma} \left(\frac {d}{d x} \left(K \frac {d \phi}{d x}\right) + Q\right) W d x = 0. \tag {2.18}
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$$
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Integrating the first term by parts results in
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$$
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\left[ W K \frac {d \phi}{d x} \right] _ {x _ {1}} ^ {x _ {2}} - \int_ {\Gamma} \left(K \frac {d W}{d x} \frac {d \phi}{d x} - Q W\right) d x = 0, \tag {2.19}
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$$
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where the limits of integration in the first term are the end points of the region $\Gamma$ . The unknown function $\phi$ may be approximated as
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$$
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\phi = \sum_ {i = 1} ^ {n} N _ {i} \phi_ {i}, \tag {2.20}
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$$
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in which n is the total number of nodes in the finite element idealisation and $N_{i}$ are the global shape functions. In the Galerkin process the number of weighting functions must equal the total number of unknown nodal values. The weighting function $W_{i}$ corresponding to node i can then be conveniently chosen such that $W_{i} = N_{i}$ . It should be noted that at nodes where the values of $\phi$ are prescribed, there is no associated unknown and consequently the weighting function for such nodes is zero. Therefore the first term in (2.19) always vanishes since at the two end points of the interval either $\phi$ is prescribed according to (2.16), in which case the weighting function for that point is zero, or $d\phi/dx$ is specified as zero according to (2.17). Substituting for $\phi$ and W in (2.19) and assembling all element contributions in the usual manner results in
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$$
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\boldsymbol {H} \varphi + \boldsymbol {f} = \mathbf {0}, \tag {2.21}
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$$
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in which typical element components are
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$$
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h _ {i j} ^ {(e)} = \int_ {\Gamma^ {(e)}} K \frac {d N _ {i} ^ {(e)}}{d x} \frac {d N _ {j} ^ {(e)}}{d x} d x, \tag {2.22}
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$$
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$$
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f _ {i} ^ {(e)} = \int_ {\Gamma^ {(e)}} Q N _ {i} ^ {(e)} d x, \tag {2.23}
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$$
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<!-- source-page: 34 -->
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where $N_{i}^{(e)}$ are the element shape functions specifying the distribution of the unknown, $\phi$ , over the element. For the specific case of a two-noded element with a linear variation in $\phi$ as shown in Fig. 2.7, the shape functions are simply
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$$
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N _ {1} ^ {(e)} = \frac {1}{2} - \frac {x}{L}, \quad N _ {2} ^ {(e)} = \frac {1}{2} + \frac {x}{L}, \tag {2.24}
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$$
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where $L$ is the length of the element.
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<details>
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<summary>text_image</summary>
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N₁(e)
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1
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L/2
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x
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N₂(e)
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2
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L/2
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</details>
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Fig. 2.7 One-dimensional two-noded element with linear variation of the unknown, $\phi$ , showing element shape functions.
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Substituting in (2.22) and (2.23), and assuming no variation of K with position in the element, gives
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$$
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\boldsymbol {H} ^ {(e)} = \frac {K}{L} \left[ \begin{array}{c c} 1 & - 1 \\ - 1 & 1 \end{array} \right], \tag {2.25}
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$$
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and
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$$
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f _ {1} ^ {(e)} = f _ {2} ^ {(e)} = \frac {Q L}{2}. \tag {2.26}
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$$
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Provided that the variation of K with $\phi$ or its derivatives is specified, the problem falls into the category discussed in the previous section and can be solved by either the method of direct iteration or the Newton–Raphson approach.
