김경종
462 lines
26 KiB
Markdown
462 lines
26 KiB
Markdown
<!-- source-page: 471 -->
|
|
|
|
5.46. Consider the four-node plate bending element in Example 5.29. Assume that the element is loaded on its top surface with the constant traction shown. Calculate the consistent nodal point forces and moments.
|
|
|
|
|
|
|
|
<details>
|
|
<summary>text_image</summary>
|
|
|
|
z
|
|
y
|
|
x
|
|
h
|
|
30°
|
|
t
|
|
t is acting
|
|
in y-z plane
|
|
</details>
|
|
|
|
$$
|
|
\mathbf {t} = \left[ \begin{array}{c} 0 \\ \sqrt {3} / 2 \\ - 1 / 2 \end{array} \right] \quad \text { force per unit area }
|
|
$$
|
|
|
|
5.47. Establish the transverse shear strain interpolation matrix $B_{\gamma}$ of the parallelogram-shaped MITC4 element shown.
|
|
|
|
|
|
|
|
<details>
|
|
<summary>text_image</summary>
|
|
|
|
y
|
|
z
|
|
x
|
|
2
|
|
2
|
|
3
|
|
60°
|
|
4
|
|
1
|
|
3
|
|
</details>
|
|
|
|
5.48. Consider the formulation of the MITC4 element and Example 4.30. Show that the MITC4 element formulation can be derived from the Hu-Washizu variational principle.
|
|
5.49. Consider the four-node shell element shown and develop the geometry and displacement interpolations (5.107) and (5.112).
|
|
|
|
|
|
|
|
<details>
|
|
<summary>text_image</summary>
|
|
|
|
z
|
|
3
|
|
2
|
|
y
|
|
40
|
|
4
|
|
1
|
|
45°
|
|
1
|
|
x
|
|
20
|
|
</details>
|
|
|
|
5.50. Show explicitly that using the general shell element formulation in (5.107) to (5.118) for a flat element is equivalent to the superposition of the Reissner-Mindlin plate element formulation in (5.88) to (5.99) and the plane stress membrane element formulation in Section 5.3.1.
|
|
|
|
<!-- source-page: 472 -->
|
|
|
|
5.51. Use a computer program to solve the problem shown in Fig. 5.36 with curved shell elements. First, use a single element, and then, use two geometrically distorted elements for the structure to study the element distortion sensitivity.
|
|
|
|
5.52. Consider the following Kirchhoff plate theory boundary conditions at the edge of a plate:
|
|
|
|
$$
|
|
w = 0; \quad \frac {\partial w}{\partial x} = \frac {\partial w}{\partial y} = 0 \tag {a}
|
|
$$
|
|
|
|
Establish a corresponding reasonable choice of boundary conditions for the Reissner-Mindlin plate theory. Also, discuss and illustrate graphically that the boundary conditions in (a) do not uniquely determine the boundary conditions for the Reissner-Mindlin plate theory.
|
|
|
|
# 5.5 NUMERICAL INTEGRATION
|
|
|
|
An important aspect of isoparametric and related finite element analysis is the required numerical integration. The required matrix integrals in the finite element calculations have been written as
|
|
|
|
$$
|
|
\int \mathbf {F} (r) d r; \int \mathbf {F} (r, s) d r d s; \int \mathbf {F} (r, s, t) d r d s d t \tag {5.131}
|
|
$$
|
|
|
|
in the one-, two-, and three-dimensional cases, respectively. It was stated that these integrals are in practice evaluated numerically using
|
|
|
|
$$
|
|
\left. \begin{array}{c} \int \mathbf {F} (r) d r = \sum_ {i} \alpha_ {i} \mathbf {F} \left(r _ {i}\right) + \mathbf {R} _ {n} \\ \int \mathbf {F} (r, s) d r d s = \sum_ {i, j} \alpha_ {i j} \mathbf {F} \left(r _ {i}, s _ {j}\right) + \mathbf {R} _ {n} \\ \int \mathbf {F} (r, s, t) d r d s d t = \sum_ {i, j, k} \alpha_ {i j k} \mathbf {F} \left(r _ {i}, s _ {j}, t _ {k}\right) + \mathbf {R} _ {n} \end{array} \right\} \tag {5.132}
|
|
$$
|
|
|
|
where the summations extend over all i, j, and k specified, the $\alpha_{i}$ , $\alpha_{ij}$ , and $\alpha_{ijk}$ are weighting factors, and $\mathbf{F}(r_{i})$ , $\mathbf{F}(r_{i}, s_{j})$ , and $\mathbf{F}(r_{i}, s_{j}, t_{k})$ are the matrices $\mathbf{F}(r)$ , $\mathbf{F}(r, s)$ , and $\mathbf{F}(r, s, t)$ evaluated at the points specified in the arguments. The matrices $R_{n}$ are error matrices, which in practice are usually not evaluated. Therefore, we use
|
|
|
|
$$
|
|
\left. \begin{array}{c} \int \mathbf {F} (r) d r = \sum_ {i} \alpha_ {i} \mathbf {F} \left(r _ {i}\right) \\ \int \mathbf {F} (r, s) d r d s = \sum_ {i, j} \alpha_ {i j} \mathbf {F} \left(r _ {i}, s _ {j}\right) \\ \int \mathbf {F} (r, s, t) d r d s d t = \sum_ {i, j, k} \alpha_ {i j k} \mathbf {F} \left(r _ {i}, s _ {j}, t _ {k}\right) \end{array} \right\} \tag {5.133}
|
|
$$
|
|
|
|
The purpose in this section is to present the theory and practical implications of numerical integrations. An important point is the integration accuracy that is needed, i.e., the number of integration points required in the element formation.
