김경종
21 KiB
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z, w Axis of symmetry 4 3 p = 1 psi ③ ④ 5 ② ① 0.5 in. 1 2 r, u r = 0.5 in. r = 1.0 in.
Figure 9–10 Discretized cylinder slice
Discretization
To illustrate the finite element solution for the cylinder, we first discretize the cylinder into four triangular elements, as shown in Figure 9–10. A horizontal slice of the cylinder represents the total cylinder behavior. Because we are performing a longhand solution, a coarse mesh of elements is used for simplicity’s sake (but without loss of generality of the method). The governing global matrix equation is
\left\{ \begin{array}{l} F _ {1 r} \\ F _ {1 z} \\ F _ {2 r} \\ F _ {2 z} \\ F _ {3 r} \\ F _ {3 z} \\ F _ {4 r} \\ F _ {4 z} \\ F _ {5 r} \\ F _ {5 z} \end{array} \right\} = [ K ] \left\{ \begin{array}{l} u _ {1} \\ w _ {1} \\ u _ {2} \\ w _ {2} \\ u _ {3} \\ w _ {3} \\ u _ {4} \\ w _ {4} \\ u _ {5} \\ w _ {5} \end{array} \right\} \tag {9.2.1}
where the ½K matrix is of order 1 0 \times 1 0 . .
Assemblage of the Stiffness Matrix
We assemble the ½K matrix in the usual manner by superposition of the individual element stiffness matrices. For simplicity’s sake, we will use the first approximation method given by Eq. (9.1.26) to evaluate the element matrices. Therefore,
[ k ] = 2 \pi \bar {r} A [ \bar {B} ] ^ {T} [ D ] [ \bar {B} ] \tag {9.2.2}
For element 1 (Figure 9–11), the coordinates are r _ { i } = 0 . 5 , z _ { i } = 0 , r _ { j } = 1 . 0 , z _ { j } = 0 \mathrm { _ { \it { i } } } r _ { m } = 0 . 7 5 , , and z _ { m } = 0 . 2 5 ~ ( i = 1 , j = 2 . , and m = 5 for element 1) for the globalcoordinate axes as set up in Figure 9–10.
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5 ① 1 2
Figure 9–11 Element 1 of the discretized cylinder
We now evaluate [ \bar { B } ] , where [ \bar { B } ] is given by Eq. (9.1.19) evaluated at the centroid of the element \begin{array} { r } { r = \bar { r } , z = \bar { z } , } \end{array} and expanded here as
[ \bar {B} ] = \frac {1}{2 A} \left[ \begin{array}{c c c c c c} \beta_ {i} & 0 & \beta_ {j} & 0 & \beta_ {m} & 0 \\ 0 & \gamma_ {i} & 0 & \gamma_ {j} & 0 & \gamma_ {m} \\ \frac {\alpha_ {i}}{\bar {r}} + \beta_ {i} + \frac {\gamma_ {i} \bar {z}}{\bar {r}} & 0 & \frac {\alpha_ {j}}{\bar {r}} + \beta_ {j} + \frac {\gamma_ {j} \bar {z}}{\bar {r}} & 0 & \frac {\alpha_ {m}}{\bar {r}} + \beta_ {m} + \frac {\gamma_ {m} \bar {z}}{\bar {r}} & 0 \\ \gamma_ {i} & \beta_ {i} & \gamma_ {j} & \beta_ {j} & \gamma_ {m} & \beta_ {m} \end{array} \right] \tag {9.2.3}
