김경종
24 KiB
The force matrix due to the pressure applied alongside j-m is determined as follows:
L _ {j - m} = \left[ (2 - 1) ^ {2} + (3 - 0) ^ {2} \right] ^ {1 / 2} = 3. 1 6 3 \text { in. }
p _ {x} = p \cos \theta = 2 0 0 0 \left(\frac {3}{3 . 1 6 3}\right) = 1 8 9 6 \mathrm{lb} / \mathrm{in} ^ {2} \tag {15.1.78}
p _ {y} = p \sin \theta = 2 0 0 0 \left(\frac {1}{3 . 1 6 3}\right) = 6 3 2 \mathrm{lb} / \mathrm{in} ^ {2}
where y is the angle measured from the x axis to the normal to surface j-m. Using Eq. (6.3.7) to evaluate the surface forces, we have
\left\{f _ {L} \right\} = \iint_ {S _ {j - m}} \left[ N _ {s} \right] ^ {T} \left\{ \begin{array}{c} p _ {x} \\ p _ {y} \end{array} \right\} d S
= \iint_ {S _ {j - m}} \left[ \begin{array}{c c} N _ {i} & 0 \\ 0 & N _ {i} \\ N _ {j} & 0 \\ 0 & N _ {j} \\ N _ {m} & 0 \\ 0 & N _ {m} \end{array} \right] \left\{ \begin{array}{l} p _ {x} \\ p _ {y} \end{array} \right\} d S = \frac {t L _ {j - m}}{2} \left[ \begin{array}{c c} 0 & 0 \\ 0 & 0 \\ 1 & 0 \\ 0 & 1 \\ 1 & 0 \\ 0 & 1 \end{array} \right] \left\{ \begin{array}{l} p _ {x} \\ p _ {y} \end{array} \right\} \tag {15.1.79}
Evaluating Eq. (15.1.79), we obtain
\left\{f _ {L} \right\} = \frac {(1 \text { in. }) (3 . 1 6 3 \text { in. })}{2} \left[ \begin{array}{l l} 0 & 0 \\ 0 & 0 \\ 1 & 0 \\ 0 & 1 \\ 1 & 0 \\ 0 & 1 \end{array} \right] \left\{ \begin{array}{l} 1 8 9 6 \\ 6 3 2 \end{array} \right\} = \left\{ \begin{array}{c} 0 \\ 0 \\ 3 0 0 0 \\ 1 0 0 0 \\ 3 0 0 0 \\ 1 0 0 0 \end{array} \right\} \text { lb } \tag {15.1.80}
Using Eqs. (15.1.76), (15.1.77), and (15.1.80), we find that the complete set of element equations is
\frac {1 \times 1 0 ^ {6}}{3} \left[ \begin{array}{c c c c c c} 7 5 & 1 5 & - 6 9 & - 3 & - 6 & - 1 2 \\ & 3 5 & 3 & - 1 9 & - 1 8 & - 1 6 \\ & & 7 5 & - 1 5 & - 6 & 1 2 \\ & & & 3 5 & 1 8 & - 1 6 \\ & & & & 1 2 & 0 \\ \text { Symmetry } & & & & & 3 2 \end{array} \right] \left\{ \begin{array}{l} u _ {i} \\ v _ {i} \\ u _ {j} \\ v _ {j} \\ u _ {m} \\ v _ {m} \end{array} \right\} = \left\{ \begin{array}{l} - 1 2, 6 0 0 \\ - 4 2 0 0 \\ 1 5, 6 0 0 \\ - 3 2 0 0 \\ 3 0 0 0 \\ 9 4 0 0 \end{array} \right\} \tag {15.1.81}
where the force matrix is \{ f _ { T } \} + \{ f _ { L } \} , obtained by adding Eqs. (15.1.77) and (15.1.80).
For the plane stress plate fixed along one edge and subjected to a uniform temperature rise of 5 0 ^ { \circ } \mathrm { C } as shown in Figure 15–11, determine the nodal displacements and the stresses in each element. Let E ¼ 210 GPa, n ¼ 0:30, t ¼ 5 mm, and \alpha = 1 2 \times 1 0 ^ { - 6 } \mathrm { ( m m / m m ) / ^ { \circ } C } .
The discretized plate is shown in Figure 15–11. We begin by evaluating the stiffness matrix of each element using Eq. (6.2.52).
