김경종
391 lines
25 KiB
Markdown
391 lines
25 KiB
Markdown
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The element body forces and surface forces should not be automatically lumped at the nodes, but for a consistent formulation (one that is formulated from the same shape functions used to formulate the stiffness matrix), Eqs. (6.3.1) and (6.3.7), respectively, should be used. (Problems 8.3 and 8.4 illustrate this concept.) These forces can be added to any concentrated nodal forces to obtain the element force matrix. Here the element force matrix is of order 12
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1 because, in general, there could be an x and a y component of force at each of the six nodes associated with the element. The element equations are then given by
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$$
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\left\{ \begin{array}{c} f _ {1 x} \\ f _ {1 y} \\ \vdots \\ f _ {6 y} \end{array} \right\} = \left[ \begin{array}{c c c} k _ {1 1} & \dots & k _ {1, 1 2} \\ k _ {2 1} & & k _ {2, 1 2} \\ \vdots & & \vdots \\ k _ {1 2, 1} & \dots & k _ {1 2, 1 2} \end{array} \right] \left\{ \begin{array}{c} u _ {1} \\ v _ {1} \\ \vdots \\ v _ {6} \end{array} \right\} \tag {8.1.18}
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$$
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# Steps 5–7
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Steps 5–7, which involve assembling the global stiffness matrix and equations, determining the unknown global nodal displacements, and calculating the stresses, are identical to those in Section 6.2 for the CST. However, instead of constant stresses in each element, we now have a linear variation of the stresses in each element. Common practice was to use the centroidal element stresses. Current practice is to use the average of the nodal element stresses.
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# 8.2 Example LST Stiffness Determination
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To illustrate some of the procedures outlined in Section 8.1 for deriving an LST stiffness matrix, consider the following example. Figure 8–3 shows a specific LST and its coordinates. The triangle is of base dimension b and height h, with midside nodes.
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<details>
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<summary>line</summary>
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| Point | x | y |
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|---|---|---|
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| 1 | 0 | 0 |
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| 2 | 2 | 0 |
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| 3 | 0 | 3 |
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| 4 | 2 | 4 |
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| 5 | 0 | 5 |
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| 6 | 2 | 6 |
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</details>
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Figure 8–3 LST triangle for evaluation of a stiffness matrix
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Using the first six equations of Eq. (8.1.5), we calculate the coefficients $a_{1}$ through $a_{6}$ by evaluating the displacement u at each of the six known coordinates of each node as follows:
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$$
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\begin{array}{l} u _ {1} = u (0, 0) = a _ {1} \\ u _ {2} = u (b, 0) = a _ {1} + a _ {2} b + a _ {4} b ^ {2} \\ u _ {3} = u (0, h) = a _ {1} + a _ {3} h + a _ {6} h ^ {2} \\ u _ {4} = u \left(\frac {b}{2}, \frac {h}{2}\right) = a _ {1} + a _ {2} \frac {b}{2} + a _ {3} \frac {h}{2} + a _ {4} \left(\frac {b}{2}\right) ^ {2} + a _ {5} \frac {b h}{4} + a _ {6} \left(\frac {h}{2}\right) ^ {2} \tag {8.2.1} \\ u _ {5} = u \left(0, \frac {h}{2}\right) = a _ {1} + a _ {3} \frac {h}{2} + a _ {6} \left(\frac {h}{2}\right) ^ {2} \\ u _ {6} = u \left(\frac {b}{2}, 0\right) = a _ {1} + a _ {2} \frac {b}{2} + a _ {4} \left(\frac {b}{2}\right) ^ {2} \\ \end{array}
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$$
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Solving Eqs. (8.2.1) simultaneously for the $a_{i}$ 's, we obtain
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$$