|
|
|
|
As presented above, in finite element analysis we integrate matrices, which means that each element of the matrix considered is integrated individually. Hence, for the derivation
|
|
|
|
<!-- source-page: 473 -->
|
|
|
|
of the numerical integration formulas we can consider a typical element of a matrix, which we denote as $F$ .
|
|
|
|
Consider the one-dimensional case first, i.e., the integration of $\int_{a}^{b} F(r) \, dr$ . In an isoparametric element calculation we would actually have a = -1 and $b = +1$ .
|
|
|
|
The numerical integration of $\int_{a}^{b} F(r) \, dr$ is essentially based on passing a polynomial $\psi(r)$ through given values of $F(r)$ and then using $\int_{a}^{b} \psi(r) \, dr$ as an approximation to $\int_{a}^{b} F(r) \, dr$ . The number of evaluations of $F(r)$ and the positions of the sampling points in the interval from a to b determine how well $\psi(r)$ approximates $F(r)$ and hence the error of the numerical integration (see, for example, C. E. Fröberg [A]).
|
|
|
|
# 5.5.1 Interpolation Using a Polynomial
|
|
|
|
Assume that $F(r)$ has been evaluated at the $(n + 1)$ distinct points $r_{0}, r_{1}, \ldots, r_{n}$ to obtain $F_{0}, F_{1}, \ldots, F_{n}$ , respectively, and that a polynomial $\psi(r)$ is to be passed through these data. Then there is a unique polynomial $\psi(r)$ given as
|
|
|
|
$$
|
|
\psi (r) = a _ {0} + a _ {1} r + a _ {2} r ^ {2} + \dots + a _ {n} r ^ {n} \tag {5.134}
|
|
$$
|
|
|
|
Using the condition $\psi(r) = F(r)$ at the $(n + 1)$ interpolating points, we have
|
|
|
|
$$
|
|
\mathbf {F} = \mathbf {V a} \tag {5.135}
|
|
$$
|
|
|
|
where
|
|
|
|
$$
|
|
\mathbf {F} = \left[ \begin{array}{c} F _ {0} \\ F _ {1} \\ \cdot \\ \cdot \\ F _ {n} \end{array} \right]; \quad \mathbf {a} = \left[ \begin{array}{c} a _ {0} \\ a _ {1} \\ \cdot \\ \cdot \\ a _ {n} \end{array} \right] \tag {5.136}
|
|
$$
|
|
|
|
and V is the Vandermonde matrix,
|
|
|
|
$$
|
|
\mathbf {V} = \left[ \begin{array}{c c c c c} 1 & r _ {0} & r _ {0} ^ {2} & \dots & r _ {0} ^ {n} \\ 1 & r _ {1} & r _ {1} ^ {2} & \dots & r _ {1} ^ {n} \\ \cdot & \cdot & \cdot & & \cdot \\ \cdot & \cdot & \cdot & & \cdot \\ 1 & r _ {n} & r _ {n} ^ {2} & \dots & r _ {n} ^ {n} \end{array} \right] \tag {5.137}
|
|
$$
|
|
|
|
Since det $\mathbf{V} \neq 0$ , provided that the $r_i$ are distinct points, we have a unique solution for $\mathbf{a}$ .