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4 3 500 mm 1 2 1 4 ③ ② ① 500 mm
Figure 15–11 Discretized plate subjected to a temperature change
Element 1
Element 1 has coordinates x1 ¼ 0, y1 ¼ 0, x2 ¼ 0:5, y2 ¼ 0, x5 ¼ 0:25, and y _ { 5 } = 0 . 2 5 . From Eqs. (6.2.10), we obtain
\begin{array}{l} \beta_ {1} = y _ {2} - y _ {5} = - 0. 2 5 \mathrm{m} \quad \beta_ {2} = y _ {5} - y _ {1} = 0. 2 5 \mathrm{m} \quad \beta_ {5} = y _ {1} - y _ {2} = 0 \\ \gamma_ {1} = x _ {5} - x _ {2} = - 0. 2 5 \mathrm{m} \quad \gamma_ {2} = x _ {1} - x _ {5} = - 0. 2 5 \mathrm{m} \quad \gamma_ {5} = x _ {2} - x _ {1} = 0. 5 \mathrm{m} \tag {15.1.82} \\ \end{array}
Using Eqs. (6.2.32) in Eq. (6.2.34), we have
\begin{array}{l} [ B ] = \frac {1}{2 A} \left[ \begin{array}{c c c c c c} \beta_ {1} & 0 & \beta_ {2} & 0 & \beta_ {5} & 0 \\ 0 & \gamma_ {1} & 0 & \gamma_ {2} & 0 & \gamma_ {5} \\ \gamma_ {1} & \beta_ {1} & \gamma_ {2} & \beta_ {2} & \gamma_ {5} & \beta_ {5} \end{array} \right] \\ = \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. 2 5 & - 0. 2 5 & - 0. 2 5 & 0. 2 5 & 0. 5 & 0 \end{array} \right] \frac {1}{\mathrm{m}} \tag {15.1.83} \\ \end{array}
For plane stress, ½D is given by
\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] = \frac {2 1 0 \times 1 0 ^ {9}}{0 . 9 1} \left[ \begin{array}{c c c} 1 & 0. 3 & 0 \\ 0. 3 & 1 & 0 \\ 0 & 0 & 0. 3 5 \end{array} \right] \frac {\mathrm{N}}{\mathrm{m} ^ {2}} \tag {15.1.84}
We obtain the element stiffness matrix using
[ k ] = t A [ B ] ^ {T} [ D ] [ B ] \tag {15.1.85}
Substituting the results of Eqs. (15.1.83) and (15.1.84) into Eq. (15.1.85) and carrying out the multiplications, we have
\underline {{{k}}} = 4. 6 1 5 \times 1 0 ^ {7} \left[ \begin{array}{c c c c c c} d _ {1 x} & d _ {1 y} & d _ {2 x} & d _ {2 y} & d _ {5 x} & d _ {5 y} \\ 8. 4 3 7 5 & 4. 0 6 2 5 & - 4. 0 6 2 5 & - 0. 3 1 2 5 & - 4. 3 7 5 & - 3. 7 5 \\ 4. 0 6 2 5 & 8. 4 3 7 5 & 0. 3 1 2 5 & 4. 0 6 2 5 & - 4. 3 7 5 & - 1 2. 5 \\ - 4. 0 6 2 5 & 0. 3 1 2 5 & 8. 4 3 7 5 & - 4. 0 6 2 5 & - 4. 3 7 5 & 3. 7 5 \\ - 0. 3 1 2 5 & 4. 0 6 2 5 & - 4. 0 6 2 5 & 8. 4 3 7 5 & 4. 3 7 5 & - 1 2. 5 \\ - 4. 3 7 5 & - 4. 3 7 5 & - 4. 3 7 5 & 4. 3 7 5 & 8. 7 5 & 0 \\ - 3. 7 5 & - 1 2. 5 & 3. 7 5 & - 1 2. 5 & 0 & 2 5 \end{array} \right] \frac {\mathrm{N}}{\mathrm{m}} \tag {15.1.86}
Element 2
For element 2, the coordinates are x_{2} = 0.5 , y_{2} = 0 , x_{3} = 0.5 , y_{3} = 0.5 , x_{5} = 0.25 , and y_{5} = 0.25 . Proceeding as for element 1, we obtain
\beta_ {2} = 0. 2 5 \mathrm{m} \quad \beta_ {3} = 0. 2 5 \mathrm{m} \quad \beta_ {5} = - 0. 5 \mathrm{m}