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\begin{array}{l} a _ {1} = u _ {1} \quad a _ {2} = \frac {4 u _ {6} - 3 u _ {1} - u _ {2}}{b} \quad a _ {3} = \frac {4 u _ {5} - 3 u _ {1} - u _ {3}}{h} \\ a _ {4} = \frac {2 (u _ {2} - 2 u _ {6} + u _ {1})}{b ^ {2}} \quad a _ {5} = \frac {4 (u _ {1} + u _ {4} - u _ {5} - u _ {6})}{b h} \tag {8.2.2} \\ a _ {6} = \frac {2 \left(u _ {3} - 2 u _ {5} + u _ {1}\right)}{h ^ {2}} \\ \end{array}
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$$
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Substituting Eqs. (8.2.2) into the displacement expression for $u$ from Eqs. (8.1.2), we have
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$$
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\begin{array}{l} u = u _ {1} + \left[ \frac {4 u _ {6} - 3 u _ {1} - u _ {2}}{b} \right] x + \left[ \frac {4 u _ {5} - 3 u _ {1} - u _ {3}}{h} \right] y + \left[ \frac {2 (u _ {2} - 2 u _ {6} + u _ {1})}{b ^ {2}} \right] x ^ {2} \\ + \left[ \frac {4 (u _ {1} + u _ {4} - u _ {5} - u _ {6})}{b h} \right] x y + \left[ \frac {2 (u _ {3} - 2 u _ {5} + u _ {1})}{h ^ {2}} \right] y ^ {2} \tag {8.2.3} \\ \end{array}
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$$
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Similarly, solving for $a_7$ through $a_{12}$ by evaluating the displacement $v$ at each of the six nodes and then substituting the results into the expression for $v$ from Eqs. (8.1.2), we obtain
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$$
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\begin{array}{l} v = v _ {1} + \left[ \frac {4 v _ {6} - 3 v _ {1} - v _ {2}}{b} \right] x + \left[ \frac {4 v _ {5} - 3 v _ {1} - v _ {3}}{h} \right] y + \left[ \frac {2 (v _ {2} - 2 v _ {6} + v _ {1})}{b ^ {2}} \right] x ^ {2} \\ + \left[ \frac {4 (v _ {1} + v _ {4} - v _ {5} - v _ {6})}{b h} \right] x y + \left[ \frac {2 (v _ {3} - 2 v _ {5} + v _ {1})}{h ^ {2}} \right] y ^ {2} \tag {8.2.4} \\ \end{array}
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$$
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Using Eqs. (8.2.3) and (8.2.4), we can express the general displacement expressions in terms of the shape functions as
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$$
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\left\{ \begin{array}{l} u \\ v \end{array} \right\} = \left[ \begin{array}{c c c c c c c c c c c c} N _ {1} & 0 & N _ {2} & 0 & N _ {3} & 0 & N _ {4} & 0 & N _ {5} & 0 & N _ {6} & 0 \\ 0 & N _ {1} & 0 & N _ {2} & 0 & N _ {3} & 0 & N _ {4} & 0 & N _ {5} & 0 & N _ {6} \end{array} \right] \left\{ \begin{array}{l} u _ {1} \\ v _ {1} \\ \vdots \\ v _ {6} \end{array} \right\} \tag {8.2.5}
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$$
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where the shape functions are obtained by collecting coefficients that multiply each $u _ { i }$ term in Eq. (8.2.3). For instance, collecting all terms that multiply by $u _ { 1 }$ in Eq. (8.2.3), we obtain $N _ { 1 }$ . These shape functions are then given by
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$$
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N _ {1} = 1 - \frac {3 x}{b} - \frac {3 y}{h} + \frac {2 x ^ {2}}{b ^ {2}} + \frac {4 x y}{b h} + \frac {2 y ^ {2}}{h ^ {2}} \quad N _ {2} = \frac {- x}{b} + \frac {2 x ^ {2}}{b ^ {2}}
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$$
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$$
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N _ {3} = \frac {- y}{h} + \frac {2 y ^ {2}}{h ^ {2}} \quad N _ {4} = \frac {4 x y}{b h} \quad N _ {5} = \frac {4 y}{h} - \frac {4 x y}{b h} - \frac {4 y ^ {2}}{h ^ {2}} \tag {8.2.6}
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$$
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$$
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N _ {6} = \frac {4 x}{b} - \frac {4 x ^ {2}}{b ^ {2}} - \frac {4 x y}{b h}
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$$
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Using Eq. (8.2.5) in Eq. (8.1.10), and performing the differentiations indicated on u and $v ,$ we obtain
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$$
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\underline {{\varepsilon}} = \underline {{B}} \underline {{d}} \tag {8.2.7}
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$$
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where $\underline { { B } }$ is of the form of $\operatorname { E q . }$ (8.1.17), with the resulting $\beta ^ { \mathrm { { * } } } { }$ and $\gamma \mathbf { \bar { s } }$ in $\operatorname { E q } .$ (8.1.17) given by