|
|
|
|
However, a more convenient way to obtain $\psi(r)$ is to use Lagrangian interpolation. First, we recall that the $(n+1)$ functions 1, $r, r^{2}, \ldots, r^{n}$ form an $(n+1)$ -dimensional vector space, say $V_{n}$ , in which $\psi(r)$ is an element (see Section 2.3). Since the coordinates $a_{0}, a_{1}, a_{2}, \ldots, a_{n}$ of $\psi(r)$ are relatively difficult to evaluate using (5.135), we seek a different basis for the space $V_{n}$ in which the coordinates of $\psi(r)$ are more easily evaluated. This basis is provided by the fundamental polynomials of Lagrangian interpolation, given as
|
|
|
|
$$
|
|
l _ {j} (r) = \frac {(r - r _ {0}) (r - r _ {1}) \cdots (r - r _ {j - 1}) (r - r _ {j + 1}) \cdots (r - r _ {n})}{(r _ {j} - r _ {0}) (r _ {j} - r _ {1}) \cdots (r _ {j} - r _ {j - 1}) (r _ {j} - r _ {j + 1}) \cdots (r _ {j} - r _ {n})} \tag {5.138}
|
|
$$
|
|
|
|
where
|
|
|
|
$$
|
|
l _ {j} (r _ {i}) = \delta_ {i j} \tag {5.139}
|
|
$$
|
|
|
|
<!-- source-page: 474 -->
|
|
|
|
where $\delta_{ij}$ is the Kronecker delta; i.e., $\delta_{ij} = 1$ for i = j, and $\delta_{ij} = 0$ for $i \neq j$ . Using the property in (5.139), the coordinates of the base vectors are simply the values of $F(r)$ , and the polynomial $\psi(r)$ is
|
|
|
|
$$
|
|
\psi (r) = F _ {0} l _ {0} (r) + F _ {1} l _ {1} (r) + \dots + F _ {n} l _ {n} (r) \tag {5.140}
|
|
$$
|
|
|
|
EXAMPLE 5.33: Establish the interpolating polynomial $\psi(r)$ for the function $F(r) = 2^r - r$ when the data at the points $r = 0, 1$ , and 3 are used. In this case $r_0 = 0$ , $r_1 = 1$ , $r_2 = 3$ , and $F_0 = 1$ , $F_1 = 1$ , $F_2 = 5$ .
|
|
|
|
In the first approach we use the relation in (5.135) to calculate the unknown coefficients $a_{0}$ , $a_{1}$ , and $a_{2}$ of the polynomial $\psi(r) = a_{0} + a_{1}r + a_{2}r^{2}$ . In this case we have
|
|
|
|
$$
|
|
\left[ \begin{array}{l l l} 1 & 0 & 0 \\ 1 & 1 & 1 \\ 1 & 3 & 9 \end{array} \right] \left[ \begin{array}{l} a _ {0} \\ a _ {1} \\ a _ {2} \end{array} \right] = \left[ \begin{array}{l} 1 \\ 1 \\ 5 \end{array} \right]
|
|
$$
|
|
|
|
The solution gives $a_0 = 1$ , $a_1 = -\frac{2}{3}$ , $a_2 = \frac{2}{3}$ , and therefore $\psi(r) = 1 - \frac{2}{3} r + \frac{2}{3} r^2$ .
|
|
|
|
If Lagrangian interpolation is employed, we use the relation in (5.140) which in this case gives
|
|
|
|
$$
|
|
\psi (r) = (1) \frac {(r - 1) (r - 3)}{(- 1) (- 3)} + (1) \frac {(r) (r - 3)}{(1) (- 2)} + (5) \frac {(r) (r - 1)}{(3) (2)}
|
|
$$
|
|
|
|
or, as before, $\psi (r) = 1 - \frac{2}{3} r + \frac{2}{3} r^2$
|
|
|
|
# 5.5.2 The Newton-Cotes Formulas (One-Dimensional Integration)
|
|
|
|
Having established an interpolating polynomial $\psi(r)$ , we can now obtain an approximation to the integral $\int_{a}^{b} F(r) \, dr$ . In Newton-Cotes integration, it is assumed that the sampling points of F are spaced at equal distances, and we define
|
|
|
|
$$
|
|
r _ {0} = a; \quad r _ {n} = b; \quad h = \frac {b - a}{n} \tag {5.141}
|
|
$$
|
|
|
|
Using Lagrangian interpolation to obtain $\psi(r)$ as an approximation to $F(r)$ , we have
|
|
|
|
$$
|
|
\int_ {a} ^ {b} F (r) d r = \sum_ {i = 0} ^ {n} \left[ \int_ {a} ^ {b} l _ {i} (r) d r \right] F _ {i} + R _ {n} \tag {5.142}
|
|
$$
|
|
|
|
or, evaluated, $\int_{a}^{b} F(r) \, dr = (b - a) \sum_{i=0}^{n} C_i^n F_i + R_n$ (5.143)
|
|
|
|
where $R_{n}$ is the remainder and the $C_i^n$ are the Newton-Cotes constants for numerical integration with $n$ intervals.
|
|
|
|
The Newton-Cotes constants and corresponding remainder terms are summarized in Table 5.5 for n = 1 to 6. The cases n = 1 and n = 2 are the well-known trapezoidal rule and Simpson formula. We note that the formulas for n = 3 and n = 5 have the same order of accuracy as the formulas for n = 2 and n = 4, respectively. For this reason, the even formulas with n = 2 and n = 4 are used in practice.