\gamma_ {2} = - 0. 2 5 \mathrm{m} \quad \gamma_ {3} = 0. 2 5 \mathrm{m} \quad \gamma_ {5} = 0
The element stiffness matrix then becomes
\underline {{k}} = 4. 6 1 5 \times 1 0 ^ {7} \left[ \begin{array}{c c c c c c} d _ {2 x} & d _ {2 y} & d _ {3 x} & d _ {3 y} & d _ {5 x} & d _ {5 y} \\ 8. 4 3 7 5 & - 4. 0 6 2 5 & 4. 0 6 2 5 & - 0. 3 1 2 5 & - 1 2. 5 & 4. 3 7 5 \\ - 4. 0 6 2 5 & 8. 4 3 7 5 & 0. 3 1 2 5 & - 4. 0 6 2 5 & 3. 7 5 & - 4. 3 7 5 \\ 4. 0 6 2 5 & 0. 3 1 2 5 & 8. 4 3 7 & 4. 0 6 2 5 & - 1 2. 5 & - 4. 3 7 5 \\ - 0. 3 1 2 5 & - 4. 0 6 2 5 & 4. 0 6 2 5 & 8. 4 3 7 5 & - 3. 7 5 & - 4. 3 7 5 \\ - 1 2. 5 & 3. 7 5 & - 1 2. 5 & - 3. 7 5 & 2 5 & 0 \\ 4. 3 7 5 & - 4. 3 7 5 & - 4. 3 7 5 & - 4. 3 7 5 & 0 & 8. 7 5 \end{array} \right] \frac {\mathrm{N}}{\mathrm{m}} \tag {15.1.87}
Element 3
For element 3, using the same steps as for element 1, we obtain the stiffness matrix as
\underline {{{k}}} = 4. 6 1 5 \times 1 0 ^ {7} \left[ \begin{array}{c c c c c c} d _ {3 x} & d _ {3 y} & d _ {4 x} & d _ {4 y} & d _ {5 x} & d _ {5 y} \\ 8. 4 3 7 & 4. 0 6 2 5 & - 4. 0 6 2 5 & - 0. 3 1 2 5 & - 4. 3 7 5 & - 3. 7 5 \\ 4. 0 6 2 5 & 8. 4 3 7 & 0. 3 1 2 5 & 4. 0 6 2 5 & - 4. 3 7 5 & - 1 2. 5 \\ - 4. 0 6 2 5 & 0. 3 1 2 5 & 8. 4 3 7 & - 4. 0 6 2 5 & - 4. 3 7 5 & 3. 7 5 \\ - 0. 3 1 2 5 & 4. 0 6 2 5 & - 4. 0 6 2 5 & 8. 4 3 7 5 & 4. 3 7 5 & - 1 2. 5 \\ - 4. 3 7 5 & - 4. 3 7 5 & - 4. 3 7 5 & 4. 3 7 5 & 8. 7 5 & 0 \\ - 3. 7 5 & - 1 2. 5 & 3. 7 5 & - 1 2. 5 & 0 & 2 5 \end{array} \right] \frac {\mathrm{N}}{\mathrm{m}} \tag {15.1.88}
Element 4
Finally, for element 4, we obtain
\underline {{{k}}} = 4. 6 1 5 \times 1 0 ^ {7} \left[ \begin{array}{c c c c c c} d _ {4 x} & d _ {4 y} & d _ {1 x} & d _ {1 y} & d _ {5 x} & d _ {5 y} \\ 8. 4 3 7 & - 4. 0 6 2 5 & 4. 0 6 2 5 & - 0. 3 1 2 5 & - 1 2. 5 & 4. 3 7 5 \\ - 4. 0 6 2 5 & 8. 4 3 7 5 & 0. 3 1 2 5 & - 4. 0 6 2 5 & 3. 7 5 & - 4. 3 7 5 \\ 4. 0 6 2 5 & 0. 3 1 2 5 & 8. 4 3 7 & 4. 0 6 2 5 & - 1 2. 5 & - 4. 3 7 5 \\ - 0. 3 1 2 5 & - 4. 0 6 2 5 & 4. 0 6 2 5 & 8. 4 3 1 & - 3. 7 5 & - 4. 3 7 5 \\ - 1 2. 5 & 3. 7 5 & - 1 2. 5 & - 3. 7 5 & 2 5 & 0 \\ 4. 3 7 5 & - 4. 3 7 5 & - 4. 3 7 5 & - 4. 3 7 5 & 0 & 8. 7 5 \end{array} \right] \frac {\mathrm{N}}{\mathrm{m}} \tag {15.1.89}
Using the direct stiffness method, we assemble the element stiffness matrices, Eqs. (15.1.86)-(15.1.89), to obtain the global stiffness matrix as
