김경종
509 lines
25 KiB
Markdown
509 lines
25 KiB
Markdown
<!-- source-page: 491 -->
|
||
|
||
From Eq. (6.1.8), the matrix $\underline { { \boldsymbol { D } } }$ is
|
||
|
||
$$
|
||
\underline {{D}} = \frac {E}{1 - v ^ {2}} \left[ \begin{array}{c c c} 1 & v & 0 \\ v & 1 & 0 \\ 0 & 0 & \frac {1 - v}{2} \end{array} \right] = \left[ \begin{array}{c c c} 3 2 & 8 & 0 \\ 8 & 3 2 & 0 \\ 0 & 0 & 1 2 \end{array} \right] \times 1 0 ^ {6} \mathrm{psi} \tag {10.5.4i}
|
||
$$
|
||
|
||
Finally, using Eq. (10.5.4b), the matrix $\underline { { k } }$ becomes
|
||
|
||
$$
|
||
\underline {{k}} = 1 0 ^ {4} \left[ \begin{array}{c c c c c c c c} 1 4 6 6 & 5 0 0 & - 8 6 6 & - 9 9 & - 7 3 3 & - 5 0 0 & 1 3 3 & 9 9 \\ 5 0 0 & 1 4 6 6 & 9 9 & 1 3 3 & - 5 0 0 & - 7 3 3 & - 9 9 & - 8 6 6 \\ - 8 6 6 & 9 9 & 1 4 6 6 & - 5 0 0 & 1 3 3 & - 9 9 & - 7 3 3 & 5 0 0 \\ - 9 9 & 1 3 3 & - 5 0 0 & 1 4 6 6 & 9 9 & - 8 6 6 & 5 0 0 & - 7 3 3 \\ - 7 3 3 & - 5 0 0 & 1 3 3 & 9 9 & 1 4 6 6 & 5 0 0 & - 8 6 6 & - 9 9 \\ - 5 0 0 & - 7 3 3 & - 9 9 & - 8 6 6 & 5 0 0 & 1 4 6 6 & 9 9 & 1 3 3 \\ 1 3 3 & - 9 9 & - 7 3 3 & 5 0 0 & - 8 6 6 & 9 9 & 1 4 6 6 & - 5 0 0 \\ 9 9 & - 8 6 6 & 5 0 0 & - 7 3 3 & - 9 9 & 1 3 3 & - 5 0 0 & 1 4 6 6 \end{array} \right]
|
||
$$
|
||
|
||
ð10:5:4jÞ
|
||
|
||
|
||
|
||
# Evaluation of Element Stresses
|
||
|
||
The stresses $\underline { { \sigma } } = \underline { { D } } \underline { { B } } \underline { { d } }$ are not constant within the quadrilateral element. Because $\underline { { B } }$ is a function of s and t coordinates, $\underline { { \sigma } }$ is also a function of s and t. In practice, the stresses are evaluated at the same Gauss points used to evaluate the sti¤ness matrix $\underline { { k } } .$ For a quadrilateral using $2 \times 2$ integration, we get four sets of stress data. To reduce the data, it is often practical to evaluate s at $s = 0 , t = 0$ instead. Another method mentioned in Section 7.4 is to evaluate the stresses in all elements at a shared (common) node and then use an average of these element nodal stresses to represent the stress at the node. Most computer programs use this method. Stress plots obtained in these programs are based on this average nodal stress method. Example 10.5 illustrates the use of Gaussian quadrature to evaluate the stress matrix at the $s = 0 , t = 0$ location of the element.
|
||
|
||
# Example 10.5
|
||
|
||
For the rectangular element shown in Figure 10–13, assume plane stress conditions with $E = 3 0 \times 1 0 ^ { 6 }$ psi, $\nu = 0 . 3 ,$ , and displacements $u _ { 1 } = 0 , v _ { 1 } = 0 , u _ { 2 } = 0 . 0 0 1$ in., $v _ { 2 } =$ 0:0015 in., $u _ { 3 } = 0 . 0 0 3$ in., $v _ { 3 } = 0 . 0 0 1 6$ in., $u _ { 4 } = 0$ , and $v _ { 4 } = 0$ . Evaluate the stresses, $\sigma _ { x } , \sigma _ { y }$ , and $\tau _ { x y }$ at $s = 0 , t = 0$ .
|
||
|
||
Using Eqs. (10.3.18)–(10.3.20), we evaluate $\underline { { B } }$ at $s = 0 , t = 0$ .
