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김경종
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<!-- source-page: 341 -->
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We use Eq. (6.2.52) to obtain the element stiffness matrix. To evaluate $\underline{k}$ , we first use Eqs. (6.2.10) to obtain the $\beta$ 's and $\gamma$ 's as follows:
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$$
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\beta_ {i} = y _ {j} - y _ {m} = 0 - 1 = - 1 \quad \gamma_ {i} = x _ {m} - x _ {j} = 0 - 2 = - 2
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$$
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$$
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\beta_ {j} = y _ {m} - y _ {i} = 1 - (- 1) = 2 \quad \gamma_ {j} = x _ {i} - x _ {m} = 0 - 0 = 0 \tag {6.2.61}
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$$
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$$
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\beta_ {m} = y _ {i} - y _ {j} = - 1 - 0 = - 1 \quad \gamma_ {m} = x _ {j} - x _ {i} = 2 - 0 = 2
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$$
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Using Eqs. (6.2.32) and (6.2.34), we obtain matrix $\underline{B}$ as
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$$
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\underline {{B}} = \frac {1}{2 (2)} \left[ \begin{array}{c c c c c c} - 1 & 0 & 2 & 0 & - 1 & 0 \\ 0 & - 2 & 0 & 0 & 0 & 2 \\ - 2 & - 1 & 0 & 2 & 2 & - 1 \end{array} \right] \tag {6.2.62}
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$$
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where we have used $A = 2$ in.2 in Eq. (6.2.62).
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Using Eq. (6.1.8) for plane stress conditions,
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$$
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\underline {{D}} = \frac {3 0 \times 1 0 ^ {6}}{1 - (0 . 2 5) ^ {2}} \left[ \begin{array}{c c c} 1 & 0. 2 5 & 0 \\ 0. 2 5 & 1 & 0 \\ 0 & 0 & \frac {1 - 0 . 2 5}{2} \end{array} \right] \text { psi } \tag {6.2.63}
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$$
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Substituting Eqs. (6.2.62) and (6.2.63) into Eq. (6.2.52), we obtain
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$$
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\begin{array}{l} \underline {{k}} = \frac {(2) 3 0 \times 1 0 ^ {6}}{4 (0 . 9 3 7 5)} \left[ \begin{array}{c c c} - 1 & 0 & - 2 \\ 0 & - 2 & - 1 \\ 2 & 0 & 0 \\ 0 & 0 & 2 \\ - 1 & 0 & 2 \\ 0 & 2 & - 1 \end{array} \right] \\ \times \left[ \begin{array}{c c c} 1 & 0. 2 5 & 0 \\ 0. 2 5 & 1 & 0 \\ 0 & 0 & 0. 3 7 5 \end{array} \right] \frac {1}{2 (2)} \left[ \begin{array}{c c c c c c} - 1 & 0 & 2 & 0 & - 1 & 0 \\ 0 & - 2 & 0 & 0 & 0 & 2 \\ - 2 & - 1 & 0 & 2 & 2 & - 1 \end{array} \right] \\ \end{array}
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$$
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Performing the matrix triple product, we have
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$$
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\underline {{{k}}} = 4. 0 \times 1 0 ^ {6} \left[ \begin{array}{c c c c c c} 2. 5 & 1. 2 5 & - 2 & - 1. 5 & - 0. 5 & 0. 2 5 \\ 1. 2 5 & 4. 3 7 5 & - 1 & - 0. 7 5 & - 0. 2 5 & - 3. 6 2 5 \\ - 2 & - 1 & 4 & 0 & - 2 & 1 \\ - 1. 5 & - 0. 7 5 & 0 & 1. 5 & 1. 5 & - 0. 7 5 \\ - 0. 5 & - 0. 2 5 & - 2 & 1. 5 & 2. 5 & - 1. 2 5 \\ 0. 2 5 & - 3. 6 2 5 & 1 & - 0. 7 5 & - 1. 2 5 & 4. 3 7 5 \end{array} \right] \frac {\mathrm{lb}}{\text {in.}} \tag {6.2.64}
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$$
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<!-- source-page: 342 -->
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To evaluate the stresses, we use Eq. (6.2.36). Substituting Eqs. (6.2.62) and (6.2.63), along with the given nodal displacements, into Eq. (6.2.36), we obtain
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$$