|
|
|
|
<!-- source-page: 475 -->
|
|
|
|
TABLE 5.5 Newton-Cotes numbers and error estimates
|
|
|
|
<table><tr><td>Number of intervals n</td><td> $C_0^n$ </td><td> $C_1^n$ </td><td> $C_2^n$ </td><td> $C_3^n$ </td><td> $C_4^n$ </td><td> $C_5^n$ </td><td> $C_6^n$ </td><td>Upper bound on error $R_n$ as a function of the derivative of F</td></tr><tr><td>1</td><td> $\frac{1}{2}$ </td><td> $\frac{1}{2}$ </td><td></td><td></td><td></td><td></td><td></td><td> $10^{-1}(b - a)^3 F^{\mathrm{II}}(r)$ </td></tr><tr><td>2</td><td> $\frac{1}{6}$ </td><td> $\frac{4}{6}$ </td><td> $\frac{1}{6}$ </td><td></td><td></td><td></td><td></td><td> $10^{-3}(b - a)^5 F^{\mathrm{IV}}(r)$ </td></tr><tr><td>3</td><td> $\frac{1}{8}$ </td><td> $\frac{3}{8}$ </td><td> $\frac{3}{8}$ </td><td> $\frac{1}{8}$ </td><td></td><td></td><td></td><td> $10^{-3}(b - a)^5 F^{\mathrm{IV}}(r)$ </td></tr><tr><td>4</td><td> $\frac{7}{90}$ </td><td> $\frac{32}{90}$ </td><td> $\frac{12}{90}$ </td><td> $\frac{32}{90}$ </td><td> $\frac{7}{90}$ </td><td></td><td></td><td> $10^{-6}(b - a)^7 F^{\mathrm{VI}}(r)$ </td></tr><tr><td>5</td><td> $\frac{19}{288}$ </td><td> $\frac{75}{288}$ </td><td> $\frac{50}{288}$ </td><td> $\frac{50}{288}$ </td><td> $\frac{75}{288}$ </td><td> $\frac{19}{288}$ </td><td></td><td> $10^{-6}(b - a)^7 F^{\mathrm{VI}}(r)$ </td></tr><tr><td>6</td><td> $\frac{41}{840}$ </td><td> $\frac{216}{840}$ </td><td> $\frac{27}{840}$ </td><td> $\frac{272}{840}$ </td><td> $\frac{27}{840}$ </td><td> $\frac{216}{840}$ </td><td> $\frac{41}{840}$ </td><td> $10^{-9}(b - a)^9 F^{\mathrm{VIII}}(r)$ </td></tr></table>
|
|
|
|
EXAMPLE 5.34: Evaluate the Newton-Cotes constants when the interpolating polynomial is of order 2; i.e., $\psi(r)$ is a parabola.
|
|
|
|
In this case we have
|
|
|
|
$$
|
|
\int_ {a} ^ {b} F (r) d r \doteq \int_ {a} ^ {b} \left[ F _ {0} \frac {(r - r _ {1}) (r - r _ {2})}{(r _ {0} - r _ {1}) (r _ {0} - r _ {2})} + F _ {1} \frac {(r - r _ {0}) (r - r _ {2})}{(r _ {1} - r _ {0}) (r _ {1} - r _ {2})} + F _ {2} \frac {(r - r _ {0}) (r - r _ {1})}{(r _ {2} - r _ {0}) (r _ {2} - r _ {1})} \right] d r
|
|
$$
|
|
|
|
Using $r_{0} = a$ , $r_{1} = a + h$ , $r_{2} = a + 2h$ , where $h = (b - a)/2$ , the evaluation of the integral gives
|
|
|
|
$$
|
|
\int_ {a} ^ {b} F (r) d r \doteq \frac {b - a}{6} (F _ {0} + 4 F _ {1} + F _ {2})
|
|
$$
|
|
|
|
Hence the Newton-Cotes constants are as given in Table 5.5 for the case n = 2.
|
|
|
|
EXAMPLE 5.35: Use Simpson's rule to integrate $\int_0^3 (2^r - r) dr$ .