|
||
|
||
$$
|
||
\underline {{B}} = \frac {1}{| \underline {{J}} |} \left[ \begin{array}{l l l l} \underline {{B}} _ {1} & \underline {{B}} _ {2} & \underline {{B}} _ {3} & \underline {{B}} _ {4} \end{array} \right] \tag {10.3.18}
|
||
$$
|
||
|
||
$$
|
||
\underline {{{B}}} (0, 0) = \frac {1}{| \underline {{{J}}} (0 , 0) |} [ \underline {{{B}}} _ {1} (0, 0) \quad \underline {{{B}}} _ {2} (0, 0) \quad \underline {{{B}}} _ {3} (0, 0) \quad \underline {{{B}}} _ {4} (0, 0) ]
|
||
$$
|
||
|
||
<!-- source-page: 492 -->
|
||
|
||
By Eq. (10.3.22), jJ j is
|
||
|
||
$$
|
||
\begin{array}{l} | \underline {{J}} (0, 0) | = \frac {1}{8} [ 3 \quad 5 \quad 5 \quad 3 ] \left[ \begin{array}{c c c c} 0 & 1 & 0 & - 1 \\ - 1 & 0 & 1 & 0 \\ 0 & - 1 & 0 & 1 \\ 1 & 0 & - 1 & 0 \end{array} \right] \left\{ \begin{array}{l} 2 \\ 2 \\ 4 \\ 4 \end{array} \right\} \\ = \frac {1}{8} [ - 2 \quad - 2 \quad 2 \quad 2 ] \left\{ \begin{array}{l} 2 \\ 2 \\ 4 \\ 4 \end{array} \right\} \\ | \underline {{J}} (0, 0) | = 1 \tag {10.5.5a} \\ \end{array}
|
||
$$
|
||
|
||
By Eq. (10.3.19), we have
|
||
|
||
$$
|
||
\underline {{B}} _ {i} = \left[ \begin{array}{c c} a N _ {i, s} - b N _ {i, t} & 0 \\ 0 & c N _ {i, t} - d N _ {i, s} \\ c N _ {i, t} - d N _ {i, s} & a N _ {i, s} - b N _ {i, t} \end{array} \right] \tag {10.5.5b}
|
||
$$
|
||
|
||
By Eq. (10.3.20), we obtain
|
||
|
||
$$
|
||
a = 1 \quad b = 0 \quad c = 1 \quad d = 0
|
||
$$
|
||
|
||
Di¤erentiating the shape functions in Eq. (10.3.5) with respect to s and t and then evaluating at s ¼ 0, t ¼ 0, we obtain
|
||
|
||
$$
|
||
\begin{array}{l} N _ {1, s} = - \frac {1}{4} \quad N _ {1, t} = - \frac {1}{4} \quad N _ {2, s} = \frac {1}{4} \quad N _ {2, t} = - \frac {1}{4} \tag {10.5.5c} \\ N _ {3, s} = \frac {1}{4} \qquad N _ {3, t} = \frac {1}{4} \qquad N _ {4, s} = - \frac {1}{4} \qquad N _ {4, t} = \frac {1}{4} \\ \end{array}
|
||
$$
|
||
|
||
Therefore, substituting Eqs. (10.5.5c) into Eq. (10.5.5b), we obtain
|
||
|
||
$$
|
||
\underline {{{B}}} _ {1} = \left[ \begin{array}{c c} - \frac {1}{4} & 0 \\ 0 & - \frac {1}{4} \\ - \frac {1}{4} & - \frac {1}{4} \end{array} \right] \quad \underline {{{B}}} _ {2} = \left[ \begin{array}{c c} \frac {1}{4} & 0 \\ 0 & - \frac {1}{4} \\ - \frac {1}{4} & \frac {1}{4} \end{array} \right] \quad \underline {{{B}}} _ {3} = \left[ \begin{array}{c c} \frac {1}{4} & 0 \\ 0 & \frac {1}{4} \\ \frac {1}{4} & \frac {1}{4} \end{array} \right] \quad \underline {{{B}}} _ {4} = \left[ \begin{array}{c c} - \frac {1}{4} & 0 \\ 0 & \frac {1}{4} \\ \frac {1}{4} & - \frac {1}{4} \end{array} \right] \tag {10.5.5d}
|
||
$$
|
||
|
||
The element stress matrix $\underline { { \sigma } }$ is then obtained by substituting Eqs. (10.5.5a) for $| \underline { { J } } | = 1$ and (10.5.5d) into Eq. (10.3.18) for $\underline { { B } }$ and the plane stress $\underline { { \boldsymbol { D } } }$ matrix from Eq. (6.1.8) into the definition for $\underline { { \sigma } }$ as
|
||
|
||
$$
|
||
\underline {{\sigma}} = \underline {{D}} \underline {{B}} \underline {{d}} = (3 0) \frac {1 0 ^ {6} \left[ \begin{array}{c c c} 1 & 0 . 3 & 0 \\ 0 . 3 & 1 & 0 \\ 0 & 0 & 0 . 3 5 \end{array} \right]}{1 - 0 . 0 9} \tag {continued}
|
||
$$
|
||
|
||
<!-- source-page: 493 -->
|
||
|
||
$$
|
||
\times \left[ \begin{array}{c c c c c c c c} - 0. 2 5 & 0 & 0. 2 5 & 0 & 0. 2 5 & 0 & - 0. 2 5 & 0 \\ 0 & - 0. 2 5 & 0 & - 0. 2 5 & 0 & 0. 2 5 & 0 & 0. 2 5 \\ - 0. 2 5 & - 0. 2 5 & - 0. 2 5 & 0. 2 5 & 0. 2 5 & 0. 2 5 & 0. 2 5 & - 0. 2 5 \end{array} \right] \left\{ \begin{array}{c} 0 \\ 0 \\ 0. 0 0 1 \\ 0. 0 0 1 5 \\ 0. 0 0 3 \\ 0. 0 0 1 6 \\ 0 \\ 0 \end{array} \right\}
|
||
$$
|
||
|
||
$$
|
||
\underline {{\sigma}} = \left\{ \begin{array}{l} 3. 3 2 1 \cdot 1 0 ^ {4} \\ 1. 0 7 1 \cdot 1 0 ^ {4} \\ 1. 4 7 1 \cdot 1 0 ^ {4} \end{array} \right\} \text { psi }
|
||
$$
|
||
|
||
# 10.6 Higher-Order Shape Functions
|
||
|
||
In general, higher-order element shape functions can be developed by adding additional nodes to the sides of the linear element. These elements result in higher-order strain variations within each element, and convergence to the exact solution thus occurs at a faster rate using fewer elements. (However, a trade-o¤ exists because a more complicated element takes up so much computation time that even with few elements in the model, the computation time can become larger than for the simple linear element model.) Another advantage of the use of higher-order elements is that curved boundaries of irregularly shaped bodies can be approximated more closely than by the use of simple straight-sided linear elements.
