김경종
449 lines
24 KiB
Markdown
449 lines
24 KiB
Markdown
<!-- source-page: 121 -->
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where, in Eqs. (3.8.1)–(3.8.5), the column labels indicate the degrees of freedom associated with each element. Also, because elements 4 and 5 lie in the plane of symmetry, half of their original areas have been used in Eqs. (3.8.4) and (3.8.5).
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We will limit the solution to determining the displacement components. Therefore, considering the boundary constraints that result in zero-displacement components, we can immediately obtain the reduced set of equations by eliminating rows and columns in each element stiffness matrix corresponding to a zero-displacement component. That is, because $d _ { 1 x } = 0$ and $d _ { \mathrm { l } y } = 0$ (owing to the pin support at node 1 in Figure 3–21) and $d _ { 2 x } = 0 , d _ { 3 x } = 0$ , and $d _ { 4 x } = 0$ (owing to the symmetry condition), we can cancel rows and columns corresponding to these displacement components in each element stiffness matrix before assembling the total stiffness matrix. The resulting set of stiffness equations is
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$$
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\frac {A E}{L} \left[ \begin{array}{c c c} 1 & 0 & - \frac {1}{2} \\ 0 & 1 & - \frac {1}{2} \\ - \frac {1}{2} & - \frac {1}{2} & 1 \end{array} \right] \left\{ \begin{array}{l} d _ {2 y} \\ d _ {3 y} \\ d _ {4 y} \end{array} \right\} = \left\{ \begin{array}{c} 0 \\ 0 \\ - P \end{array} \right\} \tag {3.8.6}
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$$
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On solving Eq. (3.8.6) for the displacements, we obtain
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$$
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d _ {2 y} = \frac {- P L}{A E} \quad d _ {3 y} = \frac {- P L}{A E} \quad d _ {4 y} = \frac {- 2 P L}{A E} \tag {3.8.7}
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$$
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The ideas presented regarding the use of symmetry should be used sparingly and cautiously in problems of vibration and buckling. For instance, a structure such as a simply supported beam has symmetry about its center but has antisymmetric vibration modes as well as symmetric vibration modes. This will be shown in Chapter 16. If only half the beam were modeled using reflective symmetry conditions, the support conditions would permit only the symmetric vibration modes.
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# 3.9 Inclined, or Skewed, Supports
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In the preceding sections, the supports were oriented such that the resulting boundary conditions on the displacements were in the global directions.
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<details>
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<summary>text_image</summary>
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y
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F_{2x} 2 3
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y'
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x'
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α
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1
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x
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</details>
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Figure 3–22 Plane truss with inclined boundary conditions at node 3
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However, if a support is inclined, or skewed, at an angle a from the global x axis, as shown at node 3 in the plane truss of Figure 3–22, the resulting boundary conditions on the displacements are not in the global x-y directions but are in the local $x ^ { \prime } { - y ^ { \prime } }$ directions. We will now describe two methods used to handle inclined supports.
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In the first method, to account for inclined boundary conditions, we must perform a transformation of the global displacements at node 3 only into the local nodal coordinate system $x ^ { \prime } { - y ^ { \prime } }$ , while keeping all other displacements in the x-y global system. We can then enforce the zero-displacement boundary condition $d _ { 3 y } ^ { \prime }$ in the force/displacement equations and, finally, solve the equations in the usual manner.