|
|
|
|
In this case $n = 2$ and $h = \frac{3}{2}$ . Therefore, $r_0 = 0$ , $r_1 = \frac{3}{2}$ , $r_2 = 3$ , and $F_0 = 1$ , $F_1 = 1.328427$ , $F_2 = 5$ , and we obtain
|
|
|
|
$$
|
|
\int_ {0} ^ {3} (2 ^ {r} - r) d r \doteq \frac {3}{6} [ (1) (1) + (4) (1. 3 2 8 4 2 7) + (1) (5) ]
|
|
$$
|
|
|
|
or $\int_0^3 (2^r - r) dr \doteq 5.656854$
|
|
|
|
The exact result is $\int_0^3 (2^r - r) dr = 5.598868$
|
|
|
|
Hence the error is $R = 0.057986$
|
|
|
|
However, using the upper bound value on the error, we have
|
|
|
|
$$
|
|
R < \frac {(3 - 0) ^ {5}}{1 0 0 0} (\ln 2) ^ {4} (2 ^ {r}) = 0. 4 4 8 7 4 3
|
|
$$
|
|
|
|
<!-- source-page: 476 -->
|
|
|
|
To obtain greater accuracy in the integration using the Newton-Cotes formulas we need to employ a smaller interval h, i.e., include more evaluations of the function to be integrated. Then we have the choice between two different strategies: we may use a higher-order Newton-Cotes formula or, alternatively, employ the lower-order formula in a repeated manner, in which case the integration procedure is referred to as a composite formula. Consider the following example.
|
|
|
|
EXAMPLE 5.36: Increase the accuracy of the integration in Example 5.35 by using half the interval spacing.
|
|
|
|
In this case we have $h = \frac{3}{4}$ , and the required function values are $F_{0} = 1$ , $F_{1} = 0.931792$ , $F_{2} = 1.328427$ , $F_{3} = 2.506828$ , and $F_{4} = 5$ . The choice now lies between using the higher-order Newton-Cotes formula with n = 4 or applying the Simpson's rule twice, i.e., to the first two intervals and then to the second two intervals. Using the Newton-Cotes formula with n = 4, we obtain
|
|
|
|
$$
|
|
\int_ {0} ^ {3} (2 ^ {r} - r) d r \doteq \frac {3}{9 0} \left(7 F _ {0} + 3 2 F _ {1} + 1 2 F _ {2} + 3 2 F _ {3} + 7 F _ {4}\right)
|
|
$$
|
|
|
|
Hence, $\int_0^3 (2^r - r) dr \doteq 5.599232$
|
|
|
|
On the other hand, using Simpson's rule twice, we have
|
|
|
|
$$
|
|
\int_ {0} ^ {3} (2 ^ {r} - r) d r = \int_ {0} ^ {3 / 2} (2 ^ {r} - r) d r + \int_ {3 / 2} ^ {3} (2 ^ {r} - r) d r
|
|
$$
|
|
|
|
The integration is performed using
|
|
|
|
$$
|
|
\int_ {0} ^ {3 / 2} (2 ^ {r} - r) d r \doteq \frac {\frac {3}{2} - 0}{6} (F _ {0} + 4 F _ {1} + F _ {2})
|
|
$$
|
|
|
|
where $F_{0}, F_{1}$ , and $F_{2}$ are the function values at $r = 0$ , $r = \frac{3}{4}$ , and $r = \frac{3}{2}$ , respectively; i.e.,
|
|
|
|
$$
|
|
F _ {0} = 1; \quad F _ {1} = 0. 9 3 1 7 9 2; \quad F _ {2} = 1. 3 2 8 4 2 7
|
|
$$
|
|
|
|
Hence we use $\int_0^{3/2}(2^r - r)dr \doteq 1.513899$ (a)
|
|
|
|
Next we need to evaluate
|
|
|
|
$$
|
|
\int_ {3 / 2} ^ {3} (2 ^ {r} - r) d r \doteq \frac {3 - \frac {3}{2}}{6} (F _ {0} + 4 F _ {1} + F _ {2})
|
|
$$
|
|
|
|
where $F_{0}, F_{1}$ , and $F_{2}$ are the function values at $r = \frac{3}{2}$ , $r = \frac{9}{4}$ , and $r = 3$ , respectively,
|
|
|
|
$$
|
|
F _ {0} = 1. 3 2 8 4 2 7; \quad F _ {1} = 2. 5 0 6 8 2 8; \quad F _ {2} = 5
|
|
$$
|
|
|
|
Hence we have $\int_{3/2}^{3}(2^r - r)dr \doteq 4.088935$ (b)
|
|
|
|
Adding the results in (a) and (b), we obtain
|
|
|
|
$$
|
|
\int_ {0} ^ {3} (2 ^ {r} - r) d r \doteq 5. 6 0 2 8 3 4
|
|
$$
|
|
|
|
<!-- source-page: 477 -->
|
|
|
|
The use of a composite formula has a number of advantages over the application of high-order Newton-Cotes formulas. A composite formula, such as the repetitive use of Simpson's rule, is easy to employ. Convergence is ensured as the interval of sampling decreases, and, in practice, a sampling interval could be used that varies from one application of the basic formula to the next. This is particularly advantageous when there are discontinuities in the function to be integrated. For these reasons, in practice, composite formulas are commonly used.
