김경종
475 lines
30 KiB
Markdown
475 lines
30 KiB
Markdown
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the square root of the round-off limit, $10^{-6}$ for round-offs of order $10^{-12}$ . If computed displacements change a large amount as a result, the patch is unstable. If this test is applied to a single element, one can detect if the element is unstable. (An unstable element does not necessarily produce an unstable mesh.)
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"Weak" Patch Test. An element that fails to display constant stress in a patch of large elements has not necessarily failed the patch test. If, as the mesh is repeatedly subdivided, elements come to display the expected state of constant stress, then the element is said to have passed a "weak" patch test, and convergence to correct results is assured.
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For example, a certain incompatible plane element fails the patch test if of arbitrary quadrilateral shape, but passes if each element of the patch is a parallelogram. If a mesh of arbitrary quadrilaterals is subdivided again and again, the newly created elements approach parallelograms in shape. Thus the element is found to pass a weak patch test. This result might be anticipated by applying the standard patch test to a mesh of parallelograms, Fig. 4.6-2, using nodal loads computed by means of Eqs. 1.6-3 and 4.1-6.
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As another example, consider an axially symmetric geometry. One uses elements whose strain fields include the circumferential strain $\epsilon_{\theta}=u/r$ (Fig. 4.6-3). As a mesh is refined, h/a and $\theta_{e}$ become small for each element. Accordingly, one applies the weak patch test to see whether $\epsilon_{\theta}$ becomes constant over each element as h/a and $\theta_{e}$ approach zero.
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Remarks. It is possible for an element that fails the patch test to give better answers in a coarse mesh than an equal number of elements that pass. Nevertheless, an element that fails cannot be trusted. An element that fails may provide convergence, but may converge to an incorrect result. This behavior is observed in certain plate-bending problems when the elements used do not permit a state of constant twist.
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“Higher-order” patch tests are possible. For example, a plane element whose displacement field includes a complete quadratic expansion for u and v (Fig. 3.6-1) should be able to represent exactly a field of pure bending. “Robustness” can also be checked [4.6]. For example, Poisson’s ratio v should have no effect on pure bending of a plane mesh; by computation, one discovers to what extent an element is insensitive to v.
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Apart from their use in testing elements, patch tests provide good example problems for learning to use an unfamiliar computer program.
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<details>
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<summary>natural_image</summary>
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Pure geometric line drawing of a 3D geometric structure with no text, numbers, or symbols
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</details>
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Figure 4.6-2. Patch of parallelogram elements, ready for application of patch-test loadings.
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<details>
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<summary>text_image</summary>
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θ,v
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θe
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r,u
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4 3
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1 2
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s
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a h
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</details>
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Figure 4.6-3. A plane element in polar coordinates. Displacements u and v are respectively radial and circumferential.
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# 4.7 STRESS CALCULATION
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Stress $\{\sigma\}$ in an element can be calculated when its nodal d.o.f. $\{\mathbf{d}\}$ are known. These d.o.f. are available after the structural equations $[\mathbf{K}]\{\mathbf{D}\} = \{\mathbf{R}\}$ have been solved. Equation 1.7-8, repeated here, is
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$$
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\{\boldsymbol {\sigma} \} = [ \mathrm{E} ] (\{\boldsymbol {\epsilon} \} - \{\boldsymbol {\epsilon} _ {0} \}) + \{\boldsymbol {\sigma} _ {0} \} \tag {4.7-1}
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$$
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in which mechanical strains $\{\epsilon\} = [B]\{d\}$ are produced by displacements of the nodes. Typically, $\{\sigma_{0}\}$ is omitted and $\{\epsilon_{0}\}$ is used to account for thermal strains. Thus, in a plane problem with isotropic material, with T the temperature relative to a stress-free temperature and $\alpha$ the coefficient of thermal expansion,
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$$
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\left\{ \begin{array}{l} \sigma_ {x} \\ \sigma_ {y} \\ \tau_ {x y} \end{array} \right\} = \frac {E}{1 - \nu^ {2}} \left[ \begin{array}{c c c} 1 & \nu & 0 \\ \nu & 1 & 0 \\ 0 & 0 & \frac {1 - \nu}{2} \end{array} \right] \left([ \mathbf {B} ] \{\mathbf {d} \} - \left\{ \begin{array}{l} \alpha T \\ \alpha T \\ 0 \end{array} \right\}\right) \tag {4.7-2}
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$$
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Matrix [B] is a function of the coordinates and must be evaluated at the location in the element where stresses are desired.