|
||
|
||
To illustrate the concept of higher order elements, we will begin with the threenoded linear strain quadratic displacement (and quadratic shape functions) shown in Figure 10–14. Figure 10–14 shows a quadratic isoparametric bar element (also called a linear strain bar) with three coordinates of nodes, $x _ { 1 } , x _ { 2 } .$ , and $x _ { 3 } .$ , in the global coordinates.
|
||
|
||
# Example 10.6
|
||
|
||
For the three-noded linear strain bar isoparametric element shown in Figure 10–14, determine (a) the shape functions, $N _ { 1 } , N _ { 2 }$ , and $N _ { 3 }$ , and (b) the strain/displacement matrix [B]. Assume the general axial displacement function to be a quadratic taken as $u = a _ { 1 } + a _ { 2 } s + a _ { 3 } s ^ { 2 }$ .
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
1
|
||
x₁
|
||
L/2
|
||
3
|
||
L/2
|
||
x₃
|
||
2
|
||
x₂
|
||
s
|
||
</details>
|
||
|
||
Figure 10–14 Three-noded linear strain bar element
|
||
|
||
<!-- source-page: 494 -->
|
||
|
||
(a) As we are formulating shape functions for an isoparametric element, we assume the following axial coordinate function for x as
|
||
|
||
$$
|
||
x = a _ {1} + a _ {2} s + a _ {3} s ^ {2} \tag {10.6.1}
|
||
$$
|
||
|
||
Evaluating the $\boldsymbol { a } _ { i } { } ^ { \prime } \boldsymbol { \mathbf { s } }$ in terms of the nodal coordinates, we obtain
|
||
|
||
$$
|
||
x (- 1) = a _ {1} - a _ {2} + a _ {3} = x _ {1} \quad \text { or } \quad x _ {1} = a _ {1} - a _ {2} + a _ {3}
|
||
$$
|
||
|
||
$$
|
||
x (0) = a _ {1} = x _ {3} \quad \text { or } \quad x _ {3} = a _ {1}
|
||
$$
|
||
|
||
$$
|
||
x (1) = a _ {1} + a _ {2} + a _ {3} = x _ {2} \quad \text { or } \quad x _ {2} = a _ {1} - a _ {2} + a _ {3} \tag {10.6.2}
|
||
$$
|
||
|
||
Substituting $a _ { 1 } = x _ { 3 }$ from the second of Eqs. (10.6.2), into the first and third of Eqs. (10.6.2), we obtain $a _ { 2 }$ and $a _ { 3 }$ as follows:
|
||
|
||
$$
|
||
x _ {1} = x _ {3} - a _ {2} + a _ {3} \tag {10.6.3}
|
||
$$
|
||
|
||
$$
|
||
x _ {2} = x _ {3} + a _ {2} + a _ {3}
|
||
$$
|
||
|
||
Adding Eqs. (10.6.3) together and solving for $a _ { 3 }$ gives the following:
|
||
|
||
$$
|
||
a _ {3} = (x _ {1} + x _ {2} - 2 x _ {3}) / 2 \tag {10.6.4}
|
||
$$
|
||
|
||
$$
|
||
x _ {1} = x _ {3} - a _ {2} + ((x _ {1} + x _ {2} - 2 x _ {3}) / 2) \tag {10.6.5}
|
||
$$
|
||
|
||
$$
|
||
a _ {2} = x _ {3} - x _ {1} + \left(\left(x _ {1} + x _ {2} - 2 x _ {3}\right) / 2\right) = \left(x _ {2} - x _ {1}\right) / 2
|
||
$$
|
||
|
||
Substituting the values for $a _ { 1 } , a _ { 2 }$ and $a _ { 3 }$ into the general equation for x given by Eq. (10.6.1), we obtain
|
||
|
||
$$
|
||
x = a _ {1} + a _ {2} s + a _ {3} s ^ {2} = x _ {3} + \frac {x _ {2} - x _ {1}}{2} s + \frac {x _ {1} + x _ {2} - 2 x _ {3}}{2} s ^ {2} \tag {10.6.6}
|
||
$$
|
||
|
||
Combining like terms in $x _ { 1 } , x _ { 2 } .$ , and $x _ { 3 }$ , from Eq. (10.6.6), we obtain the final form of x as:
|
||
|
||
$$
|
||
x = \left(\frac {s (s - 1)}{2}\right) x _ {1} + \frac {s (s + 1)}{2} x _ {2} + (1 - s ^ {2}) x _ {3} \tag {10.6.7}
|
||
$$
|
||
|
||
Recall that the function x can be expressed in terms of the shape function matrix and the axial coordinates, we have from Eq. (10.6.7)
|
||
|
||
$$
|
||