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$$
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\beta_ {1} = - 3 h + \frac {4 h x}{b} + 4 y \quad \beta_ {2} = - h + \frac {4 h x}{b} \quad \beta_ {3} = 0
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$$
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$$
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\beta_ {4} = 4 y \quad \beta_ {5} = - 4 y \quad \beta_ {6} = 4 h - \frac {8 h x}{b} - 4 y \tag {8.2.8}
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$$
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$$
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\gamma_ {1} = - 3 b + 4 x + \frac {4 b y}{h} \quad \gamma_ {2} = 0 \quad \gamma_ {3} = - b + \frac {4 b y}{h}
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$$
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$$
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\gamma_ {4} = 4 x \quad \gamma_ {5} = 4 b - 4 x - \frac {8 b y}{h} \quad \gamma_ {6} = - 4 x
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$$
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These $\beta ^ { \circ } \mathbf { s }$ and $\gamma \mathbf { \bar { s } }$ are specific to the element in Figure 8–3. Specifically, using Eqs. (8.1.1) and (8.1.17) in Eq. (8.2.7), we obtain
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$$
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\varepsilon_ {x} = \frac {1}{2 A} \left[ \beta_ {1} u _ {1} + \beta_ {2} u _ {2} + \beta_ {3} u _ {3} + \beta_ {4} u _ {4} + \beta_ {5} u _ {5} + \beta_ {6} u _ {6} \right]
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$$
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$$
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\varepsilon_ {y} = \frac {1}{2 A} \left[ \gamma_ {1} v _ {1} + \gamma_ {2} v _ {2} + \gamma_ {3} v _ {3} + \gamma_ {4} v _ {4} + \gamma_ {5} v _ {5} + \gamma_ {6} v _ {6} \right]
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$$
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$$
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\gamma_ {x y} = \frac {1}{2 A} \left[ \gamma_ {1} u _ {1} + \beta_ {1} v _ {1} + \dots + \beta_ {6} v _ {6} \right]
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$$
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The stiffness matrix for a constant-thickness element can now be obtained on substituting Eqs. (8.2.8) into Eq. (8.1.17) to obtain $\underline { { B } } ,$ then substituting $\underline { { B } }$ into
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Eq. (8.1.16) and using calculus to set up the appropriate integration. The explicit expression for the $1 2 \times 1 2$ stiffness matrix, being extremely cumbersome to obtain, is not given here. Stiffness matrix expressions for higher-order elements are found in References [1] and [2].
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# 8.3 Comparison of Elements
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For a given number of nodes, a better representation of true stress and displacement is generally obtained using the LST element than is obtained with the same number of nodes using a much finer subdivision into simple CST elements. For example, using one LST yields better results than using four CST elements with the same number of nodes (Figure 8–4) and hence the same number of degrees of freedom (except for the case when constant stress exists).
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We now present results to compare the CST of Chapter 6 with the LST of this chapter. Consider the cantilever beam subjected to a parabolic load variation acting as shown in Figure 8–5. Let $E = 3 0 \times 1 0 ^ { 6 }$ psi, $\nu = 0 . 2 5$ , and $t = 1 . 0$ in.
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Table 8–1 lists the series of tests run to compare results using the CST and LST elements. Table 8–2 shows comparisons of free-end (tip) deflection and stress $\sigma _ { x }$ for each element type used to model the cantilever beam. From Table 8–2, we can observe that the larger the number of degrees of freedom for a given type of triangular element, the closer the solution converges to the exact one (compare run A-1 to run A-2, and B-1 to B-2). For a given number of nodes, the LST analysis yields some what better results for displacement than the CST analysis (compare run A-1 to run B-1).