where, using element coordinates in Eqs. (9.1.11), we have
\alpha_ {i} = r _ {j} z _ {m} - z _ {j} r _ {m} = (1. 0) (0. 2 5) - (0. 0) (0. 7 5) = 0. 2 5 \mathrm{in} ^ {2}
\alpha_ {j} = r _ {m} z _ {i} - z _ {m} r _ {i} = (0. 7 5) (0) - (0. 2 5) (0. 5) = - 0. 1 2 5 \mathrm{in} ^ {2}
\alpha_ {m} = r _ {i} z _ {j} - z _ {i} r _ {j} = (0. 5) (0. 0) - (0) (1. 0) = 0. 0 \mathrm{in} ^ {2}
\beta_ {i} = z _ {j} - z _ {m} = 0. 0 - 0. 2 5 = - 0. 2 5 \text { in. }
\beta_ {j} = z _ {m} - z _ {i} = 0. 2 5 - 0 = 0. 2 5 \text { in. } \tag {9.2.4}
\beta_ {m} = z _ {i} - z _ {j} = 0. 0 - 0. 0 = 0. 0 \text { in. }
\gamma_ {i} = r _ {m} - r _ {j} = 0. 7 5 - 1. 0 = - 0. 2 5 \text { in. }
\gamma_ {j} = r _ {i} - r _ {m} = 0. 5 - 0. 7 5 = - 0. 2 5 \text { in. }
\gamma_ {m} = r _ {j} - r _ {i} = 1. 0 - 0. 5 = 0. 5 \text { in. }
and \begin{array} { r } { \bar { r } = 0 . 5 + \frac { 1 } { 2 } ( 0 . 5 ) = 0 . 7 5 \mathrm { { i n } . \qquad ~ \bar { z } = \frac { 1 } { 3 } ( 0 . 2 5 ) = 0 . 0 8 3 3 \mathrm { { i n } . } } } \end{array}
A = \frac {1}{2} (0. 5) (0. 2 5) = 0. 0 6 2 5 \mathrm{in} ^ {2}
Substituting the results from Eqs. (9.2.4) into Eq. (9.2.3), we obtain
[ \bar {B} ] = \frac {1}{0 . 1 2 5} \left[ \begin{array}{c c c c c c} - 0. 2 5 & 0 & 0. 2 5 & 0 & 0 & 0 \\ 0 & - 0. 2 5 & 0 & - 0. 2 5 & 0 & 0. 5 \\ 0. 0 5 5 6 & 0 & 0. 0 5 5 6 & 0 & 0. 0 5 5 6 & 0 \\ - 0. 2 5 & - 0. 2 5 & - 0. 2 5 & 0. 2 5 & 0. 5 & 0 \end{array} \right] \frac {1}{\text {in.}} \tag {9.2.5}
For the axisymmetric stress case, the matrix ½D is given in Eq. (9.1.2) as
[ D ] = \frac {E}{(1 + \nu) (1 - 2 \nu)} \left[ \begin{array}{c c c c} 1 - \nu & \nu & \nu & 0 \\ \nu & 1 - \nu & \nu & 0 \\ \nu & \nu & 1 - \nu & 0 \\ 0 & 0 & 0 & \frac {1 - 2 \nu}{2} \end{array} \right] \tag {9.2.6}
With n ¼ 0:3 and E = 3 0 \times 1 0 ^ { 6 } psi, we obtain
[ D ] = \frac {3 0 (1 0 ^ {6})}{(1 + 0 . 3) [ 1 - 2 (0 . 3) ]} \left[ \begin{array}{c c c c} 1 - 0. 3 & 0. 3 & 0. 3 & 0 \\ 0. 3 & 1 - 0. 3 & 0. 3 & 0 \\ 0. 3 & 0. 3 & 1 - 0. 3 & 0 \\ 0 & 0 & 0 & \frac {1 - 2 (0 . 3)}{2} \end{array} \right] \tag {9.2.7}
or, simplifying Eq. (9.2.7),
[ D ] = 5 7. 7 (1 0 ^ {6}) \left[ \begin{array}{l l l l} 0. 7 & 0. 3 & 0. 3 & 0 \\ 0. 3 & 0. 7 & 0. 3 & 0 \\ 0. 3 & 0. 3 & 0. 7 & 0 \\ 0 & 0 & 0 & 0. 2 \end{array} \right] \text { psi } \tag {9.2.8}