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The transformation used is analogous to that for transforming a vector from local to global coordinates. For the plane truss, we use Eq. (3.3.16) applied to node 3 as follows:
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$$
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\left\{ \begin{array}{l} d _ {3 x} ^ {\prime} \\ d _ {3 y} ^ {\prime} \end{array} \right\} = \left[ \begin{array}{c c} \cos \alpha & \sin \alpha \\ - \sin \alpha & \cos \alpha \end{array} \right] \left\{ \begin{array}{l} d _ {3 x} \\ d _ {3 y} \end{array} \right\} \tag {3.9.1}
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$$
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Rewriting Eq. (3.9.1), we have
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$$
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\{d _ {3} ^ {\prime} \} = [ t _ {3} ] \{d _ {3} \} \tag {3.9.2}
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$$
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where
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$$
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\left[ t _ {3} \right] = \left[ \begin{array}{c c} \cos \alpha & \sin \alpha \\ - \sin \alpha & \cos \alpha \end{array} \right] \tag {3.9.3}
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$$
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We now write the transformation for the entire nodal displacement vector as
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$$
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\{d ^ {\prime} \} = [ T _ {1} ] \{d \} \tag {3.9.4}
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$$
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or $\{ d \} = [ T _ { 1 } ] ^ { T } \{ d ^ { \prime } \}$ ð3:9:5Þ
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where the transformation matrix for the entire truss is the $6 \times 6$ matrix
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$$
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[ T _ {1} ] = \left[ \begin{array}{l l l} {[ I ]} & {[ 0 ]} & {[ 0 ]} \\ {[ 0 ]} & {[ I ]} & {[ 0 ]} \\ {[ 0 ]} & {[ 0 ]} & {[ t _ {3} ]} \end{array} \right] \tag {3.9.6}
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$$
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Each submatrix in Eq. (3.9.6) (the identity matrix [I ], the null matrix [0], and matrix [t3] has the same $2 \times 2$ order, that order in general being equal to the number of degrees of freedom at each node.
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To obtain the desired displacement vector with global displacement components at nodes 1 and 2 and local displacement components at node 3, we use Eq. (3.9.5) to obtain
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$$
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\left\{ \begin{array}{l} d _ {1 x} \\ d _ {1 y} \\ d _ {2 x} \\ d _ {2 y} \\ d _ {3 x} \\ d _ {3 y} \end{array} \right\} = \left[ \begin{array}{l l l} [ I ] & [ 0 ] & [ 0 ] \\ [ 0 ] & [ I ] & [ 0 ] \\ [ 0 ] & [ 0 ] & [ t _ {3} ] ^ {T} \end{array} \right] \left\{ \begin{array}{l} d _ {1 x} ^ {\prime} \\ d _ {1 y} ^ {\prime} \\ d _ {2 x} ^ {\prime} \\ d _ {2 y} ^ {\prime} \\ d _ {3 x} ^ {\prime} \\ d _ {3 y} ^ {\prime} \end{array} \right\} \tag {3.9.7}
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$$
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<!-- source-page: 123 -->
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In Eq. (3.9.7), we observe that only the node 3 global components are transformed, as indicated by the placement of the $\left[ t _ { 3 } \right] ^ { T }$ matrix. We denote the square matrix in Eq. (3.9.7) by $\dot { [ T _ { 1 } ] } ^ { T }$ . In general, we place a $2 \times 2 ~ [ t ]$ matrix in $\left[ T _ { 1 } \right]$ wherever the transformation from global to local displacements is needed (where skewed supports exist).
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Upon considering Eqs. (3.9.5) and (3.9.6), we observe that only node 3 components of $\{ d \}$ are really transformed to local (skewed) axes components. This transformation is indeed necessary whenever the local axes $x ^ { \prime } { - y ^ { \prime } }$ fixity directions are known.
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Furthermore, the global force vector can also be transformed by using the same transformation as for $\{ d ^ { \prime } \}$ :
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$$
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\{f ^ {\prime} \} = [ T _ {1} ] \{f \} \tag {3.9.8}
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$$
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In global coordinates, we then have
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$$
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\{f \} = [ K ] \{d \} \tag {3.9.9}
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$$
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Premultiplying Eq. (3.9.9) by $\left[ T _ { 1 } \right]$ , we have
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$$
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[ T _ {1} ] \{f \} = [ T _ {1} ] [ K ] \{d \} \tag {3.9.10}
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$$
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For the truss in Figure 3–22, the left side of Eq. (3.9.10) is
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$$
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\left[ \begin{array}{l l l} {[ I ]} & {[ 0 ]} & {[ 0 ]} \\ {[ 0 ]} & {[ I ]} & {[ 0 ]} \\ {[ 0 ]} & {[ 0 ]} & {[ t _ {3} ]} \end{array} \right] \left\{ \begin{array}{l} f _ {1 x} \\ f _ {1 y} \\ f _ {2 x} \\ f _ {2 y} \\ f _ {3 x} \\ f _ {3 y} \end{array} \right\} = \left\{ \begin{array}{l} f _ {1 x} \\ f _ {1 y} \\ f _ {2 x} \\ f _ {2 y} \\ f _ {3 x} ^ {\prime} \\ f _ {3 y} ^ {\prime} \end{array} \right\} \tag {3.9.11}
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$$
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where the fact that local forces transform similarly to Eq. (3.9.2) as
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$$
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\{f _ {3} ^ {\prime} \} = [ t _ {3} ] \{f _ {3} \} \tag {3.9.12}
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$$
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has been used in Eq. (3.9.11). From Eq. (3.9.11), we see that only the node 3 components of $\{ f \}$ have been transformed to the local axes components, as desired.