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<details>
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<summary>natural_image</summary>
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Geometric diagram of a large triangle subdivided into smaller triangles, labeled (a), with no text or symbols.
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</details>
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<details>
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<summary>natural_image</summary>
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Simple geometric triangle diagram with six vertices and connecting lines, labeled (b) (no text or symbols on the diagram itself)
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</details>
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Figure 8–4 Basic triangular element: (a) four-CST and (b) one-LST
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<details>
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<summary>text_image</summary>
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12 in.
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48 in.
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y
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Parabolic load = 40 kip (total)
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x
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</details>
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Figure 8–5 Cantilever beam used to compare the CST and LST elements with a 4 16 mesh
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Table 8–1 Models used to compare CST and LST results for the cantilever beam of Figure 8–5
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<table><tr><td>Series of Tests Run</td><td>Number of Nodes</td><td>Number of Degrees of Freedom, $n_d$ </td><td>Number of Triangular Elements</td></tr><tr><td>A-1 4 × 16 mesh</td><td>85</td><td>160</td><td>128 CST</td></tr><tr><td>A-2 8 × 32</td><td>297</td><td>576</td><td>512 CST</td></tr><tr><td>B-1 2 × 8</td><td>85</td><td>160</td><td>32 LST</td></tr><tr><td>B-2 4 × 16</td><td>297</td><td>576</td><td>128 LST</td></tr></table>
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Table 8–2 Comparison of CST and LST results for the cantilever beam of Figure 8–5
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<table><tr><td>Run</td><td> $n_d$ </td><td>Bandwidth $^1$ $n_b$ </td><td>Tip Deflection (in.)</td><td> $σ_x$ (ksi)</td><td>Location (in.), $x, y$ </td></tr><tr><td>A-1</td><td>160</td><td>14</td><td>-0.29555</td><td>67.236</td><td>2.250, 11.250</td></tr><tr><td>A-2</td><td>576</td><td>22</td><td>-0.33850</td><td>81.302</td><td>1.125, 11.630</td></tr><tr><td>B-1</td><td>160</td><td>18</td><td>-0.33470</td><td>58.885</td><td>4.500, 10.500</td></tr><tr><td>B-2</td><td>576</td><td>22</td><td>-0.35159</td><td>69.956</td><td>2.250, 11.250</td></tr><tr><td colspan="3">Exact solution</td><td>-0.36133</td><td>80.000</td><td>0, 12</td></tr></table>
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1 Bandwidth is described in Appendix B.4.
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However, one of the reasons that the bending stress $\sigma _ { x }$ predicted by the LST model B-1 compared to CST model A-1 is not as accurate is as follows. Recall that the stress is calculated at the centroid of the element. We observe from the table that the location of the bending stress is closer to the wall and closer to the top for the CST model A-1 compared to the LST model B-1. As the classical bending stress is a linear function with increasing positive linear stress from the neutral axis for the downward applied load in this example, we expect the largest stress to be at the very top of the beam. So the model A-1 with more and smaller elements (with eight elements through the beam depth) has its centroid closer to the top (at 0.75 in. from the top) than model B-1 with few elements (two elements through the beam depth) with centroidal stress located at 1.5 in. from the top. Similarly, comparing A-2 to B-2 we observe the same trend in the results—displacement at the top end being more accurately predicted by the LST model, but stresses being calculated at the centroid making the A-2 model appear more accurate than the LST model due to the location where the stress is reported.