Using Eqs. (9.2.5) and (9.2.8), we obtain
[ \bar {B} ] ^ {T} [ D ] = \frac {5 7 . 7 (1 0 ^ {6})}{0 . 1 2 5} \left[ \begin{array}{c c c c} - 0. 1 5 8 & - 0. 0 5 8 3 & - 0. 0 3 6 1 & - 0. 0 5 \\ - 0. 0 7 5 & - 0. 1 7 5 & - 0. 0 7 5 & - 0. 0 5 \\ 0. 1 9 2 & 0. 0 9 1 7 & 0. 1 1 4 & - 0. 0 5 \\ - 0. 0 7 5 & - 0. 1 7 5 & - 0. 0 7 5 & 0. 0 5 \\ 0. 0 1 6 7 & 0. 0 1 6 6 & 0. 0 3 8 8 & 0. 1 \\ 0. 1 5 & 0. 3 5 & 0. 1 5 & 0 \end{array} \right] \tag {9.2.9}
Substituting Eqs. (9.2.5) and (9.2.9) into Eq. (9.2.2), we obtain the stiffness matrix for element 1 as
\left[ k ^ {(1)} \right] = \left(1 0 ^ {6}\right) \left[ \begin{array}{c c c c c c} i = 1 & & j = 2 & & m = 5 \\ 5 4. 4 6 & 2 9. 4 5 & - 3 1. 6 3 & 2. 2 6 & - 2 9. 3 7 & - 3 1. 7 1 \\ 2 9. 4 5 & 6 1. 1 7 & - 1 1. 3 3 & 3 3. 9 8 & - 3 1. 7 2 & - 9 5. 1 5 \\ - 3 1. 6 3 & - 1 1. 3 3 & 7 2. 5 9 & - 3 8. 5 2 & - 2 0. 3 1 & 4 9. 8 4 \\ 2. 2 6 & 3 3. 9 8 & - 3 8. 5 2 & 6 1. 1 7 & 2 2. 6 6 & - 9 5. 1 5 \\ - 2 9. 3 7 & - 3 1. 7 2 & - 2 0. 3 1 & 2 2. 6 6 & 5 6. 7 2 & 9. 0 6 \\ - 3 1. 7 1 & - 9 5. 1 5 & 4 9. 8 4 & - 9 5. 1 5 & 9. 0 6 & 1 9 0. 3 1 \end{array} \right] \frac {\mathrm{lb}}{\text { in. }} \tag {9.2.10}
where the numbers above the columns indicate the nodal orders of degrees of freedom in the element 1 stiffness matrix.
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3 5 ② 2
Figure 9–12 Element 2 of the discretized cylinder
For element 2 (Figure 9–12), the coordinates are r_{i}=1.0 , z_{i}=0.0 , r_{j}=1.0 , z_{j}=0.5 , r_{m}=0.75 , and z_{m}=0.25 (i=2, j=3, and m=5 for element 2). Therefore,
\alpha_ {i} = (1. 0) (0. 2 5) - (0. 5) (0. 7 5) = - 0. 1 2 5 \mathrm{in} ^ {2}
\alpha_ {j} = (0. 7 5) (0. 0) - (0. 2 5) (1. 0) = - 0. 2 5 \mathrm{in} ^ {2} \tag {9.2.11}
\alpha_ {m} = (1. 0) (0. 5) - (0. 0) (1. 0) = 0. 5 \mathrm{in} ^ {2}
\beta_ {i} = 0. 5 - 0. 2 5 = 0. 2 5 \text { in. } \quad \beta_ {j} = 0. 2 5 - 0. 0 = 0. 2 5 \text { in. }
\beta_ {m} = 0. 0 - 0. 5 = - 0. 5 \text { in. } \quad \gamma_ {i} = 0. 7 5 - 1. 0 = - 0. 2 5 \text { in. }
\gamma_ {j} = 1. 0 - 0. 7 5 = 0. 2 5 \text { in. } \quad \gamma_ {m} = 1. 0 - 1. 0 = 0. 0 \text { in. }
and \bar{r}=0.9167 in. \bar{z}=0.25 in. A=0.0625 in ^{2}
Using Eqs. (9.2.11) in Eq. (9.2.2) and proceeding as for element 1, we obtain the stiffness matrix for element 2 as