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Using Eq. (3.9.5) in Eq. (3.9.10), we have
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$$
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[ T _ {1} ] \{f \} = [ T _ {1} ] [ K ] [ T _ {1} ] ^ {T} \{d ^ {\prime} \} \tag {3.9.13}
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$$
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Using Eq. (3.9.11), we find that the form of Eq. (3.9.13) becomes
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$$
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\left\{ \begin{array}{l} F _ {1 x} \\ F _ {1 y} \\ F _ {2 x} \\ F _ {2 y} \\ F _ {3 x} ^ {\prime} \\ F _ {3 y} ^ {\prime} \end{array} \right\} = [ T _ {1} ] [ K ] [ T _ {1} ] ^ {T} \left\{ \begin{array}{l} d _ {1 x} \\ d _ {1 y} \\ d _ {2 x} \\ d _ {2 y} \\ d _ {3 x} ^ {\prime} \\ d _ {3 y} ^ {\prime} \end{array} \right\} \tag {3.9.14}
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$$
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as $d _ { 1 x } = d _ { 1 x } ^ { \prime } , d _ { 1 y } = d _ { 1 y } ^ { \prime } , d _ { 2 x } = d _ { 2 x } ^ { \prime } ,$ and $d _ { 2 y } = d _ { 2 y } ^ { \prime }$ from Eq. (3.9.7). Equation (3.9.14) is the desired form that allows all known global and inclined boundary conditions to
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<!-- source-page: 124 -->
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be enforced. The global forces now result in the left side of Eq. (3.9.14). To solve Eq. (3.9.14), first perform the matrix triple product $\big [ T _ { 1 } \big ] \big [ K \big ] \big [ T _ { 1 } \big ] ^ { T }$ . Then invoke the following boundary conditions (for the truss in Figure 3–22):
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$$
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d _ {1 x} = 0 \quad d _ {1 y} = 0 \quad d _ {3 y} ^ {\prime} = 0 \tag {3.9.15}
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$$
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Then substitute the known value of the applied force $F _ { 2 x }$ along with $F _ { 2 y } = 0$ and $F _ { 3 x } ^ { \prime } = 0$ into Eq. (3.9.14). Finally, partition the equations with known displacements— here equations 1, 2, and 6 of Eq. (3.9.14)—and then simultaneously solve those associated with the unknown displacements $d _ { 2 x } , d _ { 2 y }$ , and $d _ { 3 x } ^ { \prime }$ .
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After solving for the displacements, return to Eq. (3.9.14) to obtain the global reactions $F _ { 1 x }$ and $F _ { \mathrm { l j } }$ and the inclined roller reaction $F _ { 3 y } ^ { \prime }$ .
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# Example 3.11
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For the plane truss shown in Figure 3–23, determine the displacements and reactions. Let $E = 2 1 0 \mathrm { G P a } , A = 6 . 0 0 \times 1 0 ^ { - 4 } \mathrm { m } ^ { 2 }$ for elements 1 and 2, and $A = 6 { \sqrt { 2 } } \times 1 0 ^ { - 4 } \mathrm { m } ^ { 2 }$ for element 3.
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We begin by using Eq. (3.4.23) to determine each element stiffness matrix.