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\left\{ \begin{array}{l} \sigma_ {x} \\ \sigma_ {y} \\ \tau_ {x y} \end{array} \right\} = \frac {3 0 \times 1 0 ^ {6}}{1 - (0 . 2 5) ^ {2}} \left[ \begin{array}{c c c} 1 & 0. 2 5 & 0 \\ 0. 2 5 & 1 & 0 \\ 0 & 0 & 0. 3 7 5 \end{array} \right]
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$$
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$$
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\times \frac {1}{2 (2)} \left[ \begin{array}{r r r r r r} - 1 & 0 & 2 & 0 & - 1 & 0 \\ 0 & - 2 & 0 & 0 & 0 & 2 \\ - 2 & - 1 & 0 & 2 & 2 & - 1 \end{array} \right] \left\{ \begin{array}{l} 0. 0 \\ 0. 0 0 2 5 \\ 0. 0 0 1 2 \\ 0. 0 \\ 0. 0 \\ 0. 0 0 2 5 \end{array} \right\} \tag {6.2.65}
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$$
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Performing the matrix triple product in Eq. (6.2.65), we have
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$$
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\sigma_ {x} = 1 9, 2 0 0 \mathrm{psi} \quad \sigma_ {y} = 4 8 0 0 \mathrm{psi} \quad \tau_ {x y} = - 1 5, 0 0 0 \mathrm{psi} \tag {6.2.66}
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$$
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Finally, the principal stresses and principal angle are obtained by substituting the results from Eqs. (6.2.66) into Eqs. (6.1.2) and (6.1.3) as follows:
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$$
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\begin{array}{l} \sigma_ {1} = \frac {1 9 , 2 0 0 + 4 8 0 0}{2} + \left[ \left(\frac {1 9 , 2 0 0 - 4 8 0 0}{2}\right) ^ {2} + (- 1 5, 0 0 0) ^ {2} \right] ^ {1 / 2} \\ = 2 8, 6 3 9 \mathrm{psi} \\ \end{array}
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$$
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$$
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\begin{array}{l} \sigma_ {2} = \frac {1 9 , 2 0 0 + 4 8 0 0}{2} - \left[ \left(\frac {1 9 , 2 0 0 - 4 8 0 0}{2}\right) ^ {2} + (- 1 5, 0 0 0) ^ {2} \right] ^ {1 / 2} \tag {6.2.67} \\ = - 4 6 3 9 \mathrm{psi} \\ \end{array}
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$$
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$$
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\theta_ {p} = \frac {1}{2} \tan^ {- 1} \left[ \frac {2 (- 1 5 , 0 0 0)}{1 9 , 2 0 0 - 4 8 0 0} \right] = - 3 2. 2 ^ {\circ}
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$$
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# 6.3 Treatment of Body and Surface Forces
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# Body Forces
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Using the first term on the right side of Eq. (6.2.46), we can evaluate the body forces at the nodes as
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$$
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\{f _ {b} \} = \iiint_ {V} [ N ] ^ {T} \{X \} d V \tag {6.3.1}
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$$
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<!-- source-page: 343 -->
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<details>
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<summary>text_image</summary>
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y
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m
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x
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h
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i
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b
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j
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</details>
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Figure 6–12 Element with centroidal coordinate axes
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where $\{ X \} = { \left\{ \begin{array} { l } { X _ { b } } \\ { Y _ { b } } \end{array} \right\} }$ ð6:3:2Þ
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and $X _ { b }$ and $Y _ { b }$ are the weight densities in the x and y directions in units of force/unit volume, respectively. These forces may arise, for instance, because of actual body weight (gravitational forces), angular velocity (called centrifugal body forces, as described in Chapter 9), or inertial forces in dynamics.