\left[ k ^ {(2)} \right] = \left(1 0 ^ {6}\right) \left[ \begin{array}{c c c c c c} i = 2 & j = 3 & m = 5 \\ 8 5. 7 5 & - 4 6. 0 7 & 5 2. 5 2 & 1 2. 8 4 & - 1 1 8. 9 2 & 3 3. 2 3 \\ - 4 6. 0 7 & 7 4. 7 7 & - 1 2. 8 4 & - 4 1. 5 4 & 4 5. 3 2 & - 3 3. 2 3 \\ 5 2. 5 2 & - 1 2. 8 4 & 8 5. 7 4 & 4 6. 0 7 & - 1 1 8. 9 2 & - 3 3. 2 3 \\ 1 2. 8 4 & - 4 1. 5 4 & 4 6. 0 7 & 7 4. 7 7 & - 4 5. 3 2 & - 3 3. 2 3 \\ - 1 1 8. 9 2 & 4 5. 3 2 & - 1 1 8. 9 2 & - 4 5. 3 2 & 2 1 6. 4 1 & 0 \\ 3 3. 2 3 & - 3 3. 2 3 & - 3 3. 2 3 & - 3 3. 2 3 & 0 & 6 6. 4 6 \end{array} \right] \frac {\mathrm{lb}}{\text {in.}} \tag {9.2.12}
We obtain the stiffness matrices for elements 3 and 4 in a manner similar to that used to obtain the stiffness matrices for elements 1 and 2. Thus,
\left[ k ^ {(3)} \right] = \left(1 0 ^ {6}\right) \left[ \begin{array}{c c c c c c} 7 2. 5 8 & 3 8. 5 2 & - 3 1. 6 3 & 1 1. 3 3 & - 2 0. 3 1 & - 4 9. 8 4 \\ 3 8. 5 2 & 6 1. 1 7 & - 2. 2 6 & 3 3. 9 8 & - 2 2. 6 6 & - 9 5. 1 5 \\ - 3 1. 6 3 & - 2. 2 6 & 5 4. 4 6 & - 2 9. 4 5 & - 2 9. 3 7 & 3 1. 7 2 \\ 1 1. 3 3 & 3 3. 9 8 & - 2 9. 4 5 & 6 1. 1 7 & 3 1. 7 2 & - 9 5. 1 5 \\ - 2 0. 3 1 & - 2 2. 6 6 & - 2 9. 3 7 & 3 1. 7 2 & 5 6. 7 2 & - 9. 0 6 \\ - 4 9. 8 4 & - 9 5. 1 5 & 3 1. 7 2 & - 9 5. 1 5 & - 9. 0 6 & 1 9 0. 3 1 \end{array} \right] \frac {\mathrm{lb}}{\text {in.}} \tag {9.2.13}
and
\left[ k ^ {(4)} \right] = \left(1 0 ^ {6}\right) \left[ \begin{array}{c c c c c c} 4 1. 5 3 & - 2 1. 9 0 & 2 0. 3 9 & 0. 7 5 & - 6 6. 4 5 & 2 1. 1 4 \\ - 2 1. 9 0 & 4 7. 5 7 & - 0. 7 5 & - 2 6. 4 3 & 3 6. 2 4 & - 2 1. 1 4 \\ 2 0. 3 9 & - 0. 7 5 & 4 1. 5 3 & 2 1. 9 0 & - 6 6. 4 5 & - 2 1. 1 4 \\ 0. 7 5 & - 2 6. 4 3 & 2 1. 9 0 & 4 7. 5 7 & - 3 6. 2 4 & - 2 1. 1 4 \\ - 6 6. 4 5 & 3 6. 2 4 & - 6 6. 4 5 & - 3 6. 2 4 & 1 6 9. 1 4 & 0 \\ 2 1. 1 4 & - 2 1. 1 4 & - 2 1. 1 4 & - 2 1. 1 4 & 0 & 4 2. 2 8 \end{array} \right] \frac {\mathrm{lb}}{\text {in.}} \tag {9.2.14}
Using superposition of the element stiffness matrices [Eqs. (9.2.10) and (9.2.12)-(9.2.14)], where we rearrange the elements of each stiffness matrix in order of increasing nodal degrees of freedom, we obtain the global stiffness matrix as