\{x \} = \left[ \begin{array}{l l l} N _ {1} & N _ {2} & N _ {3} \end{array} \right] \left\{ \begin{array}{l} x _ {1} \\ x _ {2} \\ x _ {3} \end{array} \right\} = \left[ \left(\frac {s (s - 1)}{2}\right) \quad \frac {s (s + 1)}{2} \quad \left(1 - s ^ {2}\right) \right] \left\{ \begin{array}{l} x _ {1} \\ x _ {2} \\ x _ {3} \end{array} \right\} \tag {10.6.8}
|
||
$$
|
||
|
||
Therefore the shape functions are
|
||
|
||
$$
|
||
N _ {1} = \frac {s (s - 1)}{2} \quad N _ {2} = \frac {s (s + 1)}{2} \quad N _ {3} = (1 - s ^ {2}) \tag {10.6.9}
|
||
$$
|
||
|
||
(b) We now determine the strain/displacement matrix [B] as follows:
|
||
|
||
From our basic definition of axial strain we have
|
||
|
||
$$
|
||
\left\{\varepsilon_ {x} \right\} = \frac {d u}{d x} = \frac {d u}{d s} \frac {d s}{d x} = [ B ] \left\{ \begin{array}{l} u _ {1} \\ u _ {2} \\ u _ {3} \end{array} \right\} \tag {10.6.10}
|
||
$$
|
||
|
||
<!-- source-page: 495 -->
|
||
|
||
Using an isoparametric formulation means the displacement function is of the same form as the axial coordinate function. Therefore, using Eq. (10.6.6), we have
|
||
|
||
$$
|
||
u = u _ {3} + \frac {u _ {2}}{2} s - \frac {u _ {1}}{2} s + \frac {u _ {1}}{2} s ^ {2} + \frac {u _ {2}}{2} s ^ {2} - \frac {2 u _ {3}}{2} s ^ {2} \tag {10.6.11}
|
||
$$
|
||
|
||
Differentiating u with respect to s, we obtain
|
||
|
||
$$
|
||
\frac {d u}{d s} = \frac {u _ {2}}{2} - \frac {u _ {1}}{2} + u _ {1} s + u _ {2} s - 2 u _ {3} s = \left(s - \frac {1}{2}\right) u _ {1} + \left(s + \frac {1}{2}\right) u _ {2} + (- 2 s) u _ {3} \tag {10.6.12}
|
||
$$
|
||
|
||
We have previously proven that $dx / ds = L / 2 = [J]$ (see Eq. (10.1.9b)). This relationship holds for the higher-order one-dimensional elements as well as for the two-noded constant strain bar element. Using this relationship and Eq. (10.6.12) in Eq. (10.6.10), we obtain
|
||
|
||
$$
|
||
\frac {d u}{d x} = \frac {d u}{d s} \frac {d s}{d x} = \left(\frac {2}{L}\right) \left(\left(s - \frac {1}{2}\right) u _ {1} + \left(s + \frac {1}{2}\right) u _ {2} + (- 2 s) u _ {3}\right) \tag {10.6.13}
|
||
$$
|
||
|
||
$$
|
||
= \left(\frac {2 s - 1}{L}\right) u _ {1} + \left(\frac {2 s + 1}{L}\right) u _ {2} + \left(\frac {- 4 s}{L}\right) u _ {3}
|
||
$$
|
||
|
||
In matrix form, Eq. (10.6.13) becomes
|
||
|
||
$$
|
||
\frac {d u}{d x} = \left[ \begin{array}{c c c} \frac {2 s - 1}{L} & \frac {2 s + 1}{L} & \frac {- 4 s}{L} \end{array} \right] \left\{ \begin{array}{l} u _ {1} \\ u _ {2} \\ u _ {3} \end{array} \right\} \tag {10.6.14}
|
||
$$
|
||
|
||
As Eq. (10.6.14) represents the axial strain, we have
|
||
|
||
$$
|
||
\left\{\varepsilon_ {x} \right\} = \frac {d u}{d x} = \left[ \frac {2 s - 1}{L} \quad \frac {2 s + 1}{L} \quad \frac {- 4 s}{L} \right] \left\{ \begin{array}{l} u _ {1} \\ u _ {2} \\ u _ {3} \end{array} \right\} = [ B ] \left\{ \begin{array}{l} u _ {1} \\ u _ {2} \\ u _ {3} \end{array} \right\} \tag {10.6.15}
|
||
$$
|
||
|
||
Therefore the $[B]$ is given by
|
||
|
||
$$
|
||
[ B ] = \left[ \frac {2 s - 1}{L} \quad \frac {2 s + 1}{L} \quad \frac {- 4 s}{L} \right] \tag {10.6.16}
|
||
$$
|
||
|
||
# Example 10.7
|
||
|
||
For the three-noded bar element shown previously in Figure 10–14, evaluate the stiffness matrix analytically. Use the $[B]$ from Example 10.6.