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The calculation $\{\epsilon\} = [\mathbf{B}]\{\mathbf{d}\}$ involves differentiation of the displacement. Accordingly, one expects that stresses will be less accurate than displacements. In low-order elements stresses are often most accurate at the element centroid, less accurate at midsides, and least accurate at corners. Elements of higher order usually display multiple points of optimal accuracy for stresses. The locations of these points depend on the element geometry and the displacement field, and can often be predicted before doing numerical calculations. Stresses at other locations are usually best found by extrapolation from the optimal points (Section 6.13).
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The most commonly used elements display only the minimum of interelement continuity. $C^0$ elements are popular for determining displacements in plane and solid elasticity, and $C^1$ elements are popular for determining displacements and slopes in plate bending. Stresses depend on displacement derivatives in $C^0$ elements (or on curvatures in commonly used $C^1$ elements). Therefore, unless a state of constant stress prevails, stresses are discontinuous across boundaries between $C^0$ elements, and $C^0$ elements that share a node do not all display the same state of stress at that node. A substantial difference in stress between elements suggests a need for mesh refinement.
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The average stress at a node is more to be trusted than the nodal stress in any one element attached to the node. A weighted nodal average—for example, with each element contribution weighted in proportion to its interior corner angle at the shared node—may be more reliable than a simple nodal average. However, averaging of stresses on either side of a physical discontinuity such as a step change in thickness should be avoided. Nodal average stresses may be used by a postprocessor in graphic display of results. A caution: a smooth stress field produced by a postprocessor may hide large stress differences in adjacent elements that indicate a need for mesh refinement.
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Imagine that the average strain along a line joining any two nodes is useful. The following trick is useful. Connect a straight bar element to the two nodes, but make its cross-sectional area very small, so that its stiffness is negligible in
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comparison with the stiffness of the existing structure. The computer program will calculate the bar stress, which the user can divide by E to obtain the strain desired. Effectively, we have attached a strain gage to the structure.
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Body Forces and Surface Tractions. In the absence of $\{\epsilon_{0}\}$ and $\{\sigma_{0}\}$ , stress is due entirely to mechanical loads. Imagine that the mechanical load is entirely body force, as in the axially loaded bar of Fig. 4.7-1. If a single bar element is used, both of its nodes are fixed, and we compute $\{d\}=\{0\}$ and $\{\sigma\}=\{0\}$ . Two-element and four-element models yield increasingly better results, as expected. The exact stress variation can be obtained from any of these models if the effect of body force within each separate element is taken into account. For example, with a one-element model of weight W, we add the stress
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$$
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\sigma_ {y} = \frac {W}{A} \left(\frac {y}{L _ {T}} - \frac {1}{2}\right) \tag {4.7-3}
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$$
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to the stresses $\{\sigma\} = [\mathbf{E}][\mathbf{B}]\{\mathbf{d}\}$ , which are zero for a one-element model. A similar formula can be written for $\sigma_y$ in an arbitrary element of a multi-element model.
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Analogous refinement can be made when calculating bending moments in a beam element. The bending moment $M = EI[B]\{d\}$ can be augmented by the bending moment produced by lateral loads on an element whose ends are completely fixed. This refinement may be worthwhile because beams are frequently analyzed and their loadings are comparatively easy to categorize and incorporate in a computer program.
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For other elements, such adjustments are ignored in practice. The adjustments are less important for plane and solid elements than for beam elements. Moreover, they are not easily formulated for an element of general shape.
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Thermal Stress. As a usual rule, for analysis of thermal stress one should construct a finite element model that permits the strain field to have about the same level of complexity as seen in the temperature field. Thus, one avoids a mismatch between $\{\epsilon\}$ and $\{\epsilon_{0}\}$ in Eq. 4.7-1. A mismatch can produce an unreliable stress prediction, especially among simpler elements, as the following example shows.
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<details>
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<summary>text_image</summary>
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y
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Gravity
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L_T
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x
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</details>
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(a)
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<details>
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<summary>text_image</summary>
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-σe
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+σ
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</details>
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(b)
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<details>
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<summary>text_image</summary>
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r_c + \frac{\sigma_c}{2}
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-\frac{\sigma_c}{2} - \frac{3\sigma_c}{4}
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</details>
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(c)
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<details>
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<summary>text_image</summary>
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- \frac{\sigma_c}{4} + \frac{3\sigma_c}{4}
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+ \frac{\sigma_c}{4}
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</details>
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(d)
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Figure 4.7-1. (a) Uniform bar of weight W and cross-sectional area A, fixed at both ends. (b) Axial stress variation, where $\sigma_{c} = W/2A$ . (c,d) Stresses predicted by two-element and four-element models.