[ K ] = (1 0 ^ {6}) \left[ \begin{array}{c c c c c c c c c c} 9 5. 9 9 & 5 1. 3 5 & - 3 1. 6 3 & 2. 2 6 & 0 & 0 & 2 0. 3 9 & - 0. 7 5 & - 9 5. 8 2 & - 5 2. 8 6 \\ 5 1. 3 5 & 1 0 8. 7 4 & - 1 1. 3 3 & 3 3. 9 8 & 0 & 0 & 0. 7 5 & - 2 6. 4 3 & - 6 7. 9 6 & - 1 1 6. 3 \\ - 3 1. 6 3 & - 1 1. 3 3 & 1 5 8. 3 4 & - 8 4. 5 9 & 5 2. 5 2 & 1 2. 8 4 & 0 & 0 & - 1 3 9. 2 & 8 3. 0 7 \\ 2. 2 6 & 3 3. 9 8 & - 8 4. 5 9 & 1 3 5. 9 4 & - 1 2. 8 4 & - 4 1. 5 4 & 0 & 0 & 6 7. 9 8 & - 1 2 8. 4 \\ 0 & 0 & 5 2. 5 2 & - 1 2. 8 4 & 1 5 8. 3 3 & 8 4. 5 9 & - 3 1. 6 3 & 1 1. 3 3 & - 1 3 9. 2 & - 8 3. 0 7 \\ 0 & 0 & 1 2. 8 4 & - 4 1. 5 4 & 8 4. 5 9 & 1 3 5. 9 4 & - 2. 2 6 & 3 3. 9 8 & - 6 7. 9 8 & - 1 2 8. 4 \\ 2 0. 3 9 & 0. 7 5 & 0 & 0 & - 3 1. 6 3 & - 2. 2 6 & 9 5. 9 9 & - 5 1. 3 5 & - 9 5. 8 2 & 5 2. 8 6 \\ - 0. 7 5 & - 2 6. 4 3 & 0 & 0 & 1 1. 3 3 & 3 3. 9 8 & - 5 1. 3 5 & 1 0 8. 7 4 & 6 7. 9 6 & - 1 1 6. 3 \\ - 9 5. 8 2 & - 6 7. 9 6 & - 1 3 9. 2 & 6 7. 9 8 & - 1 3 9. 2 & - 6 7. 9 8 & - 9 5. 8 2 & 6 7. 9 6 & 4 9 8. 9 9 & 0 \\ - 5 2. 8 6 & - 1 1 6. 3 & 8 3. 0 7 & - 1 2 8. 4 & - 8 3. 0 7 & - 1 2 8. 4 & 5 2. 8 6 & - 1 1 6. 3 & 0 & 4 8 9. 3 6 \end{array} \right] \frac {\mathrm{lb}}{\text {in.}} \tag {9.2.15}
The applied nodal forces are given by Eq. (9.1.36) as
F _ {1 r} = F _ {4 r} = \frac {2 \pi (0 . 5) (0 . 5)}{2} (1) = 0. 7 8 5 \mathrm{lb} \tag {9.2.16}
All other nodal forces are zero. Using Eq. (9.2.15) for [K] and Eq. (9.2.16) for the nodal forces in Eq. (9.2.1), and solving for the nodal displacements, we obtain
\begin{array}{l} u _ {1} = 0. 0 3 2 2 \times 1 0 ^ {- 6} \text { in. } \quad w _ {1} = 0. 0 0 1 1 5 \times 1 0 ^ {- 6} \text { in. } \\ u _ {2} = 0. 0 2 1 9 \times 1 0 ^ {- 6} \text { in. } \quad w _ {2} = 0. 0 0 2 0 6 \times 1 0 ^ {- 6} \text { in. } \\ u _ {3} = 0. 0 2 1 9 \times 1 0 ^ {- 6} \text { in. } \quad w _ {3} = - 0. 0 0 2 0 6 \times 1 0 ^ {- 6} \text { in. } \tag {9.2.17} \\ u _ {4} = 0. 0 3 2 2 \times 1 0 ^ {- 6} \text { in. } \quad w _ {4} = - 0. 0 0 1 1 5 \times 1 0 ^ {- 6} \text { in. } \\ u _ {5} = 0. 0 2 4 4 \times 1 0 ^ {- 6} \text { in. } \quad w _ {5} = 0 \\ \end{array}
The results for nodal displacements are as expected because radial displacements at the inner edge are equal (u_{1}=u_{4}) and those at the outer edge are equal (u_{2}=u_{3}) . In addition, the axial displacements at the outer nodes and inner nodes are equal but opposite in sign (w_{1}=-w_{4} and w_{2}=-w_{3}) as a result of the Poisson effect and symmetry. Finally, the axial displacement at the center node is zero (w_{5}=0) , as it should be because of symmetry.