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Although the CST element is rather poor in modeling bending, we observe from Table 8–2 that the element can be used to model a beam in bending if a sufficient number of elements are used through the depth of the beam. In general, both LST and CST analyses yield results good enough for most plane stress/strain problems, provided a sufficient number of elements are used. In fact, most commercial programs incorporate the use of CST and/or LST elements for plane stress/strain problems,
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<details>
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<summary>line</summary>
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| Degrees of freedom | u_A (in.) - Linear-strain triangle | u_A (in.) - Constant-strain triangle | u_A (in.) - Exact solution | u_A (in.) - Symmetry | u_A (in.) - Grid work |
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| ------------------ | ---------------------------------- | ------------------------------------ | -------------------------- | --------------------- | --------------------- |
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| 0 | 0.0013 | 0.0011 | 0.0013 | 0.0013 | 0.0013 |
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| 100 | 0.0013 | 0.0011 | 0.0013 | 0.0013 | 0.0013 |
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| 200 | 0.0013 | 0.0011 | 0.0013 | 0.0013 | 0.0013 |
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| 300 | 0.0013 | 0.0011 | 0.0013 | 0.0013 | 0.0013 |
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| 400 | 0.0013 | 0.0011 | 0.0013 | 0.0013 | 0.0013 |
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| 500 | 0.0013 | 0.0011 | 0.0013 | 0.0013 | 0.0013 |
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</details>
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Figure 8–6 Plates subjected to parabolically distributed edge loads; comparison of results for triangular elements. (Gallagher, R. H. Finite Element Analysis: Fundamentals, ( 1975, pp. 269, 270. Reprinted by permission of Prentice Hall, Inc., Englewood Cliffs, NJ)
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although these elements are used primarily as transition elements (usually during mesh generation). The four-sided isoparametric plane stress/strain element is most frequently used in commercial programs and is described in Chapter 10.
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Also, recall that finite element displacements will always be less than (or equal to) the exact ones, because finite element models are normally predicted to be stiffer than the actual structures when the displacement formulation of the finite element method is used. (The reason for the stiffer model was discussed in Sections 3.10 and 7.3. Proof of this assertion can be found in References [4–7].
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Finally, Figure 8-6 (from Reference [8]) illustrates a comparison of CST and LST models of a plate subjected to parabolically distributed edge loads. Figure 8–6 shows that the LST model converges to the exact solution for horizontal displacement at point A faster than does the CST model. However, the CST model is quite acceptable even for modest numbers of degrees of freedom. For example, a CST model with 100 nodes (200 degrees of freedom) often yields nearly as accurate a solution as does an LST model with the same number of degrees of freedom.
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In conclusion, the results of Table 8–2 and Figure 8–6 indicate that the LST model might be preferred over the CST model for plane stress applications when relatively small numbers of nodes are used. However, the use of triangular elements of higher order, such as the LST, is not visibly advantageous when large numbers of nodes are used, particularly when the cost of formation of the element stiffnesses, equation bandwidth, and overall complexities involved in the computer modeling are considered.
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# d References
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[1] Pederson, P., ‘‘Some Properties of Linear Strain Triangles and Optimal Finite Element Models,’’ International Journal for Numerical Methods in Engineering, Vol. 7, pp. 415–430, 1973.
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[2] Tocher, J. L., and Hartz, B. J., ‘‘Higher-Order Finite Element for Plane Stress,’’ Journal of the Engineering Mechanics Division, Proceedings of the American Society of Civil Engineers, Vol. 93, No. EM4, pp. 149–174, Aug. 1967.
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[3] Bowes, W. H., and Russell, L. T., Stress Analysis by the Finite Element Method for Practicing Engineers, Lexington Books, Toronto, 1975.
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[4] Fraeijes de Veubeke, B., ‘‘Upper and Lower Bounds in Matrix Structural Analysis,’’ Matrix Methods of Structural Analysis, AGAR-Dograph 72, B. Fraeijes de Veubeke, ed., Macmillan, New York, 1964.
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[5] McLay, R. W., Completeness and Convergence Properties of Finite Element Displacement Functions: A General Treatment, American Institute of Aeronautics and Astronautics Paper No. 67–143, AIAA 5th Aerospace Meeting, New York, 1967.
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[6] Tong, P., and Pian, T. H. H., ‘‘The Convergence of Finite Element Method in Solving Linear Elastic Problems,’’ International Journal of Solids and Structures, Vol. 3, pp. 865– 879, 1967.
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[7] Cowper, G. R., ‘‘Variational Procedures and Convergence of Finite-Element Methods,’’ Numerical and Computer Methods in Structural Mechanics, S. J. Fenves, N. Perrone, A. R. Robinson, and W. C. Schnobrich, eds., Academic Press, New York, 1973.