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In Eq. (6.3.1), ½N is a linear function of x and y; therefore, the integration must be carried out. Without lack of generality, the integration is simplified if the origin of the coordinates is chosen at the centroid of the element. For example, consider the element with coordinates shown in Figure 6–12. With the origin of the coordinate placed at the centroid of the element, we have, from the definition of the centroid, $\textstyle \int \int x d A = \int \int y d A = 0$ and therefore,
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$$
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\iint \beta_ {i} x d A = \iint \gamma_ {i} y d A = 0 \tag {6.3.3}
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$$
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and $\alpha _ { i } = \alpha _ { j } = \alpha _ { m } = \frac { 2 A } { 3 }$ ð6:3:4Þ
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Using Eqs. (6.3.2)–(6.3.4) in Eq. (6.3.1), the body force at node i is then represented by
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$$
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\{f _ {b i} \} = \left\{ \begin{array}{l} X _ {b} \\ Y _ {b} \end{array} \right\} \frac {t A}{3} \tag {6.3.5}
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$$
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Similarly, considering the j and m node body forces, we obtain the same results as in Eq. (6.3.5). In matrix form, the element body forces are
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$$
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\left\{f _ {b} \right\} = \left\{ \begin{array}{l} f _ {b i x} \\ f _ {b i y} \\ f _ {b j x} \\ f _ {b j y} \\ f _ {b m x} \\ f _ {b m y} \end{array} \right\} = \left\{ \begin{array}{l} X _ {b} \\ Y _ {b} \\ X _ {b} \\ Y _ {b} \\ X _ {b} \\ Y _ {b} \end{array} \right\} \frac {A t}{3} \tag {6.3.6}
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$$
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<!-- source-page: 344 -->
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<details>
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<summary>text_image</summary>
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y
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1
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L
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2
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①
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a
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3
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p (lb/in²)
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x
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L
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</details>
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<details>
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<summary>text_image</summary>
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1
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①
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p (lb/in.²)
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2
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3
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</details>
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(b)
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Figure 6–13 (a) Elements with uniform surface traction acting on one edge and (b) element one with uniform surface traction along edge 1–3
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From the results of Eq. (6.3.6), we can conclude that the body forces are distributed to the nodes in three equal parts. The signs depend on the directions of $X _ { b }$ and $Y _ { b }$ with respect to the positive x and y global coordinates. For the case of body weight only, because of the gravitational force associated with the y direction, we have only $Y _ { b } \left( X _ { b } = 0 \right)$ .
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# Surface Forces
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Using the third term on the right side of Eq. (6.2.46), we can evaluate the surface forces at the nodes as
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$$
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\{f _ {s} \} = \iint_ {S} [ N _ {S} ] ^ {T} \{T _ {S} \} d S \tag {6.3.7}
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$$
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We emphasize that the subscript S in $[ N _ { S } ]$ in Eq. (6.3.7) means the shape functions evaluated along the surface where the surface traction is applied.
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We will now illustrate the use of Eq. (6.3.7) by considering the example of a uniform stress p (say, in pounds per square inch) acting between nodes 1 and 3 on the edge of element 1 in Figure 6–13(b). In Eq. (6.3.7), the surface traction now becomes
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$$
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\{T _ {S} \} = \left\{ \begin{array}{l} p _ {x} \\ p _ {y} \end{array} \right\} = \left\{ \begin{array}{l} p \\ 0 \end{array} \right\} \tag {6.3.8}
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$$
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and $\left[ N _ { S } \right] ^ { T } = \left[ \begin{array} { c c } { N _ { 1 } } & { 0 } \\ { 0 } & { N _ { 1 } } \\ { N _ { 2 } } & { 0 } \\ { 0 } & { N _ { 2 } } \\ { N _ { 3 } } & { 0 } \\ { 0 } & { N _ { 3 } } \end{array} \right] _ { \mathrm { \scriptsize ~ e v a l u a t e d ~ a t ~ } x = a , \ y = y }$ ð6:3:9Þ
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As the surface traction p acts along the edge at $x = a$ and $y = y$ from $y = 0$ to $y = L$ , we evaluate the shape functions at $x = a$ and $y = y$ and integrate over the surface from 0 to L in the y direction and from 0 to t in the z direction, as shown by Eq. (6.3.10).