|
||
|
||
From example 10.6, Eq. (10.6.16), we have
|
||
|
||
$$
|
||
[ B ] = \left[ \frac {2 s - 1}{L} \quad \frac {2 s + 1}{L} \quad \frac {- 4 s}{L} \right], \quad [ J ] = \frac {L}{2} \quad \text {(see Eq. (10.1.9b))} \tag {10.6.17}
|
||
$$
|
||
|
||
<!-- source-page: 496 -->
|
||
|
||
Substituting the expression for $[B]$ into Eq. (10.1.15) for the stiffness matrix, we obtain
|
||
|
||
$$
|
||
[ k ] = \frac {L}{2} \int_ {- 1} ^ {1} [ B ] ^ {T} E [ B ] A d s = \frac {A E L}{2} \int_ {- 1} ^ {1} \left[ \begin{array}{c c c} \frac {(2 s - 1) ^ {2}}{L ^ {2}} & \frac {(2 s - 1) (2 s + 1)}{L ^ {2}} & \frac {(2 s - 1) (- 4 s)}{L ^ {2}} \\ \frac {(2 s + 1) (2 s - 1)}{L ^ {2}} & \frac {(2 s + 1) ^ {2}}{L ^ {2}} & \frac {(2 s + 1) (- 4 s)}{L ^ {2}} \\ \frac {(- 4 s) (2 s - 1)}{L ^ {2}} & \frac {(- 4 s) (2 s + 1)}{L ^ {2}} & \frac {(- 4 s) ^ {2}}{L ^ {2}} \end{array} \right] d s \tag {10.6.18}
|
||
$$
|
||
|
||
Simplifying the terms in Eq. (10.6.18) for easier integration, we have
|
||
|
||
$$
|
||
[ k ] = \frac {A E}{2 L} \int_ {- 1} ^ {1} \left[ \begin{array}{c c c} 4 s ^ {2} - 4 s + 1 & 4 s ^ {2} - 1 & - 8 s ^ {2} + 4 s \\ 4 s ^ {2} - 1 & 4 s ^ {2} + 4 s + 1 & - 8 s ^ {2} - 4 s \\ - 8 s ^ {2} + 4 s & - 8 s ^ {2} - 4 s & 1 6 s ^ {2} \end{array} \right] d s \tag {10.6.19}
|
||
$$
|
||
|
||
Upon explicit integration of Eq. (10.6.19), we obtain
|
||
|
||
$$
|
||
[ k ] = \frac {A E}{2 L} \left[ \begin{array}{c c c} \frac {4}{3} s ^ {3} - 2 s ^ {2} + s & \frac {4}{3} s ^ {3} - s & - \frac {8}{3} s ^ {3} + 2 s ^ {2} \\ \frac {4}{3} s ^ {3} - s & \frac {4}{3} s ^ {3} + 2 s ^ {2} + s & - \frac {8}{3} s ^ {3} - 2 s ^ {2} \\ - \frac {8}{3} s ^ {3} + 2 s ^ {2} & - \frac {8}{3} s ^ {3} - 2 s ^ {2} & \frac {1 6}{3} s ^ {3} \end{array} \right] \Bigg | _ {- 1} ^ {1} \tag {10.6.20}
|
||
$$
|
||
|
||
Evaluating Eq. (10.6.20) at the limits 1 and -1, we have
|
||
|
||
$$
|
||
[ k ] = \frac {A E}{2 L} \left[ \begin{array}{c c c} \frac {4}{3} - 2 + 1 & \frac {4}{3} - 1 & - \frac {8}{3} + 2 \\ \frac {4}{3} - 1 & \frac {4}{3} + 2 + 1 & - \frac {8}{3} - 2 \\ - \frac {8}{3} + 2 & - \frac {8}{3} - 2 & \frac {1 6}{3} \end{array} \right] - \left[ \begin{array}{c c c} - \frac {4}{3} - 2 - 1 & - \frac {4}{3} + 1 & \frac {8}{3} + 2 \\ - \frac {4}{3} + 1 & - \frac {4}{3} + 2 - 1 & \frac {8}{3} - 2 \\ \frac {8}{3} + 2 & \frac {8}{3} - 2 & - \frac {1 6}{3} \end{array} \right] \tag {10.6.21}
|
||
$$
|
||
|
||
Simplifying Eq. (10.6.21), we obtain the final stiffness matrix as
|
||
|
||
$$
|
||
[ k ] = \frac {A E}{2 L} \left[ \begin{array}{c c c} 4. 6 7 & 0. 6 6 7 & - 5. 3 3 \\ 0. 6 6 7 & 4. 6 7 & - 5. 3 3 \\ - 5. 3 3 & - 5. 3 3 & 1 0. 6 7 \end{array} \right] \tag {10.6.22}
|
||
$$
|
||
|
||
# Example 10.8
|
||
|
||
We now illustrate how to evaluate the stiffness matrix for the three-noded bar element shown in Figure 10–15 by using two-point Gaussian quadrature. We can then compare this result to that obtained by the explicit integration performed in Example 10.7.