|
|
|
|
EXAMPLE 5.37: Use a composite formula that employs Simpson's rule to evaluate the integral $\int_{-1}^{+13} F(r) \, dr$ of the function $F(r)$ in Fig. E5.37.
|
|
|
|
This function is best integrated by considering three intervals of integration, as follows:
|
|
|
|
$$
|
|
\int_ {- 1} ^ {1 3} F d r = \int_ {- 1} ^ {2} (r ^ {3} + 3) d r + \int_ {2} ^ {9} [ 1 0 + (r - 1) ^ {1 / 3} ] d r + \int_ {9} ^ {1 3} \left[ \frac {1}{1 2 8} (1 3 - r) ^ {5} + 4 \right] d r
|
|
$$
|
|
|
|
|
|
|
|
<details>
|
|
<summary>line</summary>
|
|
| r | F(r) |
|
|
| ---- | -------- |
|
|
| 0 | ~2 |
|
|
| 2 | 11 |
|
|
| 9 | 12 |
|
|
| 13 | 4 |
|
|
</details>
|
|
|
|
Figure E5.37 Function $F(r)$
|
|
|
|
We evaluate each of the three integrals using Simpson's rule and have
|
|
|
|
$$
|
|
\int_ {- 1} ^ {2} \left(r ^ {3} + 3\right) d r = \frac {2 - (- 1)}{6} [ (1) (2) + (4) (3. 1 2 5) + (1) (1 1) ]
|
|
$$
|
|
|
|
or $\int_{-1}^{2}(r^3 +3)dr = 12.75$
|
|
|
|
$$
|
|
\int_ {2} ^ {9} [ 1 0 + (r - 1) ^ {1 / 3} ] d r \doteq \frac {9 - 2}{6} [ (1) (1 1) + (4) (1 1. 6 5 0 9 6 4) + (1) (1 2) ]
|
|
$$
|
|
|
|
or $\int_{2}^{9}\left[10 + (r - 1)^{1 / 3}\right]dr\div 81.204498$
|
|
|
|
$$
|
|
\int_ {9} ^ {1 3} \left[ \frac {1}{1 2 8} (1 3 - r) ^ {5} + 4 \right] d r \doteq \frac {1 3 - 9}{6} [ (1) (1 2) + (4) (4. 2 5) + (1) (4) ]
|
|
$$
|
|
|
|
or $\int_{9}^{13}\left[\frac{1}{128} (13 - r)^{5} + 4\right]dr\doteq 22$
|
|
|
|
<!-- source-page: 478 -->
|
|
|
|
Hence, $\int_{-1}^{13} F \, dr \doteq 12.75 + 81.204498 + 22$
|
|
|
|
or $\int_{-1}^{13} F \, dr \doteq 115.954498$
|
|
|
|
# 5.5.3 The Gauss Formulas (One-Dimensional Integration)
|
|
|
|
The basic integration schemes that we have considered so far use equally spaced sampling points, although the basic methods could be employed to construct procedures that allow the interval of sampling to be varied; i.e., the composite formulas have been introduced. The methods discussed so far are effective when measurements of an unknown function to be integrated have been taken at certain intervals. However, in the integration of finite element matrices, a subroutine is called to evaluate the unknown function F at given points, and these points may be anywhere on the element. No additional difficulties arise if the sampling points are not equally spaced. Therefore, it seems natural to try to improve the accuracy that can be obtained for a given number of function evaluations by also optimizing the positions of the sampling points. A very important numerical integration procedure in which both the positions of the sampling points and the weights have been optimized is Gauss quadrature. The basic assumption in Gauss numerical integration is that
|
|
|
|
$$
|
|
\int_ {a} ^ {b} F (r) d r = \alpha_ {1} F (r _ {1}) + \alpha_ {2} F (r _ {2}) + \dots + \alpha_ {n} F (r _ {n}) + R _ {n} \tag {5.144}
|
|
$$
|
|
|
|
where both the weights $\alpha_{1},\ldots,\alpha_{n}$ and the sampling points $r_{1},\ldots,r_{n}$ are variables. It should be recalled that in the derivation of the Newton-Cotes formulas, only the weights were unknown, and they were determined from the integration of a polynomial $\psi(r)$ that passed through equally spaced sampling points of the function $F(r)$ . We now also calculate the positions of the sampling points and therefore have 2n unknowns to determine a higher-order integration scheme.
