김경종
517 lines
23 KiB
Markdown
517 lines
23 KiB
Markdown
<!-- source-page: 371 -->
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where $t_{ij}$ is the thickness of the element at the sampling point $(r_{i}, s_{j})$ ( $t_{ij} = 1.0$ in plane strain analysis). With the matrices $F_{ij}$ as given and the weighting factors $\alpha_{ij}$ available, the required stiffness matrix can readily be evaluated.
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For the actual implementation it should be noted that in the evaluation of $J_{ij}$ and of the matrices defining the displacement derivatives in (a) and (b), only the eight possible derivatives of the interpolation functions $h_{1}, \ldots, h_{4}$ are required. Therefore, it is expedient to calculate these derivatives corresponding to the point $(r_{i}, s_{j})$ once at the start of the evaluation of $B_{ij}$ and use them whenever they are required.
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It should also be realized that considering the specific point $(r_i, s_j)$ , the relations in (a) and (b) may be written, respectively, as
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$$
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\left. \begin{array}{l} \frac {\partial u}{\partial x} = \sum_ {i = 1} ^ {4} \frac {\partial h _ {i}}{\partial x} u _ {i} \\ \frac {\partial u}{\partial y} = \sum_ {i = 1} ^ {4} \frac {\partial h _ {i}}{\partial y} u _ {i} \end{array} \right\} \tag {c}
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$$
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and $\left. \begin{array}{l} \frac{\partial v}{\partial x} = \sum_{i=1}^{4} \frac{\partial h_i}{\partial x} v_i \\ \frac{\partial v}{\partial y} = \sum_{i=1}^{4} \frac{\partial h_i}{\partial y} v_i \end{array} \right\} \tag{d}$
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Hence, we have
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$$
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\mathbf {B} = \left[ \begin{array}{c c c c c c c c} \frac {\partial h _ {1}}{\partial x} & 0 & \frac {\partial h _ {2}}{\partial x} & 0 & \frac {\partial h _ {3}}{\partial x} & 0 & \frac {\partial h _ {4}}{\partial x} & 0 \\ 0 & \frac {\partial h _ {1}}{\partial y} & 0 & \frac {\partial h _ {2}}{\partial y} & 0 & \frac {\partial h _ {3}}{\partial y} & 0 & \frac {\partial h _ {4}}{\partial y} \\ \frac {\partial h _ {1}}{\partial y} & \frac {\partial h _ {1}}{\partial x} & \frac {\partial h _ {2}}{\partial y} & \frac {\partial h _ {2}}{\partial x} & \frac {\partial h _ {3}}{\partial y} & \frac {\partial h _ {3}}{\partial x} & \frac {\partial h _ {4}}{\partial y} & \frac {\partial h _ {4}}{\partial x} \end{array} \right] \tag {e}
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$$
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where it is implied that in (c) and (d), the derivatives are evaluated at point $(r_i, s_j)$ , and therefore in (e), we have, in fact, the matrix $\mathbf{B}_{ij}$ .
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EXAMPLE 5.6: Derive the expressions needed for the evaluation of the mass matrix of the element considered in Example 5.5.
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The mass matrix of the element is given by
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$$
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\mathbf {M} = \sum_ {i, j} \alpha_ {i j} t _ {i j} \mathbf {F} _ {i j}
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$$
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where $\mathbf{F}_{ij} = \rho_{ij}\mathbf{H}_{ij}^{T}\mathbf{H}_{ij}\det \mathbf{J}_{ij}$
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and $H_{ij}$ is the displacement interpolation matrix. The displacement interpolation functions for u and v of the four-node element have been given in Example 5.5, and we have
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$$
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\mathbf {H} _ {i j} = \frac {1}{4} \left[ \begin{array}{c c c c} (1 + r _ {i}) (1 + s _ {j}) & 0 & (1 - r _ {i}) (1 + s _ {j}) & 0 \\ 0 & (1 + r _ {i}) (1 + s _ {j}) & 0 & (1 - r _ {i}) (1 + s _ {j}) \end{array} \right.