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In the numerical examples considered later in this chapter a specific form of nonlinearity will be considered, namely
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$$
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K = K _ {0} (a + b \phi), \tag {2.27}
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$$
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in which $K_{0}$ is a reference value and a and b are known constants. For solution by the Newton–Raphson process the Jacobian matrix can be considered to be the sum of symmetric and nonsymmetric components as indicated in (2.11). The symmetric part has already been calculated in (2.25) and the nonsymmetric contribution must now be calculated according to the last
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<!-- source-page: 35 -->
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term in (2.7). From (2.7), (2.22) and (2.27) the general term is given as
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$$
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h _ {i j} ^ {\prime} = \sum_ {k = 1} ^ {2} \left(\frac {\partial h _ {i k}}{\partial \phi_ {j}}\right) \phi_ {k} = \sum_ {k = 1} ^ {2} \left\{\phi_ {k} K _ {0} \int_ {- L / 2} ^ {L / 2} \frac {\partial}{\partial \phi_ {j}} [ a + b \phi ] \frac {d N _ {i} ^ {(e)}}{d x} \frac {d N _ {k} ^ {(e)}}{d x} d x \right\}. \tag {2.28}
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$$
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Noting that $\phi$ is given by (2.20) and that the shape functions are given by (2.24), the evaluation of (2.28) results in
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$$
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\boldsymbol {H} ^ {\prime (e)} = \frac {K _ {0} b}{2 L} (\phi_ {1} - \phi_ {2}) \left[ \begin{array}{c c} 1 & 1 \\ - 1 & - 1 \end{array} \right]. \tag {2.29}
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$$
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As expected, it is seen that the derivative matrix $H'(e)$ is unsymmetric.
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# 2.4 Nonlinear elastic problems
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The simplest case of nonlinear behaviour in structural problems arises from nonlinear elastic material action. The stress/strain relationship of the material is nonlinear but the material behaviour is elastic with all deformations and displacements recoverable on unloading. For example, this type of behaviour arises in hyperelastic problems $^{(7)}$ where the stresses are functions of a strain dependent material modulus.
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The nonlinear constitutive relation may be specified, for a one-dimensional situation, as
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$$
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\sigma = \frac {d W}{d \epsilon} = E _ {0}. g (\epsilon) \tag {2.30}
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$$
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where $\sigma$ is the stress, $\epsilon$ the strain and $E_{0}$ some reference value of the material modulus. The material performance will be nonlinear according to the form of the specified strain energy function, $W(\epsilon)$ .
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<details>
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<summary>text_image</summary>
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L
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Cross sectional area A
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δ
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1
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2
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F
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F
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</details>
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Fig. 2.8 Forces and displacements for a two-node element.
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The simplest form of one-dimensional finite element is the constant stress element shown in Fig. 2.8 in which a linear displacement variation is assumed between nodes 1 and 2. The force in the element is given, from (2.30), by
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$$
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F = E _ {0} A g (\delta / L), \tag {2.31}
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$$
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<!-- source-page: 36 -->
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where A is the element cross-sectional area and $\delta$ the element extension. The tangential stiffness for the material is then
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$$
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K _ {T} = \frac {d F}{d \delta} = \frac {E _ {0} A}{L} \frac {d g}{d \epsilon} = \frac {E _ {0} A}{L} g ^ {\prime} (\epsilon). \tag {2.32}
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$$
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Or, in particular, the element tangential stiffness matrix is given by
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$$
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\boldsymbol {K} _ {T} ^ {(e)} = \frac {E _ {0} A}{L} g ^ {\prime} (\epsilon) \left[ \begin{array}{c c} 1 & - 1 \\ - 1 & 1 \end{array} \right]. \tag {2.33}
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$$
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Provided that $g'(\epsilon)$ is positive for all strain values, the tangential stiffness method of solution described in Section 2.2.3 can be employed in solution with $K_{T}^{(e)}$ being directly equivalent to $H(\varphi^{r})$ . If the tangential stiffness matrix becomes zero, the assembled stiffness equations will become singular and the inversion process required by (2.13) cannot be undertaken. Solution for situations in which the material tangential stiffness becomes non-positive can be performed by use of the initial stiffness method described in Section 2.2.4. Since the initial material stiffness is employed throughout this latter process, the assembled stiffness matrix will remain positive definite throughout the computation.
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# 2.5 Elasto-plastic problems in one dimension
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In this section the essential features of elasto-plastic material behaviour are introduced, and the basic expressions are developed in a form suitable for numerical solution by some of the methods described in the previous sections.