\underline {{{K}}} = 4. 6 1 5 \times 1 0 ^ {7} \left[ \begin{array}{c c c c} d _ {1 x} & d _ {1 y} & d _ {2 x} & d _ {2 y} \\ 1 6. 8 7 4 & 8. 1 2 5 & - 4. 0 6 2 5 & - 0. 3 1 2 5 \\ 8. 1 2 5 & 1 6. 8 7 4 & 0. 3 1 2 5 & 4. 0 6 2 5 \\ - 4. 0 6 2 5 & 0. 3 1 2 5 & 1 6. 8 7 4 & - 8. 1 2 5 \\ - 0. 3 1 2 5 & 4. 0 6 2 5 & - 8. 1 2 5 & 1 6. 8 7 5 \\ 0 & 0 & 4. 0 6 2 5 & 0. 3 1 2 5 \\ 0 & 0 & - 0. 3 1 2 5 & - 4. 0 6 2 5 \\ 4. 0 6 2 5 & - 0. 3 1 2 5 & 0 & 0 \\ 0. 3 1 2 5 & - 4. 0 6 2 5 & 0 & 0 \\ - 1 6. 8 7 5 & - 8. 1 2 5 & - 1 6. 8 7 5 & 8. 1 2 5 \\ - 8. 1 2 5 & - 1 6. 8 7 5 & 8. 1 2 5 & - 1 6. 8 7 5 \end{array} \right]
\left. \begin{array}{c c c c c c} d _ {3 x} & d _ {3 y} & d _ {4 x} & d _ {4 y} & d _ {5 x} & d _ {5 y} \\ 0 & 0 & 4. 0 6 2 5 & 0. 3 1 2 5 & - 1 6. 8 7 5 & - 8. 1 2 5 \\ 0 & 0 & - 0. 3 1 2 5 & - 4. 0 6 2 5 & - 8. 1 2 5 & - 1 6. 8 7 5 \\ 4. 0 6 2 5 & - 0. 3 1 2 5 & 0 & 0 & - 1 6. 8 7 5 & 8. 1 2 5 \\ 0. 3 1 2 5 & - 4. 0 6 2 5 & 0 & 0 & 8. 1 2 5 & - 1 6. 8 7 5 \\ 1 6. 8 7 5 & 8. 1 2 5 & - 4. 0 6 2 5 & - 0. 3 1 2 5 & - 1 6. 8 7 5 & - 8. 1 2 5 \\ 8. 1 2 5 & 1 6. 8 7 5 & 0. 3 1 2 5 & 4. 0 6 2 5 & - 8. 1 2 5 & - 1 6. 8 7 5 \\ - 4. 0 6 2 5 & 0. 3 1 2 5 & 1 6. 8 7 5 & - 8. 1 2 5 & - 1 6. 8 7 5 & 8. 1 2 5 \\ - 0. 3 1 2 5 & 4. 0 6 2 5 & - 8. 1 2 5 & 1 6. 8 7 5 & 8. 1 2 5 & - 1 6. 8 7 5 \\ - 1 6. 8 7 5 & - 8. 1 2 5 & - 1 6. 8 7 5 & 8. 1 2 5 & 6 7. 5 & 0 \\ - 8. 1 2 5 & - 1 6. 8 7 5 & 8. 1 2 5 & - 1 6. 8 7 5 & 0 & 6 7. 5 \end{array} \right] \frac {\mathrm{N}}{\mathrm{m}} \tag {15.1.90}
Next, we determine the thermal force matrices for each element by using Eq. (15.1.25) as follows:
Element 1
\begin{array}{l} \left\{f _ {T} \right\} = \frac {\alpha E t T}{2 (1 - \nu)} \left\{ \begin{array}{l} \beta_ {1} \\ \gamma_ {1} \\ \beta_ {2} \\ \gamma_ {2} \\ \beta_ {5} \\ \gamma_ {5} \end{array} \right\} = \frac {(1 2 \times 1 0 ^ {- 6}) (2 1 0 \times 1 0 ^ {9}) (0 . 0 0 5 \mathrm{m}) (5 0)}{2 (1 - 0 . 3)} \left\{ \begin{array}{c} - 0. 2 5 \\ - 0. 2 5 \\ 0. 2 5 \\ - 0. 2 5 \\ 0 \\ 0. 5 \end{array} \right\} \\ = 4 5 0, 0 0 0 \left\{ \begin{array}{c} - 0. 2 5 \\ - 0. 2 5 \\ 0. 2 5 \\ - 0. 2 5 \\ 0 \\ 0. 5 \end{array} \right\} = \left\{ \begin{array}{c} f _ {T 1 x} \\ f _ {T 1 y} \\ f _ {T 2 x} \\ f _ {T 2 y} \\ f _ {T 5 x} \\ f _ {T 5 y} \end{array} \right\} = \left\{ \begin{array}{c} - 1 1 2, 5 0 0 \\ - 1 1 2, 5 0 0 \\ 1 1 2, 5 0 0 \\ - 1 1 2, 5 0 0 \\ 0 \\ 2 2 5, 0 0 0 \end{array} \right\} \mathrm{N} \tag {15.1.91} \\ \end{array}
Element 2
\left\{f _ {T} \right\} = 4 5 0, 0 0 0 \left\{ \begin{array}{r} 0. 2 5 \\ - 0. 2 5 \\ 0. 2 5 \\ 0. 2 5 \\ - 0. 5 \\ 0 \end{array} \right\} = \left\{ \begin{array}{r} f _ {T 2 x} \\ f _ {T 2 y} \\ f _ {T 3 x} \\ f _ {T 3 y} \\ f _ {T 5 x} \\ f _ {T 5 y} \end{array} \right\} = \left\{ \begin{array}{r} 1 1 2, 5 0 0 \\ - 1 1 2, 5 0 0 \\ 1 1 2, 5 0 0 \\ 1 1 2, 5 0 0 \\ - 2 2 5, 0 0 0 \\ 0 \end{array} \right\} \mathrm{N} \tag {15.1.92}