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$$
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$$
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\left. \begin{array}{c c c c} (1 - r _ {i}) (1 - s _ {j}) & 0 & (1 + r _ {i}) (1 - s _ {j}) & 0 \\ 0 & (1 - r _ {i}) (1 - s _ {j}) & 0 & (1 + r _ {i}) (1 - s _ {j}) \end{array} \right]
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$$
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<!-- source-page: 372 -->
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The determinant of the Jacobian matrix, det $J_{ij}$ , was given in Example 5.5, and $\rho_{ij}$ is the mass density at the sampling point $(r_{i}, s_{j})$ . Therefore, all required variables for the evaluation of the mass matrix have been defined.
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EXAMPLE 5.7: Derive the expressions needed for the evaluation of the body force vector $R_{B}$ and the initial stress vector $R_{I}$ of the element considered in Example 5.5.
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These vectors are obtained using the matrices $H_{ij}$ , $B_{ij}$ , and $J_{ij}$ defined in Examples 5.5 and 5.6; i.e., we have
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$$
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\mathbf {R} _ {B} = \sum_ {i, j} \alpha_ {i j} t _ {i j} \mathbf {H} _ {i j} ^ {T} \mathbf {f} _ {i j} ^ {B} \det \mathbf {J} _ {i j}
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$$
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$$
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\mathbf {R} _ {I} = \sum_ {i, j} \alpha_ {i j} t _ {i j} \mathbf {B} _ {i j} ^ {T} \boldsymbol {\tau} _ {i j} ^ {I} \det \mathbf {J} _ {i j}
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$$
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where $f_{ij}^{B}$ and $\tau_{ij}^{I}$ are the body force vector and initial stress vector evaluated at the integration sampling points.
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EXAMPLE 5.8: Derive the expressions needed in the calculation of the surface force vector $R_{s}$ when the element edge 1-2 of the four-node isoparametric element considered in Example 5.5 is loaded as shown in Fig. E5.8.
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<details>
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<summary>text_image</summary>
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f^S_y
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Node 1
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2
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s
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r
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f^S_x
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3
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4
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y, v
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x, u
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</details>
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Figure E5.8 Traction distribution along edge 1-2 of a four-node element
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The first step is to establish the displacement interpolations. Since $s = +1$ at the edge 1-2, we have, using the interpolation functions given in Example 5.5,
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$$
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u ^ {S} = \frac {1}{2} (1 + r) u _ {1} + \frac {1}{2} (1 - r) u _ {2}
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$$
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$$
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v ^ {s} = \frac {1}{2} (1 + r) v _ {1} + \frac {1}{2} (1 - r) v _ {2}
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$$
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Hence, to evaluate $\mathbf{R}_s$ in (5.34) we can use
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$$
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\mathbf {H} ^ {s} = \left[ \begin{array}{c c c c c c c c} \frac {1}{2} (1 + r) & 0 & \frac {1}{2} (1 - r) & 0 & 0 & 0 & 0 & 0 \\ 0 & \frac {1}{2} (1 + r) & 0 & \frac {1}{2} (1 - r) & 0 & 0 & 0 & 0 \end{array} \right]
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$$
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and
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$$
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\mathbf {f} ^ {S} = \left[ \begin{array}{l} f _ {x} ^ {S} \\ f _ {y} ^ {S} \end{array} \right]
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$$
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<!-- source-page: 373 -->
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where $f_x^S$ and $f_y^S$ are the $x$ and $y$ components of the applied surface force. These components may have been given as a function of $r$ .