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<!-- source-page: 345 -->
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Using Eqs. (6.3.8) and (6.3.9), we express Eq. (6.3.7) as
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$$
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\left\{f _ {s} \right\} = \int_ {0} ^ {t} \int_ {0} ^ {L} \left[ \begin{array}{c c} N _ {1} & 0 \\ 0 & N _ {1} \\ N _ {2} & 0 \\ 0 & N _ {2} \\ N _ {3} & 0 \\ 0 & N _ {3} \end{array} \right] \left\{ \begin{array}{l} p \\ 0 \end{array} \right\} d z d y \tag {6.3.10}
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$$
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Simplifying Eq. (6.3.10), we obtain
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$$
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\left\{f _ {s} \right\} = t \int_ {0} ^ {L} \left[ \begin{array}{c} N _ {1} p \\ 0 \\ N _ {2} p \\ 0 \\ N _ {3} p \\ 0 \end{array} \right] d y \tag {6.3.11}
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$$
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Now, by Eqs. (6.2.18) (with i ¼ 1), we have
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$$
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N _ {1} = \frac {1}{2 A} (\alpha_ {1} + \beta_ {1} x + \gamma_ {1} y) \tag {6.3.12}
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$$
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For convenience, we choose the coordinate system for the element as shown in Figure 6–14. Using the definition Eqs. (6.2.10), we obtain
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$$
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\alpha_ {i} = x _ {j} y _ {m} - y _ {j} x _ {m}
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$$
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or, with i ¼ 1, j ¼ 2, and m ¼ 3,
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$$
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\alpha_ {1} = x _ {2} y _ {3} - y _ {2} x _ {3} \tag {6.3.13}
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$$
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Substituting the coordinates into Eq. (6.3.13), we obtain
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$$
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\alpha_ {1} = 0 \tag {6.3.14}
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$$
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Similarly, again using Eqs. (6.2.10), we obtain
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$$
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\beta_ {1} = 0 \quad \gamma_ {1} = a \tag {6.3.15}
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$$
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Therefore, substituting Eqs. (6.3.14) and (6.3.15) into Eq. (6.3.12), we obtain
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$$
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N _ {1} = \frac {a y}{2 A} \tag {6.3.16}
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$$
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<details>
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<summary>text_image</summary>
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y
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1
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L
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p
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2
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①
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a
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3
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x
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</details>
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Figure 6–14 Representative element subjected to edge surface traction p
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<!-- source-page: 346 -->
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Similarly, using Eqs. (6.2.18), we can show that
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$$
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N _ {2} = \frac {L (a - x)}{2 A} \quad \text { and } \quad N _ {3} = \frac {L x - a y}{2 A} \tag {6.3.17}
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$$
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On substituting Eqs. (6.3.16) and (6.3.17) for $N _ { 1 } , N _ { 2 }$ , and $N _ { 3 }$ into Eq. (6.3.11), evaluating $N _ { 1 } , N _ { 2 }$ , and $N _ { 3 }$ at $x = a$ and $y = y$ (the coordinates corresponding to the location of the surface load $p )$ , and then integrating with respect to $y ,$ we obtain
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$$
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\left\{f _ {s} \right\} = \frac {t}{2 (a L / 2)} \left\{ \begin{array}{c} a \left(\frac {L ^ {2}}{2}\right) p \\ 0 \\ 0 \\ 0 \\ \left(L ^ {2} - \frac {L ^ {2}}{2}\right) a p \\ 0 \end{array} \right\} \tag {6.3.18}
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$$
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where the shape function $N _ { 2 } = 0$ between nodes 1 and 3, as should be the case according to the definitions of the shape functions. Simplifying Eq. (6.3.18), we finally obtain
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$$
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\left\{f _ {s} \right\} = \left\{ \begin{array}{l} f _ {s 1 x} \\ f _ {s 1 y} \\ f _ {s 2 x} \\ f _ {s 2 y} \\ f _ {s 3 x} \\ f _ {s 3 y} \end{array} \right\} = \left\{ \begin{array}{c} p L t / 2 \\ 0 \\ 0 \\ 0 \\ p L t / 2 \\ 0 \end{array} \right\} \tag {6.3.19}
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$$
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Figure 6–15 illustrates the results for the surface load equivalent nodal forces for both elements 1 and 2.
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We can conclude that for a constant-strain triangle, a distributed load on an element edge can be treated as concentrated loads acting at the nodes associated with the loaded edge by making the two kinds of load statically equivalent [which is equivalent to applying Eq. (6.3.7)]. However, for higher-order elements such as the
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<details>
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<summary>text_image</summary>
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1
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pLt/2
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①
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L
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2
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3
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pLt/2 + pLt/2
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②
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L
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5
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4
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a
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pLt/2
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</details>
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Figure 6–15 Surface traction equivalent nodal forces
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<!-- source-page: 347 -->
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linear-strain triangle (discussed in Chapter 8), the load replacement should be made by using Eq. (6.3.7), which was derived by the principle of minimum potential energy. For higher-order elements, this load replacement by use of Eq. (6.3.7) is generally not equal to the apparent statically equivalent one; however, it is consistent in that this replacement results directly from the energy approach.