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<details>
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<summary>text_image</summary>
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y,v
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3
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x,u
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1
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2
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</details>
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Figure 4.7-2. A triangular element.
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Let the triangular element of Fig. 4.7-2 have the linear temperature field
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$$
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T = c _ {1} x + c _ {2} y \tag {4.7-4}
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$$
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where $c_{1}$ and $c_{2}$ are constants. If the centroid of the triangle is at $x = y = 0$ , and if the element is an isotropic constant-strain triangle, then Eq. 4.1-6 shows that nodal loads $\{\mathbf{r}_e\}$ produced by the temperature field are all zero. Therefore, $\{\mathbf{d}\} = \{\mathbf{0}\}$ , and, by Eq. 4.7-2,
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$$
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\left\{ \begin{array}{l} \sigma_ {x} \\ \sigma_ {y} \\ \tau_ {x y} \end{array} \right\} = - \frac {E \alpha}{1 - \nu} \left(c _ {1} x + c _ {2} y\right) \left\{ \begin{array}{l} 1 \\ 1 \\ 0 \end{array} \right\} \tag {4.7-5}
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$$
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These stresses can be large. However, they are spurious: theory of elasticity shows that a linear temperature field produces deformation of the triangle but zero stresses. Equation 4.7-5 would indeed yield zero stresses if evaluated at the centroid of the triangle, $x = y = 0$ . This amounts to using a constant temperature field in the constant-strain triangle when calculating stresses. Generalizing, we infer that, for the purpose of stress analysis, one should reduce (if necessary) the order of the element temperature field to the same order as the element strain field. Thus, if Fig. 4.7-2 is regarded as a quadratic triangle (three corner nodes and three midside nodes), the element has a linear strain field. The temperature field of Eq. 4.7-4 is also linear, so the element correctly yields $\sigma_x = \sigma_y = \tau_{xy} = 0$ .
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$\tau_{xy} = 0$ . As a counterexample [4.7, 4.8], imagine that a uniform beam is fixed between rigid walls (e.g., for a one-element model, set $u_{1} = u_{2} = u_{3} = u_{4} = 0$ in Fig. 4.2-4). Let $\nu = 0$ and let the temperature vary linearly with distance $y$ from the midsurface of the beam, $T = T_{0}y$ . Use of the actual temperature field $T = T_{0}y$ yields $\sigma_{x} = \sigma_{y} = -E\alpha T$ , which is correct for $\sigma_{x}$ but wrong for $\sigma_{y}$ . Use of the reduced temperature field ( $T = 0$ in this case) yields $\sigma_{x} = \sigma_{y} = 0$ , which is wrong for $\sigma_{x}$ but correct for $\sigma_{y}$ .
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for $\sigma_{x}$ but correct for $\sigma_{y}$ . Accordingly, we cannot say whether it is always better to accept a complicated temperature field or to smooth it in element by element fashion, as neither strategy will be best in all cases [4.7, 4.8].