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For the evaluation of the integral in (5.34), we also need the differential surface area dS expressed in the r, s natural coordinate system. If $t_{r}$ is the thickness, dS = $t_{r}$ dl, where dl is a differential length,
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$$
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d l = \det \mathbf {J} ^ {s} d r; \quad \det \mathbf {J} ^ {s} = \left[ \left(\frac {\partial x}{\partial r}\right) ^ {2} + \left(\frac {\partial y}{\partial r}\right) ^ {2} \right] ^ {1 / 2}
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$$
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But the derivatives $\partial x / \partial r$ and $\partial y / \partial r$ have been given in Example 5.5. Using $s = +1$ , we have, in this case,
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$$
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\frac {\partial x}{\partial r} = \frac {x _ {1} - x _ {2}}{2}; \quad \frac {\partial y}{\partial r} = \frac {y _ {1} - y _ {2}}{2}
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$$
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Although the vector $R_{s}$ could in this case be evaluated in a closed-form solution (provided that the functions used in $f^{s}$ are simple), in order to keep generality in the program that calculates $R_{s}$ , it is expedient to use numerical integration. This way, variable-number-nodes elements can be implemented in an elegant manner in one program. Thus, using the notation defined in this section, we have
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$$
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\mathbf {R} _ {S} = \sum_ {i} \alpha_ {i} t _ {r i} \mathbf {F} _ {i}
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$$
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$$
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\mathbf {F} _ {i} = \mathbf {H} _ {i} ^ {s ^ {T}} \mathbf {f} _ {i} ^ {s} \det \mathbf {J} _ {i} ^ {s}
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$$
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It is noted that in this case only one-dimensional numerical integration is required because $s$ is not a variable.
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EXAMPLE 5.9: Explain how the expressions given in Examples 5.5 to 5.7 need be modified when the element considered is an axisymmetric element.
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In this case two modifications are necessary. First, we consider 1 radian of the structure. Hence, the thickness to be employed in all integrations is that corresponding to 1 radian, which means that at an integration point the thickness is equal to the radius at that point:
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$$
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t _ {i j} = \sum_ {k = 1} ^ {4} h _ {k} \Bigg | _ {r _ {i}, s _ {j}} x _ {k} \tag {a}
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$$
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Second, it is recognized that also circumferential strains and stresses are developed (see Table 4.2). Hence, the strain-displacement matrix must be augmented by one row for the hoop strain u/R; i.e., we have
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$$
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\mathbf {B} = \left[ \begin{array}{c c c c c c c c} \dots & & & & & & & \dots \\ \frac {h _ {1}}{t} & 0 & \frac {h _ {2}}{t} & 0 & \frac {h _ {3}}{t} & 0 & \frac {h _ {4}}{t} & 0 \end{array} \right] \tag {b}
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$$
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where the first three rows have already been defined in Example 5.5 and t is equal to the radius. To obtain the strain-displacement matrix at integration point $(i, j)$ we use (a) to evaluate t and substitute into (b).
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EXAMPLE 5.10: Calculate the nodal point forces of the four-node axisymmetric finite element shown in Fig. E5.10 when the element is subjected to centrifugal loading.
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Here we want to evaluate
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$$
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\mathbf {R} _ {B} = \int_ {V} \mathbf {H} ^ {T} \mathbf {f} ^ {B} d V
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$$
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<!-- source-page: 374 -->
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<details>
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<summary>text_image</summary>
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y
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R₁
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R₀
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2
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1
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s
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r