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We now recognize the force matrix $\{ f _ { s } \}$ defined by Eq. (6.3.7), and based on the principle of minimum potential energy, to be equivalent to that based on work equivalence, which we previously used in Chapter 4 when discussing distributed loads acting on beams.
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# 6.4 Explicit Expression for the Constant-Strain Triangle Stiffness Matrix
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Although the stiffness matrix is generally formulated internally in most computer programs by performing the matrix triple product indicated by Eq. (6.4.1), it is still a valuable learning experience to evaluate the stiffness matrix explicitly for the constantstrain triangular element. Hence, we will consider the plane strain case specifically in this development.
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First, recall that the stiffness matrix is given by
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$$
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[ k ] = t A [ B ] ^ {T} [ D ] [ B ] \tag {6.4.1}
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$$
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where, for the plane strain case, ½D is given by Eq. (6.1.10) and ½B is given by Eq. (6.2.34). On substituting the matrices ½D and ½B into Eq. (6.4.1), we obtain
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$$
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[ k ] = \frac {t E}{4 A (1 + \nu) (1 - 2 \nu)} \left[ \begin{array}{c c c} \beta_ {i} & 0 & \gamma_ {i} \\ 0 & \gamma_ {i} & \beta_ {i} \\ \beta_ {j} & 0 & \gamma_ {j} \\ 0 & \gamma_ {j} & \beta_ {j} \\ \beta_ {m} & 0 & \gamma_ {m} \\ 0 & \gamma_ {m} & \beta_ {m} \end{array} \right]
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$$
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$$
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\times \left[ \begin{array}{c c c} 1 - v & v & 0 \\ v & 1 - v & 0 \\ 0 & 0 & \frac {1 - 2 v}{2} \end{array} \right] \left[ \begin{array}{c c c c c c} \beta_ {i} & 0 & \beta_ {j} & 0 & \beta_ {m} & 0 \\ 0 & \gamma_ {i} & 0 & \gamma_ {j} & 0 & \gamma_ {m} \\ \gamma_ {i} & \beta_ {i} & \gamma_ {j} & \beta_ {j} & \gamma_ {m} & \beta_ {m} \end{array} \right] \tag {6.4.2}
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$$
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On multiplying the matrices in Eq. (6.4.2), we obtain Eq. (6.4.3), the explicit constant-strain triangle stiffness matrix for the plane strain case. Note that ½k is a function of the difference in the x and y nodal coordinates, as indicated by the g’s and $\beta ^ { \circ } \mathbf { s } ,$ , of the material properties E and n, and of the thickness t and surface area A of the element.
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|
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<!-- source-page: 348 -->
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$$
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\underline {{k}} = \frac {t E}{4 A (1 + \nu) (1 - 2 \nu)}
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$$
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$$
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\times \left[ \begin{array}{c c c} \beta_ {i} ^ {2} (1 - v) + \gamma_ {i} ^ {2} \left(\frac {1 - 2 v}{2}\right) & \beta_ {i} \gamma_ {i} v + \beta_ {i} \gamma_ {i} \left(\frac {1 - 2 v}{2}\right) & \beta_ {i} \beta_ {j} (1 - v) + \gamma_ {i} \gamma_ {j} \left(\frac {1 - 2 v}{2}\right) \\ & \gamma_ {i} ^ {2} (1 - v) + \beta_ {i} ^ {2} \left(\frac {1 - 2 v}{2}\right) & \beta_ {j} \gamma_ {i} v + \beta_ {i} \gamma_ {j} \left(\frac {1 - 2 v}{2}\right) \\ & & \beta_ {j} ^ {2} (1 - v) + \gamma_ {j} ^ {2} \left(\frac {1 - 2 v}{2}\right) \\ \text {Symmetry} \end{array} \right.