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Iterative Improvement. Computed stresses can be iteratively improved as follows [4.9, 4.10]. First, solve the problem in standard fashion, and apply Eq. 4.7-1 to obtain stresses in each element. Next, at each node, use stresses from the sur-
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rounding elements to produce nodal average stresses. In a typical element, interpolate to find stresses $\{\sigma\}$ within the element,
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$$
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\{\sigma \} = [ N ] \{\overline {{\sigma}} \} \tag {4.7-6}
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$$
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where [N] is a shape function matrix and $\{\overline{\sigma}\}$ contains nodal average stresses for the element at hand. Terms may be added to the right-hand side of Eq. 4.7-6 in order to involve stresses within the element as well [4.10]. Known stress boundary conditions can be accounted for in Eq. 4.7-6. We recall from Eq. 4.1-6 that the integral of $[B]^{T}\{\sigma_{0}\}$ is a vector of nodal loads $\{r_{e}\}$ that is statically equivalent to the stress distribution $\{\sigma_{0}\}$ . If, instead of $\{\sigma_{0}\}$ , we use the total stress $\{\sigma\}$ from Eq. 4.7-6, the sum of nodal loads $\{r_{e}\}$ over all numel elements of the structure should be statically equivalent to the entire structure load vector $\{R\}$ . So we write
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$$
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[ \mathbf {K} ] \{\Delta \mathbf {D} \} = \{\mathbf {R} \} - \sum_ {n = 1} ^ {\text {numel}} \left(\int_ {V _ {e}} [ \mathbf {B} ] ^ {T} \{\boldsymbol {\sigma} \} d V\right) _ {n} \tag {4.7-7}
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$$
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where $\{\sigma\}$ comes from Eq. 4.7-6. If $\{\sigma\}$ is exact, the right-hand side of Eq. 4.7-7 is zero. Otherwise, it is a load imbalance that drives the solution toward a configuration that reduces the imbalance. We compute increments $\{\Delta D\}$ from Eq. 4.7-7. The new configuration is $\{D\}_{new} = \{D\}_{old} + \{\Delta D\}$ . New stresses are computed based on $\{D\}_{new}$ , Eqs. 4.7-6 and 4.7-7 are applied to the new stresses, and another $\{\Delta D\}$ is determined. The process repeats until convergence. Note that Eq. 4.7-7 does not require repeated construction and reduction of [K].
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Example applications appear in Fig. 4.7-3 and Table 4.7-1. The four-node elements used are described in Section 6.3. Symmetry is exploited by analyzing only one quadrant of the object. Stress at point A is obtained by extrapolation of stresses at four integration stations in the adjacent element. The “standard method” in Table 4.7-1 is the usual single-pass analysis.
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Stress Concentrations. A stress raiser, such as a hole or a notch, might be analyzed by a brute-force approach—that is, by using a profusion of elements to surround
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<details>
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<summary>text_image</summary>
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σ
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A
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(a)
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σ
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</details>
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<details>
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<summary>text_image</summary>
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P ← A → P
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(b)
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</details>
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<details>
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<summary>text_image</summary>
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2 by 2
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mesh
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(c)
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</details>
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Figure 4.7-3. Flat plate with a central hole of diameter equal to half the plate width. (a) Uniform tensile load on portions distant from the hole. (b) Point forces on a square plate. (c) Typical mesh used on quadrant shaded in parts a and b.
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TABLE 4.7-1. HORIZONTAL STRESS AT POINT A IN FIG. 4.7-3 USING THE STANDARD METHOD AND THE METHOD OF EQ. 4.7-7 (SIX ITERATIONS) [4.12].
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<table><tr><td rowspan="2">Loading Case</td><td colspan="3">2 by 2 Mesh in Quadrant</td><td colspan="3">4 by 4 Mesh in Quadrant</td></tr><tr><td>Standard</td><td>Six Iterations</td><td>Exact</td><td>Standard</td><td>Six Iterations</td><td>Exact</td></tr><tr><td>Fig. 4.7-3a</td><td>30.9</td><td>38.1</td><td>43.2</td><td>36.6</td><td>39.5</td><td>43.2</td></tr><tr><td>Fig. 4.7-3b</td><td>73.6</td><td>85.4</td><td> $112.3^a$ </td><td>95.3</td><td>96.9</td><td> $112.3^a$ </td></tr></table>
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$^{a}$ From an 8 by 8 mesh of nine-node elements.
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the discontinuity. This is both tedious and expensive. An alternative that uses a coarse mesh may be available, as follows [4.13].
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Analyze the problem of interest using the coarsest mesh that will model the geometry. Let the computed stress at the point of interest be called $\sigma_{a}$ . Next, use a locally identical mesh to analyze the most closely related condition for which results are already available—for example, in a table of stress concentration factors. For this “secondary” case, call the computed stress $\sigma_{r}$ and the tabulated stress $\sigma_{t}$ . The ratio $\sigma_{t}/\sigma_{r}$ is regarded as a correction factor that can be applied to closely related geometries. Thus the stress in the case of interest is now estimated to be $(\sigma_{t}/\sigma_{r})\sigma_{a}$ rather than $\sigma_{a}$ .
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to be $(\sigma_t / \sigma_r)\sigma_a$ rather than $\sigma_a$ . The success of this method depends on the availability of tabulated cases and on the skill of the analyst in selecting an appropriate secondary case.