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1 cm
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3
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4
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x
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ω
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Density ρ
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</details>
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Figure E5.10 Four-node axisymmetric element rotating at angular velocity $\omega$ (rad/sec)
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where $f_{x}^{B} = \rho \omega^{2}R;$ $f_{y}^{B} = 0$
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$$
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R = \frac {1}{2} (1 - r) R _ {0} + \frac {1}{2} (1 + r) R _ {1}
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$$
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$$
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\mathbf {H} = \left[ \begin{array}{c c c c c c c c} h _ {1} & 0 & h _ {2} & 0 & h _ {3} & 0 & h _ {4} & 0 \\ 0 & h _ {1} & 0 & h _ {2} & 0 & h _ {3} & 0 & h _ {4} \end{array} \right]; \quad \mathbf {J} = \left[ \begin{array}{c c} \frac {R _ {1} - R _ {0}}{2} & 0 \\ 0 & \frac {1}{2} \end{array} \right]
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$$
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and the $h_i$ are defined in Fig. 5.4. Also, considering 1 radian,
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$$
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d V = \det \mathbf {J} d r d s R = \left(\frac {R _ {1} - R _ {0}}{4}\right) d r d s \left(\frac {R _ {1} + R _ {0}}{2} + \frac {R _ {1} - R _ {0}}{2} r\right)
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$$
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Hence,
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$$
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\mathbf {R} _ {B} = \frac {\rho \omega^ {2} \left(R _ {1} - R _ {0}\right)}{6 4} \int_ {r = - 1} ^ {+ 1} \int_ {s = - 1} ^ {+ 1} \left[ \begin{array}{c c} (1 + r) (1 + s) & 0 \\ 0 & (1 + r) (1 + s) \\ (1 - r) (1 + s) & 0 \\ 0 & (1 - r) (1 + s) \\ (1 - r) (1 - s) & 0 \\ 0 & (1 - r) (1 - s) \\ (1 + r) (1 - s) & 0 \\ 0 & (1 + r) (1 - s) \end{array} \right]
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$$
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$$
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\left[ \left(R _ {1} + R _ {0}\right) + \left(R _ {1} - R _ {0}\right) r \right] ^ {2} \left[ \begin{array}{l} 1 \\ 0 \end{array} \right] d r d s
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$$
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If we let $A = R_{1} + R_{0}$ and $B = R_{1} - R_{0}$ , we have
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$$
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\mathbf {R} _ {B} = \frac {\rho \omega^ {2} B}{6 4} \left[ \begin{array}{c} \frac {2}{3} (6 A ^ {2} + 4 A B + 2 B ^ {2}) \\ 0 \\ \frac {2}{3} (6 A ^ {2} - 4 A B + 2 B ^ {2}) \\ 0 \\ \frac {2}{3} (6 A ^ {2} - 4 A B + 2 B ^ {2}) \\ 0 \\ \frac {2}{3} (6 A ^ {2} + 4 A B + 2 B ^ {2}) \\ 0 \end{array} \right]
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$$
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<!-- source-page: 375 -->
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EXAMPLE 5.11: The four-node plane stress element shown in Fig. E5.11 is subjected to the given temperature distribution. If the temperature corresponding to the stress-free state is $\theta_{0}$ , evaluate the nodal point forces to which the element must be subjected so that there are no nodal point displacements.
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<details>
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<summary>text_image</summary>
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Element thickness = 1 cm
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Young's modulus E
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Poisson's ratio v
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Thermal coefficient
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of expansion α
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s
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+20°C
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2
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1
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3 cm
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r
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+10°C
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3
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4
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+20°C
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4 cm
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R₄
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R₂
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R₃