Element 3
\left\{f _ {T} \right\} = 4 5 0, 0 0 0 \left\{ \begin{array}{c} 0. 2 5 \\ 0. 2 5 \\ - 0. 2 5 \\ 0. 2 5 \\ 0 \\ - 0. 5 \end{array} \right\} = \left\{ \begin{array}{c} f _ {T 3 x} \\ f _ {T 3 y} \\ f _ {T 4 x} \\ f _ {T 4 y} \\ f _ {T 5 x} \\ f _ {T 5 y} \end{array} \right\} = \left\{ \begin{array}{c} 1 1 2, 5 0 0 \\ 1 1 2, 5 0 0 \\ - 1 1 2, 5 0 0 \\ 1 1 2, 5 0 0 \\ 0 \\ - 2 2 5, 0 0 0 \end{array} \right\} \mathrm{N} \tag {15.1.93}
Element 4
\left\{f _ {T} \right\} = 4 5 0, 0 0 0 \left\{ \begin{array}{c} - 0. 2 5 \\ 0. 2 5 \\ - 0. 2 5 \\ - 0. 2 5 \\ 0. 5 \\ 0 \end{array} \right\} = \left\{ \begin{array}{c} f _ {T 4 x} \\ f _ {T 4 y} \\ f _ {T 1 x} \\ f _ {T 1 y} \\ f _ {T 5 x} \\ f _ {T 5 y} \end{array} \right\} = \left\{ \begin{array}{c} - 1 1 2, 5 0 0 \\ 1 1 2, 5 0 0 \\ - 1 1 2, 5 0 0 \\ - 1 1 2, 5 0 0 \\ 2 2 5, 0 0 0 \\ 0 \end{array} \right\} \mathrm{N} \tag {15.1.94}
We then obtain the global thermal force matrix by direct assemblage of the element force matrices (Eqs. (15.1.91)–(15.1.94)). The resulting matrix is
\left\{ \begin{array}{l} f _ {T 1 x} \\ f _ {T 1 y} \\ f _ {T 2 x} \\ f _ {T 2 y} \\ f _ {T 3 x} \\ f _ {T 3 y} \\ f _ {T 4 x} \\ f _ {T 4 y} \\ f _ {T 5 x} \\ f _ {T 5 y} \end{array} \right\} = \left\{ \begin{array}{c} - 2 2 5, 0 0 0 \\ - 2 2 5, 0 0 0 \\ 2 2 5, 0 0 0 \\ - 2 2 5, 0 0 0 \\ 2 2 5, 0 0 0 \\ 2 2 5, 0 0 0 \\ - 2 2 5, 0 0 0 \\ 2 2 5, 0 0 0 \\ 0 \\ 0 \end{array} \right\} \mathrm{N} \tag {15.1.95}
Using Eqs. (15.1.90) and (15.1.95) and imposing the boundary conditions d _ { 1 x } = d _ { 1 y } = d _ { 4 x } = d _ { 4 y } = 0 , we obtain the system of equations for solution as
\left\{ \begin{array}{l} f _ {T 2 x} = 2 2 5, 0 0 0 \\ f _ {T 2 y} = - 2 2 5, 0 0 0 \\ f _ {T 3 x} = 2 2 5, 0 0 0 \\ f _ {T 3 y} = 2 2 5, 0 0 0 \\ f _ {T 5 x} = 0 \\ f _ {T 5 y} = 0 \end{array} \right\} = 4. 6 1 5 \times 1 0 ^ {7}
\left[ \begin{array}{c c c c c c} 1 6. 8 7 4 & - 8. 1 2 5 & 4. 0 6 2 5 & - 0. 3 1 2 5 & - 1 6. 8 7 5 & 8. 1 2 5 \\ - 8. 1 2 5 & 1 6. 8 7 5 & 0. 3 1 2 5 & - 4. 0 6 2 5 & 8. 1 2 5 & - 1 6. 8 7 5 \\ 4. 0 6 2 5 & 0. 3 1 2 5 & 1 6. 8 7 5 & 8. 1 2 5 & - 1 6. 8 7 5 & - 8. 1 2 5 \\ - 0. 3 1 2 5 & - 4. 0 6 2 5 & 8. 1 2 5 & 1 6. 8 7 5 & - 8. 1 2 5 & - 1 6. 8 7 5 \\ - 1 6. 8 7 5 & 8. 1 2 5 & - 1 6. 8 7 5 & - 8. 1 2 5 & 6 7. 5 & 0 \\ 8. 1 2 5 & - 1 6. 8 7 5 & - 8. 1 2 5 & - 1 6. 8 7 5 & 0 & 6 7. 5 \end{array} \right] \left\{ \begin{array}{l} d _ {2 x} \\ d _ {2 y} \\ d _ {3 x} \\ d _ {3 y} \\ d _ {5 x} \\ d _ {5 y} \end{array} \right\} \tag {15.1.96}
Solving Eq. (15.1.96) for the nodal displacements, we have