By using Eq. (9.1.22), we now determine the stresses in each element as
\{\sigma \} = [ D ] [ \bar {B} ] \{d \} \tag {9.2.18}
For element 1, we use Eq. (9.2.5) for ½B , Eq. (9.2.8) for ½D , and Eq. (9.2.17) for fdg in Eq. (9.2.18) to obtain
\sigma_ {r} = - 0. 3 3 8 \mathrm{psi} \quad \sigma_ {z} = - 0. 0 1 2 6 \mathrm{psi}
\sigma_ {\theta} = 0. 9 4 2 \mathrm{psi} \quad \tau_ {r z} = - 0. 1 0 3 7 \mathrm{psi}
Similarly, for element 2, we obtain
\sigma_ {r} = - 0. 1 0 5 \mathrm{psi} \quad \sigma_ {z} = - 0. 0 7 4 7 \mathrm{psi}
\sigma_ {\theta} = 0. 6 9 0 \mathrm{psi} \quad \tau_ {r z} = 0. 0 0 0 \mathrm{psi}
For element 3, the stresses are
\sigma_ {r} = - 0. 3 3 7 \mathrm{psi} \quad \sigma_ {z} = - 0. 0 1 2 5 \mathrm{psi}
\sigma_ {\theta} = 0. 9 4 2 \mathrm{psi} \quad \tau_ {r z} = 0. 1 0 3 7 \mathrm{psi}
For element 4, the stresses are
\sigma_ {r} = - 0. 4 7 0 \mathrm{psi} \quad \sigma_ {z} = 0. 1 4 9 3 \mathrm{psi}
\sigma_ {\theta} = 1. 4 2 6 \mathrm{psi} \quad \tau_ {r z} = 0. 0 0 0 \mathrm{psi}
Figure 9–13 shows the exact solution [10] along with the results determined here and the results from Reference [5]. Observe that agreement with the exact solution is quite good except for the limited results due to the very coarse mesh used in the longhand example, and in case 1 of Reference [5]. In Reference [5], stresses have been plotted at the center of the quadrilaterals and were obtained by averaging the stresses in the four connecting triangles. 9
9.3 Applications of Axisymmetric Elements
Numerous structural (and nonstructural) systems can be classified as axisymmetric. Some typical structural systems whose behavior is modeled accurately using the axisymmetric element developed in this chapter are represented in Figures 9–14, 9–15, and 9–17.
Figure 9–14 illustrates the finite element model of a steel-reinforced concrete pressure vessel. The vessel is a thick-walled cylinder with flat heads. An axis of symmetry (the z axis) exists such that only one-half of the r-z plane passing through the middle of the structure need be modeled. The concrete was modeled by using the axisymmetric triangular element developed in this chapter. The steel elements were laid out along the boundaries of the concrete elements so as to maintain continuity
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Axis of symmetry p Case 1 Case 2 Case 3 r, u (a) Finite element models (from Reference [5])
line
| Radius | Exact solution [10] | Case 1 | Case 2 | Case 3 | Section 9.2 average-of-four-triangle stresses | Section 9.2 element centroid stresses |
|---|---|---|---|---|---|---|
| 0.5 | 1.8 | 1.6 | 1.5 | 1.4 | 1.4 | 1.4 |
| 0.6 | 1.6 | 1.4 | 1.3 | 1.2 | 1.2 | 1.2 |
| 0.7 | 1.4 | 1.2 | 1.1 | 1.0 | 1.0 | 1.0 |
| 0.8 | 1.2 | 1.0 | 0.9 | 0.8 | 0.8 | 0.8 |
| 0.9 | 1.0 | 0.8 | 0.7 | 0.6 | 0.6 | 0.6 |
| 1.0 | 0.8 | 0.6 | 0.5 | 0.4 | 0.4 | 0.4 |
Figure 9–13 Finite element analysis of a thick-walled cylinder under internal pressure
(or perfect bond assumption) between the concrete and the steel. The vessel was then subjected to an internal pressure as shown in the figure. Note that the nodes along the axis of symmetry should be supported by rollers preventing motion perpendicular to the axis of symmetry.