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R₁
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R₅
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R₇
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R₆
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R₈
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</details>
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Figure E5.11 Nodal point forces due to initial temperature distribution
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In this case we have for the total stresses, due to total strains $\epsilon$ and thermal strains $\epsilon^{\text{th}}$ ,
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$$
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\boldsymbol {\tau} = \mathbf {C} (\boldsymbol {\epsilon} - \boldsymbol {\epsilon} ^ {\mathrm{th}}) \tag {a}
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$$
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where $\epsilon_{xx}^{\mathrm{th}} = \alpha(\theta - \theta_{0})$ , $\epsilon_{yy}^{\mathrm{th}} = \alpha(\theta - \theta_{0})$ , $\gamma_{xy}^{th} = 0$ . If the nodal point displacements are zero, we have $\epsilon = 0$ , and the stresses due to the thermal strains can be thought of as initial stresses. Thus, the nodal point forces are
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$$
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\begin{array}{r l} \mathbf {R} _ {I} & = \int_ {V} \mathbf {B} ^ {T} \boldsymbol {\tau} ^ {I} d V \\ \boldsymbol {\tau} ^ {I} & = - \frac {E \alpha}{1 - \nu^ {2}} \left[ \begin{array}{c c c} 1 & \nu & 0 \\ \nu & 1 & 0 \\ 0 & 0 & \frac {1 - \nu}{2} \end{array} \right] \left[ \begin{array}{l} 1 \\ 1 \\ 0 \end{array} \right] \left\{\left(\sum_ {i = 1} ^ {4} h _ {i} \theta_ {i}\right) - \theta_ {0} \right\} \end{array}
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$$
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and the $h_{i}$ are the interpolation functions defined in Fig. 5.4. Also,
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$$
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\mathbf {B} = \left[ \begin{array}{c c c c c c c c} \frac {1 + s}{8} & 0 & - \frac {1 + s}{8} & 0 & - \frac {1 - s}{8} & 0 & \frac {1 - s}{8} & 0 \\ 0 & \frac {1 + r}{6} & 0 & \frac {1 - r}{6} & 0 & - \frac {1 - r}{6} & 0 & - \frac {1 + r}{6} \\ \frac {1 + r}{6} & \frac {1 + s}{8} & \frac {1 - r}{6} & - \frac {1 + s}{8} & - \frac {1 - r}{6} & - \frac {1 - s}{8} & - \frac {1 + r}{6} & \frac {1 - s}{8} \end{array} \right]
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$$
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<!-- source-page: 376 -->
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Hence,
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$$
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\mathbf {R} _ {I} = \int_ {- 1} ^ {+ 1} \int_ {- 1} ^ {+ 1} - \left[ \begin{array}{c c c} \frac {1 + s}{8} & 0 & \frac {1 + r}{6} \\ 0 & \frac {1 + r}{6} & \frac {1 + s}{8} \\ - \frac {1 + s}{8} & 0 & \frac {1 - r}{6} \\ 0 & \frac {1 - r}{6} & - \frac {1 + s}{8} \\ - \frac {1 - s}{8} & 0 & - \frac {1 - r}{6} \\ 0 & - \frac {1 - r}{6} & - \frac {1 - s}{8} \\ \frac {1 - s}{8} & 0 & - \frac {1 + r}{6} \\ 0 & - \frac {1 + r}{6} & \frac {1 - s}{8} \end{array} \right] \left[ \begin{array}{l} 1 + \nu \\ 1 + \nu \\ 0 \end{array} \right] \frac {E \alpha}{1 - \nu^ {2}}
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$$
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$$
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[ 2. 5 (s + 3) (r + 3) - \theta_ {0} ] 3 d r d s
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$$
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$$
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\mathbf {R} _ {I} = - \frac {E \alpha}{(1 - \nu)} \left[ \begin{array}{c} 3 7. 5 - 1. 5 \theta_ {0} \\ 5 0 - 2 \theta_ {0} \\ - 3 7. 5 + 1. 5 \theta_ {0} \\ 4 0 - 2 \theta_ {0} \\ - 3 0 + 1. 5 \theta_ {0} \\ - 4 0 + 2 \theta_ {0} \\ + 3 0 - 1. 5 \theta_ {0} \\ - 5 0 + 2 \theta_ {0} \end{array} \right]
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$$
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The calculation of the initial stress force vector as performed here is a typical step in a thermal stress analysis. In a complete thermal stress analysis the temperatures are calculated as described in Section 7.2, the element load vectors due to the thermal effects are evaluated as illustrated in this example, and the solution of the equilibrium equations (4.17) of the complete element assemblage then yields the nodal point displacements. The element total strains $\epsilon$ are evaluated from the nodal point displacements and then, using (a), the final element stresses are calculated.
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EXAMPLE 5.12: Consider the elements in Fig. E5.12. Evaluate the consistent nodal point forces corresponding to the surface loading (assuming that the nodal point forces are positive when acting in the direction of the pressure).