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<details>
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<summary>text_image</summary>
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1000 kN
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2
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1 m
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①
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1 m
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y
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②
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③
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45°
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x
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y'
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x'
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</details>
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Figure 3–23 Plane truss with inclined support
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# Element 1
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$$
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\cos \theta = 0 \quad \sin \theta = 1
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$$
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$$
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\underline {{k}} ^ {(1)} = \frac {(6 . 0 \times 1 0 ^ {- 4} \mathrm{m} ^ {2}) (2 1 0 \times 1 0 ^ {9} \mathrm{N} / \mathrm{m} ^ {2})}{1 \mathrm{m}} \left[ \begin{array}{c c c c} d _ {1 x} & d _ {1 y} & d _ {2 x} & d _ {2 y} \\ 0 & 0 & 0 & 0 \\ & 1 & 0 & - 1 \\ & & 0 & 0 \\ \text { Symmetry } & & & 1 \end{array} \right] \tag {3.9.16}
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$$
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<!-- source-page: 125 -->
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# Element 2
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$$
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\cos \theta = 1 \quad \sin \theta = 0
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$$
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$$
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\underline {{k}} ^ {(2)} = \frac {(6 . 0 \times 1 0 ^ {- 4} \mathrm{m} ^ {2}) (2 1 0 \times 1 0 ^ {9} \mathrm{N} / \mathrm{m} ^ {2})}{1 \mathrm{m}} \left[ \begin{array}{c c c c} d _ {2 x} & d _ {2 y} & d _ {3 x} & d _ {3 y} \\ 1 & 0 & - 1 & 0 \\ & 0 & 0 & 0 \\ & & 1 & 0 \\ \text { Symmetry } & & & 0 \end{array} \right] \tag {3.9.17}
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$$
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# Element 3
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$$
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\cos \theta = \frac {\sqrt {2}}{2} \quad \sin \theta = \frac {\sqrt {2}}{2}
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$$
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$$
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\underline {{{{k}}}} ^ {(3)} = \frac {(6 \sqrt {2} \times 1 0 ^ {- 4} \mathrm{m} ^ {2}) (2 1 0 \times 1 0 ^ {9} \mathrm{N} / \mathrm{m} ^ {2})}{\sqrt {2} \mathrm{m}} \left[ \begin{array}{c c c c} d _ {1 x} & d _ {1 y} & d _ {3 x} & d _ {3 y} \\ 0. 5 & 0. 5 & - 0. 5 & - 0. 5 \\ & 0. 5 & - 0. 5 & - 0. 5 \\ & & 0. 5 & 0. 5 \\ \text {Symmetry} & & & 0. 5 \end{array} \right] \tag {3.9.18}
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$$
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Using the direct stiffness method on Eqs. (3.9.16)–(3.9.18), we obtain the global $\underline { { K } }$ matrix as
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$$
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\underline {{{K}}} = 1 2 6 0 \times 1 0 ^ {5} \mathrm{N} / \mathrm{m} \left[ \begin{array}{c c c c c c} 0. 5 & 0. 5 & 0 & 0 & - 0. 5 & - 0. 5 \\ & 1. 5 & 0 & - 1 & - 0. 5 & - 0. 5 \\ & & 1 & 0 & - 1 & 0 \\ & & & 1 & 0 & 0 \\ & & & & 1. 5 & 0. 5 \\ \text { Symmetry } & & & & & 0. 5 \end{array} \right] \tag {3.9.19}
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$$
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Next we obtain the transformation matrix $\underline { { T } } _ { 1 }$ using Eq. (3.9.6) to transform the global displacements at node 3 into local nodal coordinates $x ^ { \prime } { - y ^ { \prime } }$ . In using Eq. (3.9.6), the angle a is $4 5 ^ { \circ }$ .