\left\{ \begin{array}{l} d _ {2 x} \\ d _ {2 y} \\ d _ {3 x} \\ d _ {3 y} \\ d _ {5 x} \\ d _ {5 y} \end{array} \right\} = \left\{ \begin{array}{c} 3. 3 2 7 \times 1 0 ^ {- 4} \\ - 1. 9 1 1 \times 1 0 ^ {- 4} \\ 3. 3 2 7 \times 1 0 ^ {- 4} \\ 1. 9 1 1 \times 1 0 ^ {- 4} \\ 2. 1 2 3 \times 1 0 ^ {- 4} \\ 6. 6 5 4 \times 1 0 ^ {- 9} \end{array} \right\} \mathrm{m} \tag {15.1.97}
We now use Eq. (15.1.64) to obtain the stresses in each element. Using Eqs. (6.2.36) and (15.1.65), we write Eq. (15.1.64) as
\{\sigma \} = [ D ] [ B ] \{d \} - [ D ] \{\varepsilon_ {T} \} \tag {15.1.98}
Element 1
\begin{array}{l} \left\{ \begin{array}{l} \sigma_ {x} \\ \sigma_ {y} \\ \tau_ {x y} \end{array} \right\} = \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] \frac {1}{2 A} \left\{ \begin{array}{c c c c c c} \beta_ {1} & 0 & \beta_ {2} & 0 & \beta_ {5} & 0 \\ 0 & \gamma_ {1} & 0 & \gamma_ {2} & 0 & \gamma_ {5} \\ \gamma_ {1} & \beta_ {1} & \gamma_ {2} & \beta_ {2} & \gamma_ {5} & \beta_ {5} \end{array} \right\} \left\{ \begin{array}{l} d _ {1 x} \\ d _ {1 y} \\ d _ {2 x} \\ d _ {2 y} \\ d _ {5 x} \\ d _ {5 y} \end{array} \right\} \\ - \frac {E}{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} \alpha T \\ \alpha T \\ 0 \end{array} \right\} \tag {15.1.99} \\ \end{array}
Using Eqs. (15.1.82) and (15.1.97) along with the mechanical properties E, n, and a in Eq. (15.1.99), we obtain
\left\{ \begin{array}{l} \sigma_ {x} \\ \sigma_ {y} \\ \tau_ {x y} \end{array} \right\} = \frac {2 1 0 \times 1 0 ^ {9}}{0 . 9 1} \left[ \begin{array}{c c c} 1 & 0. 3 & 0 \\ 0. 3 & 1 & 0 \\ 0 & 0 & 0. 3 5 \end{array} \right]
\times \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. 2 5 & - 0. 2 5 & - 0. 2 5 & 0. 2 5 & 0. 5 & 0 \end{array} \right] \left\{ \begin{array}{c} 0 \\ 0 \\ 3. 3 2 7 \times 1 0 ^ {- 4} \\ - 1. 9 1 1 \times 1 0 ^ {- 4} \\ 2. 1 2 3 \times 1 0 ^ {- 4} \\ 6. 6 5 4 \times 1 0 ^ {- 9} \end{array} \right\}
- \frac {2 1 0 \times 1 0 ^ {9}}{0 . 9 1} \left[ \begin{array}{c c c} 1 & 0. 3 & 0 \\ 0. 3 & 1 & 0 \\ 0 & 0 & 0. 3 5 \end{array} \right] \left\{ \begin{array}{c} (1 2 \times 1 0 ^ {- 6}) (5 0) \\ (1 2 \times 1 0 ^ {- 6}) (5 0) \\ 0 \end{array} \right\} \tag {15.1.100}
Simplifying Eq. (15.1.100) yields
\left\{ \begin{array}{l} \sigma_ {x} \\ \sigma_ {y} \\ \tau_ {x y} \end{array} \right\} = \left\{ \begin{array}{c} 1. 8 0 0 \times 1 0 ^ {8} \\ 1. 3 4 2 \times 1 0 ^ {8} \\ - 1. 6 0 0 \times 1 0 ^ {7} \end{array} \right\} - \left\{ \begin{array}{c} 1. 8 \times 1 0 ^ {8} \\ 1. 8 \times 1 0 ^ {8} \\ 0 \end{array} \right\} = \left\{ \begin{array}{c} 0 \\ - 4. 5 7 \times 1 0 ^ {7} \\ - 1. 6 0 \times 1 0 ^ {7} \end{array} \right\} \mathrm{Pa} \tag {15.1.101}