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Here we want to evaluate
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$$
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\mathbf {R} _ {S} = \int_ {S} \mathbf {H} ^ {s ^ {\mathrm{T}}} \mathbf {f} ^ {s} d S
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$$
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<!-- source-page: 377 -->
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<details>
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<summary>text_image</summary>
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p2
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2
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5
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1
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p1
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s
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8
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Thickness
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r
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2 cm
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6
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7
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3
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4
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2 cm
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</details>
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(a) Two-dimensional element subjected to linearly varying pressure along one side
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<details>
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<summary>text_image</summary>
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1 cm
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p
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2
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s
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5
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p
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1
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p
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6
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p
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8
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r
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2 cm
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3
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7
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4
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2 cm
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</details>
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(b) Flat surface of three-dimensional element subjected to constant pressure p
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Figure E5.12 Two- and three-dimensional elements subjected to pressure loading
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Consider first the two-dimensional element. Since $s = +1$ at the edge 1-2, we have, using the interpolation functions for the eight-node element (see Fig. 5.4),
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$$
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h _ {5} = \frac {1}{2} (1 - r ^ {2}) (1 + s) | _ {s = + 1} = 1 - r ^ {2}
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$$
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$$
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h _ {1} = \frac {1}{4} (1 + r) (1 + s) (r + s - 1) | _ {s = + 1} = \frac {1}{2} r (1 + r)
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$$
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$$
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h _ {2} = \frac {1}{4} (1 - r) (1 + s) (s - r - 1) | _ {s = + 1} = - \frac {1}{2} r (1 - r)
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$$
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which are equal to the interpolation functions of the three-node bar in Fig. E5.2. Hence
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$$
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\left[ \begin{array}{c} u ^ {s} \\ v ^ {s} \end{array} \right] = \left[ \begin{array}{c c c c c c} \frac {1}{2} r (1 + r) & 0 & - \frac {1}{2} r (1 - r) & 0 & (1 - r ^ {2}) & 0 \\ 0 & \frac {1}{2} r (1 + r) & 0 & - \frac {1}{2} r (1 - r) & 0 & (1 - r ^ {2}) \end{array} \right] \left[ \begin{array}{c} u _ {1} \\ v _ {1} \\ u _ {2} \\ v _ {2} \\ u _ {5} \\ v _ {5} \end{array} \right]
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$$
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Also, $\mathbf{f}^S = \begin{bmatrix} f_r^S \\ f_s^S \end{bmatrix} = \begin{bmatrix} 0 \\ \frac{1}{2}(1 + r)p_1 + \frac{1}{2}(1 - r)p_2 \end{bmatrix}; \quad \det \mathbf{J}^S = 1$
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Hence,
|
||
|
||
$$
|
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\mathbf {R} _ {s} = \int_ {- 1} ^ {+ 1} \frac {t}{2} \left[ \begin{array}{c c} r (1 + r) & 0 \\ 0 & r (1 + r) \\ - r (1 - r) & 0 \\ 0 & - r (1 - r) \\ 2 (1 - r ^ {2}) & 0 \\ 0 & 2 (1 - r ^ {2}) \end{array} \right] \frac {1}{2} \left[ \begin{array}{c} 0 \\ (1 + r) p _ {1} + (1 - r) p _ {2} \end{array} \right] d r
|
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$$
|
||
|
||
<!-- source-page: 378 -->
|
||
|
||
$$
|
||
\mathbf {R} _ {s} = \frac {1}{3} \left[ \begin{array}{c c c} 0 & & \\ p _ {1} & & \\ 0 & & \\ p _ {2} & & \\ 0 & & \\ 2 (p _ {1} + p _ {2}) & & \end{array} \right] \tag {a}
|
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$$
|
||
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||
For the three-dimensional element we proceed similarly. Since the surface is flat and the loading is normal to it, only the nodal point forces normal to the surface are nonzero [see also (a)]. Also, by symmetry, we know that the forces at nodes 1, 2, 3, 4 and 5, 6, 7, 8 are equal, respectively. Using the interpolation functions of Fig. 5.4, we have for the force at node 1,
|
||
|
||
$$
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R _ {1} = p \int_ {- 1} ^ {+ 1} \int_ {- 1} ^ {+ 1} \frac {1}{4} (1 + r) (1 + s) (r + s - 1) d r d s = - \frac {1}{3} p
|
||
$$
|
||
|
||
and for the force at node 5,
|
||
|
||
$$
|
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R _ {5} = p \int_ {- 1} ^ {+ 1} \int_ {- 1} ^ {+ 1} \frac {1}{2} (1 - r ^ {2}) (1 + s) d r d s = \frac {4}{3} p
|
||
$$
|
||
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||
The total pressure loading on the surface is 4p, which, as a check, is equal to the sum of all the nodal point forces. However, it should be noted that the consistent nodal point forces at the corners of the element act in the direction opposite that of the pressure!