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As an example, consider the determination of the peak stress at A in the point-loaded plate of Fig. 4.7-3b. As the secondary case we take the tensile strip of Fig. 4.7-3a. The tabulated stress concentration factor for the tensile strip is 2.16, which yields the stress at point A as 43.2 for the load applied. Thus, from the 2 by 2 and 4 by 4 mesh results in Table 4.7-1, respectively, we estimate that the stress at point A in the point-loaded plate is
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$$
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\frac {4 3 . 2}{3 0 . 9} 7 3. 6 = 1 0 3 \quad \text { and } \quad \frac {4 3 . 2}{3 6 . 6} 9 5. 3 = 1 1 2 \tag {4.7-8}
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$$
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Both of these results are better than the results given by six iterations in Table 4.7-1.
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4.7-1.
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If the case of interest and the secondary case are identical, one of course obtains the tabulated stress. This means only that a finite element analysis is unnecessary, except perhaps as used for the entire structure in order to obtain the load applied to the portion of interest or the average stress field in a certain region (e.g., to obtain stress $\sigma$ in Fig. 4.7-3a).
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# 4.8 OTHER FORMULATION METHODS
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The present chapter is devoted to elements whose properties are based on assumed displacement fields. This type of element is the most popular. The reader should be aware that several other element types are possible and are in use. If an alternatively derived element has displacement d.o.f., it can be used in combination with displacement-based elements. Indeed, the user of an analysis program may be unaware that some elements in the program are not displacement-based.
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Displacement fields and strain energy expressions provide only one of many approaches to formulation of finite elements.
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A type of “hybrid” element makes use of an assumed stress distribution within the element and assumed displacements along its edges. The formulation procedure yields an element with displacement d.o.f. A “mixed” element has both displacement d.o.f. and force d.o.f. Its characteristic matrix is not a stiffness matrix; rather, it contains a flexibility submatrix and a submatrix that couples the displacement and force d.o.f. Both hybrid elements and mixed elements can be formulated from appropriate functionals.
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For some physical problems, such as certain problems in fluid mechanics, no variational principle exits. Finite element formulations can yet be developed by weighted residual methods, chief among which is the Galerkin method.
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# PROBLEMS
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# Section 4.1
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4.1 If element d.o.f. are given virtual (i.e., small imaginary) displacements $\{\delta \mathbf{d}\}$ , strains are changed in the amount $\{\delta \boldsymbol{\epsilon}\} = [\mathbf{B}]\{\delta \mathbf{d}\}$ . Loads acting on the structure do work that is stored as the strain energy $\delta U = \int \{\delta \boldsymbol{\epsilon}\}^T \{\boldsymbol{\sigma}\} dV$ . Complete this virtual work argument to obtain the structure equations $[\mathbf{K}]\{\mathbf{D}\} = \{\mathbf{R}\}$ . In the process, identify formulas for $[\mathbf{k}]$ and $\{\mathbf{r}_e\}$ .
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4.2 Derive the stiffness matrix of a bar, Fig. 4.2-1, by use of the $a$ -basis method, Eqs. 4.1-17 and 4.1-18.
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# Section 4.2
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4.3 The three-node bar element shown is uniform and is allowed only axial displacement u. Its nodal d.o.f. are $u_{1}$ , $u_{2}$ , and $u_{3}$ . Its shape functions are given by Eq. 4.3-10. Determine the element stiffness matrix.
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<details>
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<summary>text_image</summary>
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y
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L/2 L/2
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1 2 3 x,u
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→ u₁ → u₂ → u₃
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</details>
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Problem 4.3
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<details>
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<summary>text_image</summary>
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y,v
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L
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v₂
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u₂
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2
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u₁
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1
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β
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x,u
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v₁
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</details>
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Problem 4.4
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4.4. For the plane truss element shown, write an expression for axial displacement as a function of the four nodal d.o.f., angle $\beta$ , and an axial coordinate along the bar. Use this expression to determine the 4 by 4 element stiffness matrix. Check your result against Eq. 2.4-3.
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4.5 Imagine that a beam element has positive directions for nodal d.o.f. and nodal load as shown in the sketch, rather than the directions shown in Fig. 4.2-2. How does this sign convention change [k] of Eq. 4.2-5? What is awkward about this sign convention?
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<!-- source-page: 158 -->
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<details>
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<summary>text_image</summary>
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w₁,F₁
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θ₁,M₁
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w₂,F₂
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θ₂,M₂
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L
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</details>
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Problem 4.5
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4.6 (a) Verify the correctness of the beam element stiffness matrix, Eq. 4.2-5. Use either Eq. 4.2-5 or Eq. 4.2-7.