heatmap
| Load Case | Maximum Value (N/(m²)) |
|---|---|
| 1 of 1 | 3.06549e+007 |
| 1 of 1 | -6.0536e-009 |
Figure 15–12 Discretized plate showing displaced plate superimposed with maximum principal stress plot in Pa
Similarly, we obtain the stresses in the other elements as follows:
Element 2
\left\{ \begin{array}{l} \sigma_ {x} \\ \sigma_ {y} \\ \tau_ {x y} \end{array} \right\} = \left\{ \begin{array}{c} 1. 6 4 0 \times 1 0 ^ {8} \\ 2. 0 9 7 \times 1 0 ^ {8} \\ - 2 1 5 0 \end{array} \right\} - \left\{ \begin{array}{c} 1. 8 \times 1 0 ^ {8} \\ 1. 8 \times 1 0 ^ {8} \\ 0 \end{array} \right\} = \left\{ \begin{array}{c} - 1. 6 \times 1 0 ^ {7} \\ 2. 9 7 3 \times 1 0 ^ {7} \\ - 2 1 5 0 \end{array} \right\} \mathrm{Pa} \tag {15.1.102}
The clamped plate subjected to uniform heating (see the longhand solution, Example 15.5) was also solved using the Algor computer program from Reference [1]. The plate was discretized using the ‘‘automesh’’ feature of [1]. These results are similar to those obtained from the longhand solution of Example 15.5 using the very coarse mesh. The computer program solution with 342 elements is naturally more accurate than the longhand solution with only 4 elements. Figure 15–12 shows the discretized plate with resulting displacement superimposed on the maximum principal stress plot. 9
Reference
[1] Linear Stress and Dynamics Reference Division, Docutech On-line Documentation, Algor, Inc., Pittsburgh, PA.
Problems
15.1 For the one-dimensional steel bar fixed at the left end, free at the right end, and subjected to a uniform temperature rise T = 5 0 ^ { \circ } \mathrm { F } as shown in Figure P15–1, determine the free-end displacement, the displacement 60 in. from the fixed end, the reactions at the fixed end, and the axial stress. Let E = 3 0 \times 1 0 ^ { 6 } \mathrm { p s i } , A = 4 \mathrm { i n } ^ { 2 } , and \alpha = 7 . 0 \times 1 0 ^ { - 6 } ( \mathrm { i n . / i n . } ) / ^ { \circ } \mathrm { F } .
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T = 50°F 120 in.
Figure P15–1
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T = -20°C 3 m
Figure P15–2
15.2 For the one-dimensional steel bar fixed at each end and subjected to a uniform temperature drop of T = 2 0 ^ { \circ } \mathrm { C } as shown in Figure P15–2, determine the reactions at the fixed ends and the stress in the bar. Let E ¼ 210 GPa, A = 1 \times 1 0 ^ { - 2 } \mathrm { m } ^ { 2 } ; , and \alpha = 1 1 . 7 \times 1 0 ^ { - 6 } (mm/mm)/ C.