|
||
|
||
EXAMPLE 5.13: Calculate the deflection $u_{A}$ of the structural model shown in Fig. E5.13.
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
Z
|
||
U4
|
||
A
|
||
U2
|
||
U3
|
||
6 cm
|
||
U6
|
||
Bar with cross-sectional area = 1 cm²
|
||
U8
|
||
U7 = uA
|
||
U5
|
||
Y
|
||
P = 6000 N
|
||
6 cm
|
||
A
|
||
8 cm
|
||
E = 30 × 10⁶ N/cm²
|
||
v = 0.3
|
||
0.1 cm
|
||
0.5 cm² each
|
||
0.1 cm
|
||
Section AA
|
||
</details>
|
||
|
||
Figure E5.13 A simple structural model
|
||
|
||
<!-- source-page: 379 -->
|
||
|
||
Because of the symmetry and boundary conditions, we need to evaluate only the stiffness coefficient corresponding to $u_{A}$ . Here we have for the four-node element,
|
||
|
||
$$
|
||
\mathbf {J} = \left[ \begin{array}{c c} 4 & 0 \\ 0 & 3 \end{array} \right]; \quad \mathbf {B} = \frac {1}{4 8} \left[ \begin{array}{c c c} & 3 (1 - s) \\ \dots & 0 & \dots \\ & - 4 (1 + r) \end{array} \right]
|
||
$$
|
||
|
||
$$
|
||
k _ {7 7} = \int_ {- 1} ^ {+ 1} \int_ {- 1} ^ {+ 1} \left(\frac {1}{4 8}\right) ^ {2} \frac {E}{1 - \nu^ {2}} [ 3 (1 - s) \mid 0 \mid - 4 (1 + r) ] \left[ \begin{array}{c} 3 (1 - s) \\ 3 \nu (1 - s) \\ - 2 (1 - \nu) (1 + r) \end{array} \right] \tag {12} (0. 1) d r d s
|
||
$$
|
||
|
||
or $k_{77} = 1,336,996.34\mathrm{N / cm}$
|
||
|
||
Also, the stiffness of the truss is AE/L, or
|
||
|
||
$$
|
||
k = \frac {(1) (3 0 \times 1 0 ^ {6})}{8} = 3, 7 5 0, 0 0 0 \mathrm{N/cm}
|
||
$$
|
||
|
||
Hence $k_{\mathrm{total}} = 6.424 \times 10^{6} \mathrm{~N/cm}$
|
||
|
||
and $u_{A} = 9.34\times 10^{-4}\mathrm{cm}$
|
||
|
||
EXAMPLE 5.14: Consider the five-node element in Fig. E5.14. Evaluate the consistent nodal point forces corresponding to the stresses given.
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
4 cm
|
||
1 cm
|
||
1 cm
|
||
5
|
||
Thickness = 1 cm
|
||
3
|
||
4
|
||
y
|
||
τxx = 0
|
||
τyy = 10 N/cm²
|
||
τxy = τyx = 20 N/cm²
|
||
x
|
||
</details>
|
||
|
||
Figure E5.14 Five-node element with stresses given
|
||
|
||
Using the interpolation functions in Fig. 5.4, we can evaluate the strain-displacement matrix of the element:
|
||
|
||
$$
|
||
\mathbf {B} = \frac {1}{8} \left[ \begin{array}{c c c c c} (1 + s) & 0 & - s (1 + s) & 0 & s (1 - s) \\ 0 & 2 (1 + r) & 0 & 2 (1 - r) (1 + 2 s) & 0 \\ 2 (1 + r) & (1 + s) & 2 (1 - r) (1 + 2 s) & - s (1 + s) & - 2 (1 - r) (1 - 2 s) \end{array} \right.