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$$
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\left[ T _ {1} \right] = \left[ \begin{array}{c c c c c c} 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & \sqrt {2} / 2 & \sqrt {2} / 2 \\ 0 & 0 & 0 & 0 & - \sqrt {2} / 2 & \sqrt {2} / 2 \end{array} \right] \tag {3.9.20}
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$$
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<!-- source-page: 126 -->
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Next we use Eq. (3.9.14) (in general, we would use Eq. (3.9.13)) to express the assembled equations. First define $\underline { { K } } ^ { * } = \underline { { T } } _ { 1 } \underline { { K } } \underline { { T } } _ { 1 } ^ { T }$ and evaluate in steps as follows:
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$$
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\underline {{{T _ {1}}}} \underline {{{K}}} = 1 2 6 0 \times 1 0 ^ {5} \left[ \begin{array}{c c c c c c} 0. 5 & 0. 5 & 0 & 0 & - 0. 5 & - 0. 5 \\ 0. 5 & 1. 5 & 0 & - 1 & - 0. 5 & - 0. 5 \\ 0 & 0 & 1 & 0 & - 1 & 0 \\ 0 & - 1 & 0 & 1 & 0 & 0 \\ - 0. 7 0 7 & - 0. 7 0 7 & - 0. 7 0 7 & 0 & 1. 4 1 4 & 0. 7 0 7 \\ 0 & 0 & 0. 7 0 7 & 0 & - 0. 7 0 7 & 0 \end{array} \right] \tag {3.9.21}
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$$
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and
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$$
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\underline {{{T _ {1}}}} \underline {{{K}}} \underline {{{T _ {1}}}} ^ {T} = 1 2 6 0 \times 1 0 ^ {5} \mathrm{N} / \mathrm{m} \left[ \begin{array}{c c c c c c} 0. 5 & 0. 5 & 0 & 0 & - 0. 7 0 7 & 0 \\ 0. 5 & 1. 5 & 0 & - 1 & - 0. 7 0 7 & 0 \\ 0 & 0 & 1 & 0 & - 0. 7 0 7 & 0. 7 0 7 \\ 0 & - 1 & 0 & 1 & 0 & 0 \\ - 0. 7 0 7 & - 0. 7 0 7 & - 0. 7 0 7 & 0 & 1. 5 0 0 & - 0. 5 0 0 \\ 0 & 0 & 0. 7 0 7 & 0 & - 0. 5 0 0 & 0. 5 0 0 \end{array} \right] \tag {3.9.22}
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$$
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Applying the boundary conditions, $d _ { 1 x } = d _ { 1 y } = d _ { 2 y } = d _ { 3 \nu } ^ { \prime } = 0$ , to Eq. (3.9.22), we obtain
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$$
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\left\{ \begin{array}{l} F _ {2 x} = 1 0 0 0 \mathrm{kN} \\ F _ {3 x} ^ {\prime} = 0 \end{array} \right\} = (1 2 6 \times 1 0 ^ {3} \mathrm{kN/m}) \left[ \begin{array}{c c} 1 & - 0. 7 0 7 \\ - 0. 7 0 7 & 1. 5 0 \end{array} \right] \left\{ \begin{array}{l} d _ {2 x} \\ d _ {3 x} ^ {\prime} \end{array} \right\} \tag {3.9.23}
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$$
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Solving Eq. (3.9.23) for the displacements yields
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$$
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d _ {2 x} = 1 1. 9 1 \times 1 0 ^ {- 3} \mathrm{m} \tag {3.9.24}
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$$
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$$
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d _ {3 x} ^ {\prime} = 5. 6 1 3 \times 1 0 ^ {- 3} \mathrm{m}
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$$
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Postmultiplying the known displacement vector times Eq. (3.9.22) (see Eq. (3.9.14), we obtain the reactions as
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$$
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F _ {1 x} = - 5 0 0 \mathrm{kN}
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$$
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$$
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F _ {1 y} = - 5 0 0 \mathrm{kN} \tag {3.9.25}
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$$
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$$
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F _ {2 y} = 0
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$$
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$$
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F _ {3 y} ^ {\prime} = 7 0 7 \mathrm{kN}
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$$
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The free-body diagram of the truss with the reactions is shown in Figure 3–24. You can easily verify that the truss is in equilibrium.
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In the second method used to handle skewed boundary conditions, we use a boundary element of large stiffness to constrain the desired displacement. This is the method used in some computer programs [9].
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<!-- source-page: 127 -->
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<details>
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<summary>text_image</summary>
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1000 kN
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2
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3
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707 kN
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1
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500 kN
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500 kN
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</details>
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Figure 3–24 Free-body diagram of the truss of Figure 3–23
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Boundary elements are used to specify nonzero displacements and rotations to nodes. They are also used to evaluate reactions at rigid and flexible supports. Boundary elements are two-node elements. The line defined by the two nodes specifies the direction along which the force reaction is evaluated or the displacement is specified. In the case of moment reaction, the line specifies the axis about which the moment is evaluated and the rotation is specified.