|
||
|
||
<!-- source-page: 497 -->
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
0 | → s
|
||
1 L/2 3 L/2 2
|
||
x₁ * s₁ x₃ * s₂ x₂
|
||
</details>
|
||
|
||
Figure 10–15 Three–noded bar with two Gauss points
|
||
|
||
Starting with Eq. (10.6.18), we have for the sti¤ness matrix
|
||
|
||
$$
|
||
[ k ] = \frac {L}{2} \int_ {- 1} ^ {1} [ B ] ^ {T} E [ B ] A d s = \frac {A E L}{2} \int_ {- 1} ^ {1} \left[ \begin{array}{c c c} \frac {(2 s - 1) ^ {2}}{L ^ {2}} & \frac {(2 s - 1) (2 s + 1)}{L ^ {2}} & \frac {(2 s - 1) (- 4 s)}{L ^ {2}} \\ \frac {(2 s + 1) (2 s - 1)}{L ^ {2}} & \frac {(2 s + 1) ^ {2}}{L ^ {2}} & \frac {(2 s + 1) (- 4 s)}{L ^ {2}} \\ \frac {(- 4 s) (2 s - 1)}{L ^ {2}} & \frac {(- 4 s) (2 s + 1)}{L ^ {2}} & \frac {(- 4 s) ^ {2}}{L ^ {2}} \end{array} \right] d s \tag {10.6.23}
|
||
$$
|
||
|
||
Using two-point Guassian quadrature, we evaluate the sti¤ness matrix at the two points shown in Figure 10–15 given by
|
||
|
||
$$
|
||
s _ {1} = - 0. 5 7 7 3 5, \quad s _ {2} = 0. 5 7 7 3 5 \tag {10.6.24}
|
||
$$
|
||
|
||
with weights given by
|
||
|
||
$$
|
||
W _ {1} = 1, \quad W _ {2} = 1 \tag {10.6.25}
|
||
$$
|
||
|
||
We then evaluate each term in the integrand of Eq. (10.6.23) at each Gauss point and multiply each term by its weight (here each weight is 1). We then add those Gauss point evaluations together to obtain the final term for each element of the sti¤ness matrix. For two-point evaluation, there will be two terms added together to obtain each element of the sti¤ness matrix. We proceed to evaluate the sti¤ness matrix term by term as follows:
|
||
|
||
The one–one element:
|
||
|
||
$$
|
||
\sum_ {i = 1} ^ {2} W _ {i} (2 s _ {i} - 1) ^ {2} = (1) [ 2 (- 0. 5 7 7 3 5) - 1 ] ^ {2} + (1) [ 2 (0. 5 7 7 3 5) - 1 ] ^ {2} = 4. 6 6 6 7
|
||
$$
|
||
|
||
The one–two element:
|
||
|
||
$$
|
||
\begin{array}{l} \sum_ {i = 1} ^ {2} W _ {i} (2 s _ {i} - 1) (2 s _ {i} + 1) = (1) [ (2) (- 0. 5 7 7 3 5) - 1 ] [ (2) (- 0. 5 7 7 3 5) + 1 ] \\ + (1) [ (2) (0. 5 7 7 3 5) - 1 ] [ (2) (0. 5 7 7 3 5) + 1 ] = 0. 6 6 6 7 \\ \end{array}
|
||
$$
|
||
|
||
The one–three element:
|
||
|
||
$$
|
||
\begin{array}{l} \sum_ {i = 1} ^ {2} W _ {i} \left(- 4 s _ {i} \left(2 s _ {i} - 1\right)\right) = (1) (- 4) (- 0. 5 7 7 3 5) [ (2) (- 0. 5 7 7 3 5) - 1 ] \\ + (1) (- 4) (0. 5 7 7 3 5) [ (2) (0. 5 7 7 3 5) - 1 ] = - 5. 3 3 3 3 \\ \end{array}
|
||
$$
|
||
|
||
<!-- source-page: 498 -->
|
||
|
||
The two–two element:
|
||
|
||
$$
|
||
\sum_ {i = 1} ^ {2} W _ {i} (2 s _ {i} + 1) ^ {2} = (1) [ (2) (- 0. 5 7 7 3 5) + 1 ] ^ {2} + (1) [ (2) (0. 5 7 7 3 5) + 1 ] ^ {2} = 4. 6 6 6 7
|
||
$$
|
||
|
||
The two–three element:
|
||
|
||
$$
|
||
\begin{array}{l} \sum_ {i = 1} ^ {2} W _ {i} \left[ - 4 s _ {i} \left(2 s _ {i} + 1\right) \right] = (1) (- 4) (- 0. 5 7 7 3 5) [ (2) (- 0. 5 7 7 3 5) + 1 ] \\ + (1) (- 4) (0. 5 7 7 3 5) [ (2) (0. 5 7 7 3 5) + 1 ] = - 5. 3 3 3 3 \\ \end{array}
|
||
$$
|
||
|
||
The three–three element:
|
||
|
||
$$
|
||
\sum_ {i = 1} ^ {2} W _ {i} \left(1 6 s _ {i} ^ {2}\right) = (1) (1 6) (- 0. 5 7 7 3 5) ^ {2} + (1) (1 6) (0. 5 7 7 3 5) ^ {2} = 1 0. 6 6 6 7
|
||
$$
|
||
|
||
By symmetry, the two–one element equals the one–two element, etc. Therefore, from the evaluations of the terms above, the final sti¤ness matrix is
|
||
|
||
$$
|
||
\underline {{k}} = \frac {A E}{2 L} \left[ \begin{array}{c c c} 4. 6 7 & 0. 6 6 7 & - 5. 3 3 \\ 0. 6 6 7 & 4. 6 7 & - 5. 3 3 \\ - 5. 3 3 & - 5. 3 3 & 1 0. 6 7 \end{array} \right] \tag {10.6.26}
|
||
$$
|
||
|
||
Equation (10.6.26) is identical to Eq. (10.6.22) obtained analytically by direct explicit integration of each term in the sti¤ness matrix.