|
|
|
|
In analogy with the derivation of the Newton-Cotes formulas, we use an interpolating polynomial $\psi(r)$ of the form given in (5.140),
|
|
|
|
$$
|
|
\psi (r) = \sum_ {j = 1} ^ {n} F _ {j} l _ {j} (r) \tag {5.145}
|
|
$$
|
|
|
|
where n samplings points are now considered, $r_{1}, \ldots, r_{n}$ , which are still unknown. For the determination of the values $r_{1}, \ldots, r_{n}$ , we define a function $P(r)$ ,
|
|
|
|
$$
|
|
P (r) = (r - r _ {1}) (r - r _ {2}) \cdot \cdot \cdot (r - r _ {n}) \tag {5.146}
|
|
$$
|
|
|
|
which is a polynomial of order $n$ . We note that $P(r) = 0$ at the sampling points $r_1, \ldots, r_n$ . Therefore, we can write
|
|
|
|
$$
|
|
F (r) = \psi (r) + P (r) (\beta_ {0} + \beta_ {1} r + \beta_ {2} r ^ {2} + \dots) \tag {5.147}
|
|
$$
|
|
|
|
Integrating $F(r)$ , we obtain
|
|
|
|
$$
|
|
\int_ {a} ^ {b} F (r) d r = \sum_ {j = 1} ^ {n} F _ {j} \left[ \int_ {a} ^ {b} l _ {j} (r) d r \right] + \sum_ {j = 0} ^ {\infty} \beta_ {j} \left[ \int_ {a} ^ {b} r ^ {j} P (r) d r \right] \tag {5.148}
|
|
$$
|
|
|
|
<!-- source-page: 479 -->
|
|
|
|
where it should be noted that in the first integral on the right in (5.148), functions of order $(n - 1)$ and lower are integrated, and in the second integral the functions that are integrated are of order n and higher. The unknown values $r_{j}, j = 1, 2, \ldots, n$ , can now be determined using the conditions
|
|
|
|
$$
|
|
\int_ {a} ^ {b} P (r) r ^ {k} d r = 0 \quad k = 0, 1, 2, \dots , n - 1 \tag {5.149}
|
|
$$
|
|
|
|
Then, since the polynomial $\psi(r)$ passes through $n$ sampling points of $F(r)$ , and $P(r)$ vanishes at these points, the conditions in (5.149) mean that the required integral $\int_{a}^{b} F(r) \, dr$ is approximated by integrating a polynomial of order $(2n - 1)$ instead of $F(r)$ .
|
|
|
|
In summary, using the Newton-Cotes formulas, we use $(n + 1)$ equally spaced sampling points and integrate exactly a polynomial of order at most n. On the other hand, in Gauss quadrature we require n unequally spaced sampling points and integrate exactly a polynomial of order at most $(2n - 1)$ . Polynomials of orders less than n and $(2n - 1)$ , respectively, for the two cases are also integrated exactly.
|
|
|
|
To determine the sampling points and the integration weights, we realize that they depend on the interval a to b. However, to make the calculations general, we consider a natural interval from -1 to +1 and deduce the sampling points and weights for any interval. Namely, if $r_{i}$ is a sampling point and $\alpha_{i}$ is the weight for the interval -1 to +1, the corresponding sampling point and weight in the integration from a to b are
|
|
|
|
$$
|
|
\frac {a + b}{2} + \frac {b - a}{2} r _ {i} \quad \text { and } \quad \frac {b - a}{2} \alpha_ {i}
|
|
$$
|
|
|
|
respectively.
|
|
|
|
Hence, consider an interval from -1 to +1. The sampling points are determined from (5.149) with a = -1 and b = +1. To calculate the integration weights we substitute for $F(r)$ in (5.144) the interpolating polynomial $\psi(r)$ from (5.145) and perform the integration. It should be noted that because the sampling points have been determined, the polynomial $\psi(r)$ is known, and hence
|
|
|
|
$$
|
|
\alpha_ {j} = \int_ {- 1} ^ {+ 1} l _ {j} (r) d r; \quad j = 1, 2, \dots , n \tag {5.150}
|
|
$$
|
|
|
|
TABLE 5.6 Sampling points and weights in Gauss-Legendre numerical integration (interval -1 to +1)
|
|
<table><tr><td>n</td><td colspan="3"> $r_i$ </td><td colspan="3"> $\alpha_i$ </td></tr><tr><td>1</td><td>0.</td><td colspan="2">(15 zeros)</td><td>2.</td><td colspan="2">(15 zeros)</td></tr><tr><td>2</td><td>±0.57735</td><td>02691</td><td>89626</td><td>1.00000</td><td>00000</td><td>00000</td></tr><tr><td>3</td><td>±0.77459</td><td>66692</td><td>41483</td><td>0.55555</td><td>55555</td><td>55556</td></tr><tr><td></td><td>0.00000</td><td>00000</td><td>00000</td><td>0.88888</td><td>88888</td><td>88889</td></tr><tr><td>4</td><td>±0.86113</td><td>63115</td><td>94053</td><td>0.34785</td><td>48451</td><td>37454</td></tr><tr><td></td><td>±0.33998</td><td>10435</td><td>84856</td><td>0.65214</td><td>51548</td><td>62546</td></tr><tr><td>5</td><td>±0.90617</td><td>98459</td><td>38664</td><td>0.23692</td><td>68850</td><td>56189</td></tr><tr><td></td><td>±0.53846</td><td>93101</td><td>05683</td><td>0.47862</td><td>86704</td><td>99366</td></tr><tr><td></td><td>0.00000</td><td>00000</td><td>00000</td><td>0.56888</td><td>88888</td><td>88889</td></tr><tr><td>6</td><td>±0.93246</td><td>95142</td><td>03152</td><td>0.17132</td><td>44923</td><td>79170</td></tr><tr><td></td><td>±0.66120</td><td>93864</td><td>66265</td><td>0.36076</td><td>15730</td><td>48139</td></tr><tr><td></td><td>±0.23861</td><td>91860</td><td>83197</td><td>0.46791</td><td>39345</td><td>72691</td></tr></table>
|
|
|
|
<!-- source-page: 480 -->
|
|
|
|
The sampling points and weights for the interval -1 to +1 have been published by A. N. Lowan, N. Davids, and A. Levenson [A] and are reproduced in Table 5.6 for n = 1 to 6.