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[8] Gallagher, R., Finite Element Analysis Fundamentals, Prentice Hall, Englewood Cliffs, NJ, 1975.
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[9] Zienkiewicz, O. C., The Finite Element Method, 3rd ed., McGraw-Hill, New York, 1977.
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# d Problems
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8.1 Evaluate the shape functions given by Eq. (8.2.6). Sketch the variation of each function over the surface of the triangular element shown in Figure 8–3.
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8.2 Express the strains $\varepsilon _ { x } , \varepsilon _ { y } ,$ and $\gamma _ { x y }$ for the element of Figure 8–3 by using the results given in Section 8.2. Evaluate these strains at the centroid of the element; then evaluate the stresses at the centroid in terms of E and n. Assume plane stress conditions apply.
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<details>
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<summary>text_image</summary>
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y
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3
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p
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5
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</details>
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Figure P8–3
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8.3 For the element of Figure 8–3 (shown again as Figure P8–3) subjected to the uniform pressure shown acting over the vertical side, determine the nodal force replacement system using Eq. (6.3.7). Assume an element thickness of t.
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8.4 For the element of Figure 8–3 (shown as Figure P8–4) subjected to the linearly varying line load shown acting over the vertical side, determine the nodal force replacement system using Eq. (6.3.7). Compare this result to that of Problem 6.9. Are these results expected? Explain.
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<details>
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<summary>text_image</summary>
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p₀
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3
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1
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y
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</details>
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Figure P8–4
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8.5 For the linear-strain elements shown in Figure P8–5, determine the strains $\varepsilon _ { x } , \varepsilon _ { y }$ , and $\gamma _ { x y }$ . Evaluate the stresses $\sigma _ { x } , \sigma _ { y } .$ , and $\tau _ { x y }$ at the centroids. The coordinates of the nodes are shown in units of inches. Let $E = \dot { 3 } 0 \times 1 0 ^ { 6 }$ psi, $\nu = 0 . 2 5$ , and $t = 0 . 2 5$ in. for both elements. Assume plane stress conditions apply. The nodal displacements are given as
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$$
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u _ {1} = 0. 0 \quad v _ {1} = 0. 0
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$$
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$$
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u _ {2} = 0. 0 0 1 \text { in. } \quad v _ {2} = 0. 0 0 2 \text { in. }
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$$
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$$
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u _ {3} = 0. 0 0 0 5 \text { in. } \quad v _ {3} = 0. 0 0 0 2 \text { in. }
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$$
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$$
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u _ {4} = 0. 0 0 0 2 \text { in. } \quad v _ {4} = 0. 0 0 0 1 \text { in. }
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$$
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$$
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u _ {5} = 0. 0 \quad v _ {5} = 0. 0 0 0 1 \text { in. }
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$$
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$$
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u _ {6} = 0. 0 0 0 5 \text { in. } \quad v _ {6} = 0. 0 0 1 \text { in. }
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$$
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(Hint: Use the results of Section 8.2.)
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<details>
|
||
<summary>line</summary>
|
||
| Point | x | y |
|
||
|---|---|---|
|
||
| 1 | 0 | 0 |
|
||
| 2 | 4 | 0 |
|
||
| 3 | 0 | 6 |
|
||
| 4 | 2 | 3 |
|
||
| 5 | 0 | 3 |
|
||
| 6 | 2 | 0 |
|
||
</details>
|
||
|
||
(a)
|
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|
||
|
||
|
||
<details>
|
||
<summary>line</summary>
|
||
|
||
| Point | x | y |
|
||
|---|---|---|
|
||
| 1 | 0 | 0 |
|
||
| 2 | 6 | 0 |
|
||
| 3 | 0 | 4 |
|
||
| 4 | 3 | 2 |
|
||
| 5 | 0 | 2 |
|
||
| 6 | 3 | 0 |
|
||
</details>
|
||
|
||
(b)
|
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Figure P8–5
|
||
|
||
<!-- source-page: 429 -->
|
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|
||
8.6 For the linear-strain element shown in Figure P8–6, determine the strains $\varepsilon _ { x } , \varepsilon _ { y } .$ and $\gamma _ { x y } .$ Evaluate these strains at the centroid of the element; then evaluate the stresses $\sigma _ { x } , \sigma _ { y }$ , and $\tau _ { x y }$ at the centroid. The coordinates of the nodes are shown in units of millimeters. Let E ¼ 210 GPa, $\nu = 0 . 2 5 ,$ , and t ¼ 10 mm. Assume plane stress conditions apply. Use the nodal displacements given in Problem 8.5 (converted to millimeters). Note that the $\beta ^ { \mathrm { { * } } } { }$ and $\gamma \mathbf { \bar { s } }$ from the example in Section 8.2 cannot be used here as the element in Figure P8–6 is oriented differently than the one in Figure 8–3.