|
||
|
||
To further illustrate the concept of higher-order elements, we will consider the quadratic and cubic element shape functions as described in Reference [3]. Figure 10–16 shows a quadratic isoparametric element with four corner nodes and four additional midside nodes. This eight-noded element is often called a ‘‘Q8’’ element.
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
Edge s = -1
|
||
(-1, 1)
|
||
(-1, 0)
|
||
8
|
||
(-1, -1)
|
||
4
|
||
(0, 1)
|
||
7
|
||
Edge t = +1
|
||
3
|
||
(1, 1)
|
||
6
|
||
(1, 0)
|
||
s
|
||
Edge s = +1
|
||
2
|
||
(1, -1)
|
||
5
|
||
Edge t = -1
|
||
x
|
||
y
|
||
</details>
|
||
|
||
Figure 10–16 Quadratic isoparametric element
|
||
|
||
<!-- source-page: 499 -->
|
||
|
||
The shape functions of the quadratic element are based on the incomplete cubic polynomial such that coordinates x and y are
|
||
|
||
$$
|
||
x = a _ {1} + a _ {2} s + a _ {3} t + a _ {4} s t + a _ {5} s ^ {2} + a _ {6} t ^ {2} + a _ {7} s ^ {2} t + a _ {8} s t ^ {2} \tag {10.6.27}
|
||
$$
|
||
|
||
$$
|
||
y = a _ {9} + a _ {1 0} s + a _ {1 1} t + a _ {1 2} s t + a _ {1 3} s ^ {2} + a _ {1 4} t ^ {2} + a _ {1 5} s ^ {2} t + a _ {1 6} s t ^ {2}
|
||
$$
|
||
|
||
These functions have been chosen so that the number of generalized degrees of freedom (2 per node times 8 nodes equals 16) are identical to the total number of $\boldsymbol { a } _ { i } { } ^ { \prime } \boldsymbol { \mathbf { s } }$ . The literature also refers to this eight-noded element as a ‘‘serendipity’’ element as it is based on an incomplete cubic, but it yields good results in such cases as beam bending. We are also reminded that because we are considering an isoparametric formulation, displacements u and v are of identical form as x and $y ,$ respectively, in Eq. (10.6.27).
|
||
|
||
To describe the shape functions, two forms are required—one for corner nodes and one for midside nodes, as given in Reference [3]. For the corner nodes ði ¼ 1; 2; 3; 4Þ,
|
||
|
||
$$
|
||
N _ {1} = \frac {1}{4} (1 - s) (1 - t) (- s - t - 1)
|
||
$$
|
||
|
||
$$
|
||
N _ {2} = \frac {1}{4} (1 + s) (1 - t) (s - t - 1) \tag {10.6.28}
|
||
$$
|
||
|
||
$$
|
||
N _ {3} = \frac {1}{4} (1 + s) (1 + t) (s + t - 1)
|
||
$$
|
||
|
||
$$
|
||
N _ {4} = \frac {1}{4} (1 - s) (1 + t) (- s + t - 1)
|
||
$$
|
||
|
||
or, in compact index notation, we express Eqs. (10.6.28) as
|
||
|
||
$$
|
||
N _ {i} = \frac {1}{4} (1 + s s _ {i}) (1 + t t _ {i}) (s s _ {i} + t t _ {i} - 1) \tag {10.6.29}
|
||
$$
|
||
|
||
where i is the number of the shape function and
|
||
|
||
$$
|
||
s _ {i} = - 1, 1, 1, - 1 \quad (i = 1, 2, 3, 4) \tag {10.6.30}
|
||
$$
|
||
|
||
$$
|
||
t _ {i} = - 1, - 1, 1, 1 \quad (i = 1, 2, 3, 4)
|
||
$$
|
||
|
||
For the midside nodes ði ¼ 5; 6; 7; 8Þ,
|
||
|
||
$$
|
||
N _ {5} = \frac {1}{2} (1 - t) (1 + s) (1 - s)
|
||
$$
|
||
|
||
$$
|
||
N _ {6} = \frac {1}{2} (1 + s) (1 + t) (1 - t) \tag {10.6.31}
|
||
$$
|
||
|
||
$$
|
||
N _ {7} = \frac {1}{2} (1 + t) (1 + s) (1 - s)
|
||
$$
|
||
|
||
$$
|
||
N _ {8} = \frac {1}{2} (1 - s) (1 + t) (1 - t)
|
||
$$
|
||
|
||
or, in index notation,
|
||
|
||
$$
|
||