Figure 9–15 shows a finite element model of a high-strength steel die used in a thin-plastic-film-making process [7]. The die is an irregularly shaped disk. An axis of symmetry with respect to geometry and loading exists as shown. The die was modeled by using simple quadrilateral axisymmetric elements. The locations of high stress were
heatmap
| E value | Value (psi) |
|---|---|
| ν_conc. | 0.15 |
| ν_steel | 0.25 |
| E_conc. | 5 × 10⁶ psi |
| E_liner | 30 × 10⁶ psi |
| E_reinf. | 30 × 10⁶ psi |
| E_rest. | 28 × 10⁶ psi |
| E_reduced | 3/3 E_conc. |
(a） Two-dimensional view of a finite element idealization for a prestressed concrete reactor vessel (PCRV)
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Concrete MERIDIONAL STEEL REINFORCEMENT Circumferential steel reinforcement
(b）Axisymmetric idealization of the steel reinforcement
Figure 9–14 Model of steel-reinforced concrete pressure vessel (from Reference [4], North Holland Physics Publishing, Amsterdam)
of primary concern. Figure 9–16 shows a plot of the von Mises stress contours for the die of Figure 9–15. The von Mises (or equivalent, or effective) stress [8] is often used as a failure criterion in design. Notice the artificially high stresses at the location of load F as explained in Section 7.1.
(Recall that the failure criterion based on the maximum distortion energy theory for ductile materials subjected to static loading predicts that a material will fail if the von Mises stress reaches the yield strength of the material.) Also recall from Eqs. (6.5.37) and (6.5.38), the von Mises stress \sigma _ { v m } is related to the principal stresses by the expression
\sigma_ {v m} = \frac {1}{\sqrt {2}} \sqrt {\left(\sigma_ {1} - \sigma_ {2}\right) ^ {2} + \left(\sigma_ {2} - \sigma_ {3}\right) ^ {2} + \left(\sigma_ {3} - \sigma_ {1}\right) ^ {2}} \tag {9.3.1}
other
| Dimension | Value |
|---|---|
| Fixed edge | 2.646 in |
| Total Width | 2.75 in |
| Total Height | 1.126 in |
| Fixed Edge Width | 0.25 in |
| Fixed Edge Width | 0.50 in |
| Fixed Edge Width | 0.875 in |
| Fixed Edge Width | 1.0 in |
| Total Width | 45,750 lb |
| Total Height | 1.126 in |
Figure 9–15 Model of a high-strength steel die (924 nodes and 830 elements)
Axis of symmetry
other
| Peak Position (psi) |
|---|
| 20,712 |
| 33,779 |
| 46,846 |
| 59,914 |
| 72,981 |
| 86,049 |
| 99,116 |
| 7644 |
Figure 9–16 von Mises stress contour plot of axisymmetric model of Figure 9–15 (also producing a radial inward deflection of about 0.015 in.)
Figure 9–17 (a) Stepped shaft subjected to axial load and (b) the discretized model
where the principal stresses are given by \sigma _ { 1 } , \sigma _ { 2 } , and \sigma _ { 3 } . These results were obtained from the commercial computer code ANSYS [12].
Other dies with modifications in geometry were also studied to evaluate the most suitable die before the construction of an expensive prototype. Confidence in the acceptability of the prototype was enhanced by doing these comparison studies. Finally, Figure 9–17 shows a stepped 4130 steel shaft with a fillet radius subjected to an axial pressure of 1000 psi in tension. Fatigue analysis for reversed axial loading required an accurate stress concentration factor to be applied to the average axial stress of 1000 psi. The stress concentration factor for the geometry shown was to be determined. Therefore, locations of highest stress were necessary. Figure 9–18 shows the resulting maximum principal stress plot using a computer program [11]. The largest principal stress was 1932.5 psi at the fillet. Other examples of the use of the axisymmetric element can be found in References [2]–[6].
In this chapter, we have shown the finite element analysis of axisymmetric systems using a simple three-noded triangular element to be analogous to that of the two-dimensional plane stress problem using three-noded triangular elements as