|
||
|
||
|
||
|
||
<details>
|
||
<summary>line</summary>
|
||
|
||
| Point | x | y |
|
||
|---|---|---|
|
||
| 1 | 0 | 0 |
|
||
| 2 | 6 | 0 |
|
||
| 3 | 6 | 6 |
|
||
| 4 | 6 | 3 |
|
||
| 5 | 3 | 3 |
|
||
| 6 | 3 | 0 |
|
||
</details>
|
||
|
||
Figure P8–6
|
||
|
||
8.7 Evaluate the shape functions for the linear-strain triangle shown in Figure P8–7. Then evaluate the $\underline { { B } }$ matrix. Units are millimeters.
|
||
|
||
|
||
|
||
<details>
|
||
<summary>line</summary>
|
||
|
||
| Point | x-coordinate | y-coordinate |
|
||
|---|---|---|
|
||
| 1 | 0 | 0 |
|
||
| 2 | 60 | 0 |
|
||
| 3 | 60 | 60 |
|
||
| 4 | 60 | 4 |
|
||
| 5 | 60 | 5 |
|
||
</details>
|
||
|
||
Figure P8–7
|
||
|
||
8.8 Use the LST element to solve Example 7.2. Compare the results.
|
||
8.9 Write a computer program to solve plane stress problems using the LST element.
|
||
|
||
<!-- source-page: 430 -->
|
||
|
||
# Axisymmetric Elements
|
||
|
||
# Introduction
|
||
|
||
In previous chapters, we have been concerned with line or one-dimensional elements (Chapters 2–5) and two-dimensional elements (Chapters 6–8). In this chapter, we consider a special two-dimensional element called the axisymmetric element. This element is quite useful when symmetry with respect to geometry and loading exists about an axis of the body being analyzed. Problems that involve soil masses subjected to circular footing loads or thick-walled pressure vessels can often be analyzed using the element developed in this chapter.
|
||
|
||
We begin with the development of the stiffness matrix for the simplest axisymmetric element, the triangular torus, whose vertical cross section is a plane triangle.
|
||
|
||
We then present the longhand solution of a thick-walled pressure vessel to illustrate the use of the axisymmetric element equations. This is followed by a description of some typical large-scale problems that have been modeled using the axisymmetric element.
|
||
|
||
# 9.1 Derivation of the Stiffness Matrix
|
||
|
||
In this section, we will derive the stiffness matrix and the body and surface force matrices for the axisymmetric element. However, before the development, we will first present some fundamental concepts prerequisite to the understanding of the derivation. Axisymmetric elements are triangular tori such that each element is symmetric with respect to geometry and loading about an axis such as the z axis in Figure 9–1. Hence, the z axis is called the axis of symmetry or the axis of revolution. Each vertical cross section of the element is a plane triangle. The nodal points of an axisymmetric triangular element describe circumferential lines, as indicated in Figure 9–1.
|
||
|
||
In plane stress problems, stresses exist only in the x-y plane. In axisymmetric problems, the radial displacements develop circumferential strains that induce stresses $\sigma _ { r } , \sigma _ { \theta } , \sigma _ { z }$ , and $\tau _ { r z }$ , where $r , \theta ,$ and z indicate the radial, circumferential, and longitudinal
|