|
|
|
|
The coefficients in Table 5.6 can be calculated directly using (5.149) and (5.150) (see Example 5.38). However, for larger n the solution becomes cumbersome, and it is expedient to use Legendre polynomials to solve for the coefficients, which are thus referred to as Gauss-Legendre coefficients.
|
|
|
|
EXAMPLE 5.38: Derive the sampling points and weights for two-point Gauss quadrature. In this case $P(r) = (r - r_{1})(r - r_{2})$ and (5.149) gives the two equations
|
|
|
|
$$
|
|
\begin{array}{l} \int_ {- 1} ^ {+ 1} (r - r _ {1}) (r - r _ {2}) d r = 0 \\ \int_ {- 1} ^ {+ 1} (r - r _ {1}) (r - r _ {2}) r d r = 0 \\ \end{array}
|
|
$$
|
|
|
|
Solving, we obtain
|
|
|
|
$$
|
|
r _ {1} r _ {2} = - \frac {1}{3}
|
|
$$
|
|
|
|
and
|
|
|
|
$$
|
|
r _ {1} + r _ {2} = 0
|
|
$$
|
|
|
|
Hence $r_1 = -\frac{1}{\sqrt{3}}; \quad r_2 = +\frac{1}{\sqrt{3}}$
|
|
|
|
The corresponding weights are obtained using (5.150), which in this case gives
|
|
|
|
$$
|
|
\begin{array}{l} \alpha_ {1} = \int_ {- 1} ^ {+ 1} \frac {r - r _ {2}}{r _ {1} - r _ {2}} d r \\ \alpha_ {2} = \int_ {- 1} ^ {+ 1} \frac {r - r _ {1}}{r _ {2} - r _ {1}} d r \\ \end{array}
|
|
$$
|
|
|
|
Since $r_2 = -r_1$ , we obtain $\alpha_1 = \alpha_2 = 1.0$ .
|
|
|
|
EXAMPLE 5.39: Use two-point Gauss quadrature to evaluate the integral $\int_{0}^{3}(2^{r}-r)dr$ considered in Examples 5.35 and 5.36.
|
|
|
|
Using two-point Gauss quadrature, we obtain from (5.144),
|
|
|
|
$$
|
|
\int_ {0} ^ {3} (2 ^ {r} - r) d r \doteq \alpha_ {1} F (r _ {1}) + \alpha_ {2} F (r _ {2}) \tag {a}
|
|
$$
|
|
|
|
where $\alpha_{1}, \alpha_{2}$ and $r_{1}, r_{2}$ are weights and sampling points, respectively. Since the interval is from 0 to 3, we need to determine the values $\alpha_{1}, \alpha_{2}, r_{1}$ , and $r_{2}$ from the values given in Table 5.6,
|
|
|
|
$$
|
|
\begin{array}{l} \alpha_ {1} = \frac {3}{2} (1); \quad \alpha_ {2} = \frac {3}{2} (1) \\ r _ {1} = \frac {3}{2} \left(1 - \frac {1}{\sqrt {3}}\right); \quad r _ {2} = \frac {3}{2} \left(1 + \frac {1}{\sqrt {3}}\right) \\ \end{array}
|
|
$$
|
|
|
|
where $1 / \sqrt{3} = 0.5773502692$ . Thus,
|
|
|
|
$$
|
|
F (r _ {1}) = 0. 9 1 7 8 5 9 7 8; \quad F (r _ {2}) = 2. 7 8 9 1 6 3 8 9
|
|
$$
|
|
|
|
and (a) gives $\int_0^3 (2^r - r) dr \doteq 5.56053551$
|