15.3 For the plane truss shown in Figure P15–3, bar element 2 is subjected to a uniform temperature rise of T = 5 0 ^ { \circ } \mathrm { F } . Let E = 3 0 \times 1 0 ^ { 6 } psi, A = 2 ~ \mathrm { i n } ^ { 2 } , and \alpha = 7 . 0 \times 1 0 ^ { - 6 } ( \mathrm { i n . / i n . } ) / ^ { \circ } \mathrm { F } . The lengths of the truss elements are shown in the figure. Determine the stresses in each bar. [Hint: See Eqs. (3.6.4) and (3.6.6) in Example 3.5 for the global and reduced \underline { { K } } matrices.]
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10 ft ① 45° ② 45° ③ 10 ft 3 4
Figure P15–3
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10 ft 4 3 2 ③ ② ① 30° 30° 1
Figure P15–4
15.4 For the plane truss shown in Figure P15–4, bar element 1 is subjected to a uniform temperature rise of 3 0 ^ { \circ } \mathrm { F } . Let E = 3 0 \times 1 0 ^ { 6 } psi, A = 2 \ \mathrm { i n } ^ { 2 } . , and \alpha = 7 . 0 \times 1 0 ^ { - 6 } (in./ \mathrm { i n . } ) / { } ^ { \circ } \mathrm { F } . . The lengths of the truss elements are shown in the figure. Determine the stresses in each bar. (Hint: Use Problem 3.21 for \underline { { K } } . )
15.5 For the structure shown in Figure P15–5, bar element 1 is subjected to a uniform temperature rise of T = 2 0 ^ { \circ } \mathrm { C } . . Let E = 2 1 0 ~ \mathrm { \ G P a } , A = 2 \times \mathrm { \bar { 1 0 } } ^ { - 2 } ~ \mathrm { ~ m } ^ { 2 } , and { \mathfrak { X } } = 1 2 \times 1 0 ^ { - 6 } \mathrm { m m / m m } ) / ^ { \circ } \mathrm { C } . . Determine the stresses in each bar.
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3 2 3 Rigid bar 2 2 2 1 1 4 3 m 1 m
Figure P15–5
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4 3 2 ③ ② ① 30° 30° 1 3 m
Figure P15–6
15.6 For the plane truss shown in Figure P15–6, bar element 2 is subjected to a uniform temperature drop of T = 2 0 ^ { \circ } \mathrm { C } . . Let E = 7 0 \ \mathrm { \ G P a } , A = 4 \times \mathrm { i } 0 ^ { - 2 } ~ \mathrm { m } ^ { 2 } . , and \varnothing = 2 3 \times 1 0 ^ { - 6 } (mm/mm)/ C. Determine the stresses in each bar and the displacement of node 1.
15.7 For the bar structure shown in Figure P15–7, element 1 is subjected to a uniform temperature rise of T = 3 0 ^ { \circ } \mathrm { C } . Let E ¼ 210 GPa, A = 3 \times \mathrm { { \bar { 1 0 ^ { - 2 } } } ~ m ^ { 2 } } , and \varnothing = 1 2 \times 1 0 ^ { - 6 } (mm/mm)/ C. Determine the displacement of node 1 and the stresses in each bar.
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2 ① 1 m 1 ② 1 m 3
Figure P15–7
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Steel Brass Steel 1 2
Figure P15–8
15.8 A bar assemblage consists of two outer steel bars and an inner brass bar. The threebar assemblage is then heated to raise the temperature by an amount T = 4 0 ^ { \circ } \mathrm { F } . Let all cross-sectional areas be A = 2 \mathrm { i n } ^ { 2 } and L = 6 0 in., E _ { \mathrm { s t e e l } } = 3 0 \times 1 0 ^ { 6 } psi, E _ { \mathrm { b r a s s } } = 1 5 \times 1 0 ^ { 6 } \mathrm { p s i } , \alpha _ { \mathrm { s t e e l } } = 6 . 5 \times 1 0 ^ { - 6 } / { } ^ { \circ } \mathrm { F } , and \alpha _ { \mathrm { b r a s s } } = 1 0 \times 1 0 ^ { - 6 } / { } ^ { \circ } \mathrm { F } . Determine (a) the displacement of node 2 and (b) the stress in the steel and brass bars. See Figure P15–8.
15.9 It has been a practice on some trucks to have an intake manifold made of an aluminum alloy ( \alpha = 2 2 . 7 \times 1 0 ^ { - 6 } / ^ { \circ } \mathrm { C } ) bolted to a plate made of steel (not shown)