|
||||
$$
|
||||
|
||||
$$
|
||||
\begin{array}{l} \beta_ {i} \gamma_ {j} v + \beta_ {j} \gamma_ {i} \left(\frac {1 - 2 v}{2}\right) \quad \beta_ {i} \beta_ {m} (1 - v) + \gamma_ {i} \gamma_ {m} \left(\frac {1 - 2 v}{2}\right) \quad \beta_ {i} \gamma_ {m} v + \beta_ {m} \gamma_ {i} \left(\frac {1 - 2 v}{2}\right) \\ \gamma_ {i} \gamma_ {j} (1 - v) + \beta_ {i} \beta_ {j} \left(\frac {1 - 2 v}{2}\right) \quad \beta_ {m} \gamma_ {i} v + \beta_ {i} \gamma_ {m} \left(\frac {1 - 2 v}{2}\right) \quad \gamma_ {i} \gamma_ {m} (1 - v) + \beta_ {i} \beta_ {m} \left(\frac {1 - 2 v}{2}\right) \\ \beta_ {j} \gamma_ {j} v + \beta_ {j} \gamma_ {j} \left(\frac {1 - 2 v}{2}\right) \quad \beta_ {j} \beta_ {m} (1 - v) + \gamma_ {j} \gamma_ {m} \left(\frac {1 - 2 v}{2}\right) \quad \beta_ {j} \gamma_ {m} v + \gamma_ {j} \beta_ {m} \left(\frac {1 - 2 v}{2}\right) \\ \gamma_ {j} ^ {2} (1 - v) + \beta_ {j} ^ {2} \left(\frac {1 - 2 v}{2}\right) \quad \beta_ {m} \gamma_ {j} v + \beta_ {j} \gamma_ {m} \left(\frac {1 - 2 v}{2}\right) \quad \gamma_ {j} \gamma_ {m} (1 - v) + \beta_ {j} \beta_ {m} \left(\frac {1 - 2 v}{2}\right) \\ \beta_ {m} ^ {2} (1 - v) + \gamma_ {m} ^ {2} \left(\frac {1 - 2 v}{2}\right) \quad \gamma_ {m} \beta_ {m} v + \beta_ {m} \gamma_ {m} \left(\frac {1 - 2 v}{2}\right) \\ \gamma_ {m} ^ {2} (1 - \nu) + \beta_ {m} ^ {2} \left(\frac {1 - 2 \nu}{2}\right) \\ \end{array}
|
||||
$$
|
||||
|
||||
(6.4.3)
|
||||
|
||||
For the plane stress case, we need only replace $1 - \nu$ by 1, $(1 - 2\nu)/2$ by $(1 - \nu)/2$ , and $(1 + \nu)(1 - 2\nu)$ outside the brackets by $1 - \nu^2$ in Eq. (6.4.3).
|
||||
|
||||
Finally, it should be noted that for Poisson's ratio $\nu$ approaching 0.5, as in rubberlike materials and plastic solids, for instance, a material becomes incompressible [2]. For plane strain, as $\nu$ approaches 0.5, the denominator becomes zero in the material property matrix [see Eq. (6.1.10)] and hence in the stiffness matrix, Eq. (6.4.3).
|
||||
|
||||
<!-- source-page: 349 -->
|
||||
|
||||
A value of n near 0.5 can cause ill-conditioned structural equations. A special formulation (called a penalty formulation [3]) has been used in this case.
|
||||
|
||||
# 6.5 Finite Element Solution of a Plane Stress Problem
|
||||
|
||||
To illustrate the finite element method for a plane stress problem, we now present a detailed solution.
|
||||
|
||||
# Example 6.2
|
||||
|
||||
For a thin plate subjected to the surface traction shown in Figure 6–16, determine the nodal displacements and the element stresses. The plate thickness t ¼ 1 in., $E = 3 0 \times$ $1 0 ^ { 6 }$ psi, and $\nu = 0 . 3 0$ .
|
||||
|
||||
# Discretization
|
||||
|
||||
To illustrate the finite element method solution for the plate, we first discretize the plate into two elements, as shown in Figure 6–17. It should be understood that the coarseness of the mesh will not yield as true a predicted behavior of the plate as would a finer mesh, particularly near the fixed edge. However, since we are performing a longhand solution, we will use a coarse discretization for simplicity (but without loss of generality of the method).