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Elasto-plastic behaviour is characterised by an initial elastic material response on to which a plastic deformation is superimposed after a certain level of stress has been reached. $^{(8)}$ Plastic deformation is essentially irreversible on unloading and is incompressible in nature. The onset of plastic deformation (or yielding) is governed by a yield criterion and post-yield deformation generally occurs at a greatly reduced material stiffness. Basic theoretical expressions for a general continuum are provided in Chapter 7.
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For one-dimensional situations, the material parameters required to completely define elasto-plastic behaviour are most conveniently obtained from a uniaxial tension test. Figure 2.9 shows an idealised stress-strain curve for a material and identical behaviour is assumed in tension and compression. The material initially deforms according to the elastic modulus, E, until the stress level reaches a value $\sigma_{Y}$ designated the uniaxial yield stress. On increasing the load further, the material is assumed to exhibit linear strain-hardening, characterised by the tangential modulus, $E_{T}$ .
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At some stage after initial yielding, consider a further load application resulting in an incremental increase of stress, $d\sigma$ , accompanied by a change of strain, $d\epsilon$ . Assuming that the strain can be separated into elastic and plastic
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<!-- source-page: 37 -->
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<details>
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<summary>line</summary>
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| Strain, ε | Stress, σ | Elastic behaviour Slope, E |
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| --------- | --------- | -------------------------- |
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| 0 | 0 | 0 |
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| ε | ε | dε_e |
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| ε | ε | dε_p |
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| ε | ε | dε_p |
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| ε | ε | dε_p |
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| ε | ε | dε_p |
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| ε | ε | dε_p |
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| ε | ε | dε_p |
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| ε | ε | dε_p |
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| ε | ε | dε_p |
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| ε | ε | dε_p |
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| σ_Y | σ_Y | Elastic behaviour Slope, E |
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| σ_σ | σ_σ | Elastic behaviour Slope, E |
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| SloE E | SloE E | Elastic response Slope, E_T |
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| SloE E_T | SloE E_T | Elastic response Slope, E_T |
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</details>
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Fig. 2.9 Elastic, linear strain-hardening stress–strain behaviour for the uniaxial case.
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components, so that
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$$
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d \epsilon = d \epsilon_ {e} + d \epsilon_ {p}, \tag {2.34}
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$$
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we define a strain-hardening parameter, $H'$ , as
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$$
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H ^ {\prime} = \frac {d \sigma}{d \epsilon_ {p}}. \tag {2.35}
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$$
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This can be interpreted as the slope of the strain-hardening portion of the stress–strain curve after removal of the elastic strain component. Thus
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$$
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H ^ {\prime} = \frac {d \sigma}{d \epsilon - d \epsilon_ {e}} = \frac {E _ {T}}{1 - E _ {T} / E}. \tag {2.36}
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$$
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With reference to Fig. 2.8, consider the behaviour of a linear displacement element, which has a cross-sectional area A, when it is subjected to a gradually increasing axial force, F, which results in an extension, $\delta$ . Provided that F/A is less than or equal to the uniaxial yield stress, $\sigma_{Y}$ , the material behaviour will be elastic, exhibiting a stiffness of
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$$
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K _ {e} = \frac {F}{\delta} = \frac {E A}{L}, \tag {2.37}
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$$
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<!-- source-page: 38 -->
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then the element stiffness matrix is simply
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$$
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\boldsymbol {K} _ {e} ^ {(e)} = \frac {E A}{L} \left[ \begin{array}{c c} 1 & - 1 \\ - 1 & 1 \end{array} \right]. \tag {2.38}
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$$
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Suppose F is increased until the material has yielded. Consider a further incremental increase in load dF which causes an additional element extension, $d\delta$ . Then
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$$
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d \delta = (d \epsilon_ {e} + d \epsilon_ {p}) L, \tag {2.39}
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$$
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where $L$ is the element length. Also, on use of (2.35)
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$$
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d F = d \sigma A = A H ^ {\prime} d \epsilon_ {p}. \tag {2.40}
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$$
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The tangential stiffness for the material is then
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$$
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K _ {e p} = \frac {d F}{d \delta} = \frac {A H ^ {\prime} d \epsilon_ {p}}{L \left(d \sigma / E + d \epsilon_ {p}\right)}. \tag {2.41}
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$$
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Or, using (2.35) and rearranging
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$$
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K _ {e p} = \frac {E A}{L} \left(1 - \frac {E}{E + H ^ {\prime}}\right). \tag {2.42}
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$$
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Finally, the element stiffness for elasto-plastic material behaviour is given by\*
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$$
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\boldsymbol {K} _ {e p} ^ {(e)} = \frac {E A}{L} \left(1 - \frac {E}{E + H ^ {\prime}}\right) \left[ \begin{array}{c c} 1 & - 1 \\ - 1 & 1 \end{array} \right]. \tag {2.43}
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$$
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In (2.42) it can be seen that the first term represents the elastic stiffness, as given by (2.38). The second term accounts for the reduction in stiffness from the elastic value due to yielding.