|
||
$$
|
||
|
||
$$
|
||
\left. \begin{array}{c c c c c} 0 & (1 - s) & 0 & - 2 (1 - s ^ {2}) & 0 \\ - 2 (1 - r) (1 - 2 s) & 0 & - 2 (1 + r) & 0 & - 8 (1 - r) s \\ s (1 - s) & - 2 (1 + r) & (1 - s) & - 8 (1 - r) s & - 2 (1 - s ^ {2}) \end{array} \right]
|
||
$$
|
||
|
||
<!-- source-page: 380 -->
|
||
|
||
where we used
|
||
|
||
$$
|
||
\mathbf {J} = \left[ \begin{array}{l l} 2 & 0 \\ 0 & 1 \end{array} \right]
|
||
$$
|
||
|
||
The required nodal point forces can now be evaluated using (5.35); hence,
|
||
|
||
$$
|
||
\mathbf {R} _ {I} = \int_ {- 1} ^ {+ 1} \int_ {- 1} ^ {+ 1} \mathbf {B} ^ {T} \left[ \begin{array}{l} 0 \\ 1 0 \\ 2 0 \end{array} \right] (2) d r d s
|
||
$$
|
||
|
||
which gives
|
||
|
||
$$
|
||
\mathbf {R} _ {i} ^ {T} = \left[ \begin{array}{l l l l l l l l l l l} 4 0 & 4 0 & 4 0 & \frac {4 0}{3} & - 4 0 & - \frac {8 0}{3} & - 4 0 & 0 & 0 & - \frac {8 0}{3} \end{array} \right]
|
||
$$
|
||
|
||
It should be noted that the forces in this vector are also equal to the nodal point consistent forces that correspond to the (constant) surface tractions, which are in equilibrium with the internal stresses given in Fig. E.5.14.
|
||
|
||
Earlier we mentioned briefly the possible use of subparametric elements: here the geometry is interpolated to a lower degree than the displacements. In the above examples, the nodes corresponding to the higher-order interpolation functions (nodes 5 and higher for the two-dimensional elements) were always placed at their “natural” positions so that the Jacobian matrix would be the same if, for the geometry interpolation, only the “basic” lower-order functions were used. Hence, in this case the subparametric two-dimensional element, using only the four corner nodes for the interpolation of the geometry, gives the same element matrices as the isoparametric element. For instance, in Example 5.14, the Jacobian matrix J would be the same using only the basic four-node interpolation functions, and hence the vector $R_{i}$ for the subparametric element (using the four corner nodes for the geometry interpolation and the five nodes for the displacement interpolation) would be the same as for the isoparametric five-node element.
|
||
|
||
However, while the use of subparametric elements decreases somewhat the computational effort, such use also limits the generality of the finite element discretization and in addition complicates the solution procedures considerably in geometrically nonlinear analysis (where the new geometry of an element is obtained by adding the displacements to the previous geometry; see Chapter 6).
|
||
|
||
# 5.3.2 Triangular Elements
|
||
|
||
In the previous section we discussed quadrilateral isoparametric elements that can be used to model very general geometries. However, in some cases the use of triangular or wedge elements may be attractive. Triangular elements can be formulated using different approaches, which we briefly discuss in this section.
|
||
|
||
# Triangular Elements Formulated by Collapsing Quadrilateral Elements
|
||
|
||
Since the elements discussed in Section 5.3.1 can be distorted, as shown for example in Fig. 5.2, a natural way of generating triangular elements appears to be to simply distort the basic quadrilateral element into the required triangular form (see Fig. 5.7). This is achieved in practice by assigning the same global node to two corner nodes of the element. We demonstrate this procedure in the following example.
|