|
||||
|
||||
In Figure 6–17, the original tensile surface traction in Figure 6–16 has been converted to nodal forces as follows:
|
||||
|
||||
$$
|
||||
F = \frac {1}{2} T A
|
||||
$$
|
||||
|
||||
$$
|
||||
F = \frac {1}{2} (1 0 0 0 \mathrm{psi}) (1 \text { in. } \times 1 0 \text { in. })
|
||||
$$
|
||||
|
||||
$$
|
||||
F = 5 0 0 0 \mathrm{lb}
|
||||
$$
|
||||
|
||||
|
||||
|
||||
<details>
|
||||
<summary>text_image</summary>
|
||||
|
||||
1 in.
|
||||
20 in.
|
||||
10 in.
|
||||
T = 1000 psi
|
||||
</details>
|
||||
|
||||
Figure 6–16 Thin plate subjected to tensile stress
|
||||
|
||||
<!-- source-page: 350 -->
|
||||
|
||||
|
||||
|
||||
<details>
|
||||
<summary>text_image</summary>
|
||||
|
||||
2
|
||||
①
|
||||
y
|
||||
1
|
||||
x
|
||||
3
|
||||
5000 lb
|
||||
②
|
||||
4
|
||||
5000 lb
|
||||
</details>
|
||||
|
||||
Figure 6–17 Discretized plate
|
||||
|
||||
In general, for higher-order elements, Eq. (6.3.7) should be used to convert distributed surface tractions to nodal forces. However, for the CST element, we have shown in Section 6.3 that a statically equivalent force replacement can be used directly, as has been done here.
|
||||
|
||||
The governing global matrix equation is
|
||||
|
||||
$$
|
||||
\{F \} = [ K ] \{d \} \tag {6.5.1}
|
||||
$$
|
||||
|
||||
Expanding matrices in Eq. (6.5.1), we obtain
|
||||
|
||||
$$
|
||||
\left\{ \begin{array}{l} F _ {1 x} \\ F _ {1 y} \\ F _ {2 x} \\ F _ {2 y} \\ F _ {3 x} \\ F _ {3 y} \\ F _ {4 x} \\ F _ {4 y} \end{array} \right\} = \left\{ \begin{array}{l} R _ {1 x} \\ R _ {1 y} \\ R _ {2 x} \\ R _ {2 y} \\ 5 0 0 0 \\ 0 \\ 5 0 0 0 \\ 0 \end{array} \right\} = [ K ] \left\{ \begin{array}{l} d _ {1 x} \\ d _ {1 y} \\ d _ {2 x} \\ d _ {2 y} \\ d _ {3 x} \\ d _ {3 y} \\ d _ {4 x} \\ d _ {4 y} \end{array} \right\} = [ K ] \left\{ \begin{array}{l} 0 \\ 0 \\ 0 \\ 0 \\ d _ {3 x} \\ d _ {3 y} \\ d _ {4 x} \\ d _ {4 y} \end{array} \right\} \tag {6.5.2}
|
||||
$$
|
||||
|
||||
where ½K is an $8 \times 8$ matrix (two degrees of freedom per node with four nodes) before deleting rows and columns to account for the fixed boundary support conditions at nodes 1 and 2.
|
||||
|
||||
# Assemblage of the Stiffness Matrix
|
||||
|
||||
We assemble the global stiffness matrix by superposition of the individual element stiffness matrices. By Eq. (6.2.52), the stiffness matrix for an element is
|
||||
|
||||
$$
|
||||
[ k ] = t A [ B ] ^ {T} [ D ] [ B ] \tag {6.5.3}
|
||||
$$
|
||||
|
||||
In Figure $_ { 6 - 1 8 }$ for element 1, we have coordinates $x _ { i } = 0 , y _ { i } = 0 , x _ { j } = 2 0 , y _ { j } = 1 0$ , $x _ { m } = 0 ;$ , and $y _ { m } = 1 0$ , since the global coordinate axes are set up at node 1, and
|
||||
|
||||
$$
|
||||
\begin{array}{l} A = \frac {1}{2} b h \\ A = \left(\frac {1}{2}\right) (2 0) (1 0) = 1 0 0 \mathrm{in} ^ {2} \\ \end{array}
|
||||
$$
|
||||
Reference in New Issue
Block a user