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\* The element stiffness matrix can be written in the standard finite element form
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$$
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\boldsymbol {K} _ {e} ^ {(e)} = \int_ {V} \boldsymbol {B} ^ {T} \boldsymbol {D} \boldsymbol {B} d V = A \int_ {0} ^ {L} \boldsymbol {B} ^ {T} \boldsymbol {D} \boldsymbol {B} d x,
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$$
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where integration is made over the volume of the element. For this one-dimensional application, D = E and
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$$
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\boldsymbol {B} = \left[ \frac {d N _ {1} ^ {(e)}}{d x}, \quad \frac {d N _ {2} ^ {(e)}}{d x} \right] = \left[ - \frac {1}{L}, \quad \frac {1}{L} \right],
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$$
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where $N_{1}^{(e)}$ and $N_{2}^{(e)}$ are given by (2.24). The tangential stiffness matrix for elastoplastic material behaviour is obtained by replacing D by
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$$
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\boldsymbol {D} _ {e p} = E \left(1 - \frac {E}{E + H ^ {\prime}}\right).
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$$
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<!-- source-page: 39 -->
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For a perfectly plastic material behaviour, after initial yielding equation (2.36) implies that $H' = 0$ and it is then evident from (2.43) that $\boldsymbol{K}_{ep}^{(e)} = 0$ . This implies that the tangential (elasto-plastic) stiffness matrix for such a material is singular and the tangential stiffness method cannot generally be employed in solution. If a significant number of elements in the structure has yielded, the assembled tangential stiffness matrix will be singular, and the inversion or reduction demanded by (2.13) cannot be performed. This difficulty can be avoided by use of the initial stiffness method in which the elastic element stiffnesses are employed at every stage of the computation, thereby ensuring a positive definite assembled stiffness matrix.
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# 2.6 Problems
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In this section some tasks are set for the reader which illustrate some further points in connection with the topics discussed in the chapter.
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2.1 Use the direct iteration method to solve the following one degree of freedom problem, $H\phi + f = 0$ where $f = 10$ and $H$ depends on $\phi$ according to $H = 10(1 + e^{3\phi})$ .
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2.2 Repeat Problem 2.1 using the Newton–Raphson method. Compare the solutions and the computational effort required in each.
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2.3 Solve the following one degree of freedom problem by both the tangential stiffness and initial stiffness method. Apply the total load f as two equal increments
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$$
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H \phi + f = 0, \quad f = 1 0, \quad H = 2 0 (1 - \phi).
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$$
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2.4 The more general form of the boundary condition (2.17) in Section 2.3 is $d\phi / dx + q + \alpha \cdot \phi = 0$ , where $q$ and $\alpha$ are constants and $\phi$ is the undetermined value of the unknown at the boundary point. Repeat the Galerkin process of Section 2.3 to include these additional terms. In particular, determine the additional nodal force contribution and the discrete 'external' nodal stiffness which arise.
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2.5 For the two-noded element with linear variation in $\phi$ with shape functions as given by (2.24), evaluate the element stiffness matrix when $K$ is a function of $x$ . Assume that the spatial variation of $K$ within the element is linear and obtained by interpolation of the specified nodal values by use of the element shape functions.