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(b) Verify that $[\mathbf{B}]\{\mathbf{d}\}$ and $[\mathbf{k}]\{\mathbf{d}\}$ are both zero for a small rigid-body rotation of the beam element about its left end.
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4.7 Imagine that a pin-jointed plane truss is modeled by plane frame elements. If rotational d.o.f. $\theta$ are suppressed at all nodes, is the truss correctly modeled? Explain.
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4.8 (a) Imagine that the uniform bar element of Fig. 4.2-1 is to have two d.o.f. at each node—namely, axial displacement $u$ and axial strain $\epsilon_x$ . Thus $\{\mathbf{d}\} = \left[u_1 \quad \epsilon_{x1} \quad u_2 \quad \epsilon_{x2}\right]^T$ . Derive the 4 by 4 element stiffness matrix. (b) Should this element be used to model a bar that has abrupt changes in cross section, as in Fig. 1.1-1b? Explain.
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4.9 Consider a plane grillage, which consists of bars that all occupy the $xy$ plane. Let each bar be uniform and parallel to either the $x$ axis or the $y$ axis. Each bar can resist bending, in either the $yz$ or the $zx$ plane, and torsion about its axis. Write the 6 by 6 stiffness matrix of an element that lies parallel to the $y$ axis. Express your answer in terms of $E$ , $I_x$ , $G$ , $J$ , and element length $L$ . Indicate the d.o.f. on which [k] operates, arrayed in order consistent with [k].
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4.10 The cantilever beam shown is modeled by an extremely coarse mesh of constant-strain triangles. The beam is loaded by an end moment. Qualitatively plot the variation of $\sigma_{x}$ versus $x$ along the $x$ axis.
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4.11 (a) For the particular triangular element shown, evaluate the strain-displacement matrix [B] in terms of dimensions $a$ and $b$ . Let d.o.f. in $\{\mathbf{d}\}$ have the ordering $\lfloor u_1, v_1, u_2, v_2, u_3, v_3 \rfloor^T$ .
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(b) Let $\{\mathbf{d}\}$ contain displacements consistent with Eq. 4.2-9—that is, $u_{1} = a_{1}, u_{2} = a_{1} + a_{2}a$ , and so on. Show that $[\mathbf{B}]\{\mathbf{d}\}$ yields the strains of Eq. 4.2-10.
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4.2-10. (c) Determine [k] in terms of $a, b, E$ , and $t$ if thickness $t$ is constant and Poisson's ratio is zero.
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(d) Fix nodes 1 and 2, apply a $y$ -direction force $P$ to node 3, and solve for $u_{3}$ and $v_{3}$ . Also compute stresses in the element.
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<details>
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<summary>text_image</summary>
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y
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c
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x
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F
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M = 2cF
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c
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F
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</details>
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Problem 4.10
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<details>
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<summary>text_image</summary>
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y,v
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a
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3
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b
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1
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2
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x,u
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</details>
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Problem 4.11
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<details>
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<summary>text_image</summary>
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y, v
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E = 10^7
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v = 0
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t = 1
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4
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3
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j
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x, u
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1
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2
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2
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</details>
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<details>
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<summary>text_image</summary>
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20
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31
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j
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19
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30
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</details>
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Problem 4.12
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4.12 Element $j$ is of the type shown in Fig. 4.2-4. It has stiffness matrix $[\mathbf{k}]$ and nodal displacement vector $\{\mathbf{d}\} = \lfloor u_1, v_1, u_2, v_2, \ldots, v_4 \rfloor^T$ . Element $j$ is to be attached to nodes 19, 20, 30, and 31 of the structure, a fragment of which is shown. The structure stiffness matrix $[\mathbf{K}]$ is to be stored in full (not banded) format, and no d.o.f. are as yet removed by imposing support conditions. What is the numerical contribution of element $j$ to the single coefficient in $[\mathbf{K}]$ at the intersection of
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(a) row 48 and column 39?
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(b) row 37 and column 37?
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(c) row 59 and column 61?
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4.13 Consider the plane bilinear element of Fig. 4.2-4 and the linear displacement field $u = a_{1} + a_{2}x + a_{3}y$ , $v = a_{4} + a_{5}x + a_{6}y$ . Let nodal d.o.f. $\{\mathbf{d}\}$ be consistent with this field; that is, let $u_{1} = a_{1} - a_{2}a - a_{3}b$ , and so on.