N _ {i} = \frac {1}{2} (1 - s ^ {2}) (1 + t t _ {i}) \quad t _ {i} = - 1, 1 \quad (i = 5, 7) \tag {10.6.32}
|
||
$$
|
||
|
||
$$
|
||
N _ {i} = \frac {1}{2} (1 + s s _ {i}) (1 - t ^ {2}) \qquad s _ {i} = 1, - 1 \qquad (i = 6, 8)
|
||
$$
|
||
|
||
<!-- source-page: 500 -->
|
||
|
||
We can observe from Eqs. (10.6.28) and (10.6.31) that an edge (and displacement) can vary with $s ^ { 2 }$ (along t constant) or with $t ^ { \dot { 2 } }$ (along s constant). Furthermore, $N _ { i } = 1$ a t node i and $N _ { i } = 0$ at the other nodes, as it must be according to our usual definition of shape functions.
|
||
|
||
The displacement functions are given by
|
||
|
||
$$
|
||
\begin{array}{l} \left\{ \begin{array}{c} u \\ v \end{array} \right\} = \left[ \begin{array}{c c c c c c c c c c c c c c c c} N _ {1} & 0 & N _ {2} & 0 & N _ {3} & 0 & N _ {4} & 0 & N _ {5} & 0 & N _ {6} & 0 & N _ {7} & 0 & N _ {8} & 0 \\ 0 & N _ {1} & 0 & N _ {2} & 0 & N _ {3} & 0 & N _ {4} & 0 & N _ {5} & 0 & N _ {6} & 0 & N _ {7} & 0 & N _ {8} \end{array} \right] \\ \times \left\{ \begin{array}{c} u _ {1} \\ v _ {1} \\ u _ {2} \\ v _ {2} \\ \vdots \\ v _ {8} \end{array} \right\} \tag {10.6.33} \\ \end{array}
|
||
$$
|
||
|
||
and the strain matrix is now
|
||
|
||
$$
|
||
\underline {{\varepsilon}} = \underline {{D}} ^ {\prime} \underline {{N}} \underline {{d}}
|
||
$$
|
||
|
||
with $\underline { { B } } = \underline { { D } } ^ { \prime } \underline { { N } }$
|
||
|
||
We can develop the matrix $\underline { { B } }$ using Eq. (10.3.17) with $\underline { { \boldsymbol { D } } } ^ { \prime }$ from Eq. (10.3.16) and with $\underline { { N } }$ now the $2 \times 1 6$ matrix given in Eq. (10.6.33), where the $N { \mathrm { : } } $ are defined in explicit form by Eq. (10.6.28) and (10.6.31).
|
||
|
||
To evaluate the matrix $\underline { { B } }$ and the matrix $\underline { { k } }$ for the eight-noded quadratic isoparametric element, we now use the nine-point Gauss rule (often described as a $3 \times 3$ rule). Results using $2 \times 2$ and $3 \times 3$ rules have shown significant di¤erences, and the $3 \times 3$ rule is recommended by Bathe and Wilson [7]. Table 10–1 indicates the locations of points and the associated weights. The $3 \times 3$ rule is shown in Figure 10–17.
|
||
|
||
By adding a ninth node at $s = 0 , t = 0$ in Figure 10–16, we can create an element called $\mathrm { ~ a ~ } ^ { \bullet \bullet } \mathrm { Q } ^ { 9 . \bullet }$ This is an internal node that is not connected to any other nodes. We then add the $a _ { 1 7 } s ^ { 2 } t ^ { 2 }$ and $a _ { 1 8 } s ^ { 2 } t ^ { 2 }$ terms to x and $y ,$ respectively in Eq. (10.6.27) and to u and $y .$ . The element is then called a Lagrange element as the shape functions can be derived using Lagrange interpolation formulas. For more on this subject consult [8].
|
||
|
||
|
||
|
||
<details>
|
||
<summary>scatter</summary>
|
||
| t | s |
|
||
|----|------|
|
||
| 1 | -0.7745 |
|
||
| 2 | -0.7745 |
|
||
| 3 | -0.7745 |
|
||
| 4 | -0.7745 |
|
||
| 5 | -0.7745 |
|
||
| 6 | -0.7745 |
|
||
| 7 | -0.7745 |
|
||
| 8 | 0.7745 |
|
||
| 9 | 0.7745 |
|
||
</details>
|
||
|
||
Figure 10–17 3 3 rule in two dimensions
|