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2.6 Suppose that a heat loss also occurs by convection from the surface area of an element, which is given by $h.\phi$ where h is the convection coefficient. If C is the circumference of the element, determine the additional contribution to $H^{(e)}$ resulting from this. $^{(9)}$
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2.7 Determine the nonlinear portion, $H^{(e)}$ , of the Jacobian matrix for a material dependence $K = K_0(1 + e^{b\phi})$ . Assume a two-noded linear element.
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2.8 Evaluate the stiffness matrix $H^{(e)}$ for a three-noded element for a heat conduction problem. Assume that the element has shape functions
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|
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<!-- source-page: 40 -->
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|
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$$
|
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N _ {1} ^ {(e)} = - \frac {2 x}{L ^ {2}} \left(\frac {L}{2} - x\right), \quad N _ {2} ^ {(e)} = \frac {4}{L ^ {2}} \left(\frac {L}{2} - x\right) \left(\frac {L}{2} + x\right),
|
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$$
|
||||
|
||||
$$
|
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N _ {3} ^ {(e)} = \frac {2 x}{L ^ {2}} \left(\frac {L}{2} + x\right),
|
||||
$$
|
||||
|
||||
and also that $K = K_0(a + b\phi)$ where $K_0, a$ and $b$ are constants.
|
||||
|
||||
2.9 Repeat.Problem 2.8 for the case where $K_0$ is additionally a function of $x$ . Assume that the nodal values of $K_0$ are given.
|
||||
2.10 Solve the nonlinear elastic problem of Fig. 2.10 by hand calculation. Use the tangential stiffness method and assume the total load to be applied in two equal increments.
|
||||
|
||||
|
||||
|
||||
<details>
|
||||
<summary>text_image</summary>
|
||||
|
||||
A = 1.0
|
||||
σ = 20(ε - ε²)
|
||||
1
|
||||
2
|
||||
P = 4.8
|
||||
5
|
||||
</details>
|
||||
|
||||
Fig. 2.10 Nonlinear elastic example—Problem 2.10.
|
||||
|
||||
2.11 Solve Problem 2.10 if the structure is loaded by incrementally increasing the prescribed value of displacement at node 2. Increase the applied displacement in two equal increments up to a maximum value of $\phi_{2} = 3.0$ . Since the element stiffnesses become negative at the higher increment, use the initial stiffness method.
|
||||
2.12 A locking material is one in which the stiffness increases with increasing strains. For example, if $g(\epsilon) = \epsilon^{2}$ can both the tangential stiffness and the initial stiffness methods be used to solve such material problems?
|
||||
|
||||
|
||||
|
||||
<details>
|
||||
<summary>text_image</summary>
|
||||
|
||||
1
|
||||
2
|
||||
20
|
||||
E = 1000
|
||||
A = 1.0
|
||||
σY = 10
|
||||
H' = 100
|
||||
10
|
||||
</details>
|
||||
|
||||
Fig. 2.11 Elasto-plastic example—Problem 2.13.
|
||||
|
||||
2.13 Determine the nodal displacement of node 2 of the structure shown in Fig. 2.11 as the applied load is increased to 10 units in two equal increments. Assume elasto-plastic material behaviour and use the tangential stiffness approach for solution.
|
||||
|
||||
|
||||
|
||||
<details>
|
||||
<summary>text_image</summary>
|
||||
|
||||
1
|
||||
I
|
||||
2
|
||||
20
|
||||
II
|
||||
3
|
||||
10
|
||||
10
|
||||
</details>
|
||||
<table><tr><td></td><td>Element I</td><td>Element II</td></tr><tr><td>E</td><td>1000</td><td>1000</td></tr><tr><td>A</td><td>1.0</td><td>1.0</td></tr><tr><td> $\sigma_{Y}$ </td><td>5.0</td><td>5.0</td></tr><tr><td>H'</td><td>200</td><td>-100</td></tr></table>
|
||||
|
||||
Fig. 2.12 Bimaterial elasto-plastic example—Problem 2.14.
|
||||
Reference in New Issue
Block a user