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(a) Show that Eq. 4.2-13 yields the given $u$ and $v$ fields.
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(b) Show that Eq. 4.2-14 yields the constant-strain state $\epsilon_{x} = a_{2}, \epsilon_{y} = a_{6}$ , and $\gamma_{xy} = a_{3} + a_{5}$ .
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4.14 The best approximation of a state of pure bending that a plane bilinear element can display is $u = \overline{u}xy / ab$ and $v = 0$ , where $\overline{u}$ is the magnitude of a corner displacement.
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(a) Sketch the deformed element. Show nodal forces $F$ that produce the deformation state.
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(b) If Poisson's ratio is taken as zero, strain energy per unit volume is $(E\epsilon_x^2 + E\epsilon_y^2 + G\gamma_{xy}^2)/2$ . Use this information to determine $F$ as a function of $a, b, E, G, \overline{u}$ , and element thickness $t$ .
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(c) What is the correct value of $F$ , according to elementary beam theory?
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(d) In parts (b) and (c), one can define a stiffness measure $S$ as $S = F / \overline{u}$ . Give a physical explanation as to why $S_{(b)} > S_{(c)}$ .
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(e) Show that the ratio $S_{(b)} / S_{(c)}$ approaches unity only as $a / b$ approaches zero. What happens as $a / b$ becomes large?
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4.15 For the rectangular solid element of Fig. 4.2-5, write out the first three columns of [N] and the first three columns of [B].
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4.16 The block of material shown is loaded by axial force $P = \sigma_c bt$ , which produces axial deflection $D$ . Axial stiffness is $k = P / D$ .
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(a) Show that $k$ is inversely proportional to $L$ if cross-sectional area $A = bt$ remains constant.
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(b) Show that $k$ is independent of $b$ and $L$ if $t$ remains constant and the aspect ratio $b / L$ is not changed.
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<details>
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<summary>text_image</summary>
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|
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b
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||
L
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t
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σc
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||
</details>
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Problem 4.16
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(c) Show that $k$ is directly proportional to a linear dimension if the shape of the element is not changed. (These behaviors are in fact observed in axial, plane, and solid elements, respectively.)
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# Section 4.3
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4.17 Show that integrals containing $\{\epsilon_0\}$ and $\{\sigma_0\}$ in Eq. 4.1-6 yield nodal loads $\{\mathbf{r}_e\}$ that are self-equilibrating—that is, they sum to zero net force on an element. For simplicity, restrict your argument to a $C^0$ element, such as a plane bilinear element.
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4.18 Write a work equation analogous to Eq. 4.3-3, but include $\{\epsilon_0\}$ and $\{\sigma_0\}$ rather than $\{\Phi\}$ . Interpret the result: that is, can loads $\{\mathbf{r}_e\}$ still be called work-equivalent?
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4.19 (a) A two-node bar element is uniformly heated an amount $T$ . The cross-sectional area of the bar varies linearly from $A_{1}$ at $x = 0$ to $A_{2}$ at $x = L$ . What loads $\{\mathbf{r}_e\}$ are predicted by Eq. 4.1-6? Continue to use the linear shape functions given in Fig. 4.3-2. (b) Use the methods of elementary mechanics of materials to compute the forces this bar would apply to rigid walls if the bar were placed between the walls and uniformly heated an amount $T$ . Why do these forces differ from loads $\{\mathbf{r}_e\}$ of part (a)?
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4.20 A three-node bar element is subjected to a temperature change that varies linearly with $x$ , as shown. What axially directed nodal loads appear? Shape functions are given by Eq. 4.3-10.
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|
||
|
||
<details>
|
||
<summary>text_image</summary>
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|
||
1
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2
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T₃
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L/2
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L/2
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3
|
||
x
|
||
</details>
|
||
|
||
Problem 4.20
|
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|
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4.21 Let temperature $T$ in the plane bilinear element of Fig. 4.2-4 vary as $T = T_0x$ , where $T_0$ is a constant. Material properties $E, \nu,$ and $\alpha$ are constant over the element, as is thickness $t$ . What nodal loads $\{\mathbf{r}_e\}$ are produced by $T$ ?
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4.22 Imagine that a uniform prestress $\sigma_x = \overline{\sigma}$ exists throughout the trilinear element of Fig. 4.2-5. All other stresses are zero. What nodal loads $\{\mathbf{r}_e\}$ are produced?
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