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We consider boundary elements that are used to obtain reaction forces (rigid boundary elements) or specify translational displacements (displacement boundary elements) as truss elements with only one nonzero translational stiffness. Boundary elements used to either evaluate reaction moments or specify rotations behave like beam elements with only one nonzero stiffness corresponding to the rotational stiffness about the specified axis.
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The elastic boundary elements are used to model flexible supports and to calculate reactions at skewed or inclined boundaries. Consult Reference [9] for more details about using boundary elements.
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# 3.10 Potential Energy Approach to Derive Bar Element Equations
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We now present the principle of minimum potential energy to derive the bar element equations. Recall from Section 2.6 that the total potential energy $\pi _ { p }$ was defined as the sum of the internal strain energy U and the potential energy of the external forces W:
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$$
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\pi_ {p} = U + \Omega \tag {3.10.1}
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$$
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To evaluate the strain energy for a bar, we consider only the work done by the internal forces during deformation. Because we are dealing with a one-dimensional bar, the internal force doing work is given in Figure 3–25 as $\sigma _ { x } ( \Delta y ) ( \Delta z )$ , due only to normal stress $\sigma _ { x } .$ . The displacement of the x face of the element is $\Delta x ( \varepsilon _ { x } )$ ; the displacement of the $x + \Delta x$ face is $\Delta x ( \varepsilon _ { x } + d \varepsilon _ { x } )$ . The change in displacement is then
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<!-- source-page: 128 -->
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<details>
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<summary>text_image</summary>
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y
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x
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z
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Δy
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σx(Δy)(Δz)
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Δz
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Δx
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Δx dεx
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</details>
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Figure 3–25 Internal force in a one-dimensional bar
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$\Delta x d \varepsilon _ { x } ,$ where $d \varepsilon _ { x }$ is the differential change in strain occurring over length Dx. The differential internal work (or strain energy) dU is the internal force multiplied by the displacement through which the force moves, given by
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$$
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d U = \sigma_ {x} (\Delta y) (\Delta z) (\Delta x) d \varepsilon_ {x} \tag {3.10.2}
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$$
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Rearranging and letting the volume of the element approach zero, we obtain, from Eq. (3.10.2),
|
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$$
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d U = \sigma_ {x} d \varepsilon_ {x} d V \tag {3.10.3}
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$$
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For the whole bar, we then have
|
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$$
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U = \iiint_ {V} \left\{\int_ {0} ^ {\varepsilon_ {x}} \sigma_ {x} d \varepsilon_ {x} \right\} d V \tag {3.10.4}
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$$
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||
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Now, for a linear-elastic (Hooke’s law) material as shown in Figure 3–26, we see that $\sigma _ { x } = E \varepsilon _ { x }$ . Hence substituting this relationship into Eq. (3.10.4), integrating with respect to $\varepsilon _ { x } .$ , and then resubstituting $\sigma _ { x }$ for $E \varepsilon _ { x }$ , we have
|
||
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||
$$
|
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U = \frac {1}{2} \iiint_ {V} \sigma_ {x} \varepsilon_ {x} d V \tag {3.10.5}
|
||
$$
|
||
|
||
as the expression for the strain energy for one-dimensional stress.
|
||
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||
The potential energy of the external forces, being opposite in sign from the external work expression because the potential energy of external forces is lost when the
|
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|
||
|
||
|
||
<details>
|
||
<summary>line</summary>
|
||
| ε_x | σ_x |
|
||
| --- | --- |
|
||
| 0 | 0 |
|
||
| 1 | 1 |
|
||
| 2 | 2 |
|
||
| 3 | 3 |
|
||
| 4 | 4 |
|
||
| 5 | 5 |
|
||
| 6 | 6 |
|
||
| 7 | 7 |
|
||
| 8 | 8 |
|
||
| 9 | 9 |
|
||
| 10 | 10 |
|
||
</details>
|
||
|
||
Figure 3–26 Linear-elastic (Hooke’s law) material
|
||
|
||
<!-- source-page: 129 -->
|
||
|
||
work is done by the external forces, is given by
|
||
|
||
$$
|
||
\Omega = - \iint_ {V} \int \hat {X} _ {b} \hat {u} d V - \iint_ {S _ {1}} \hat {T} _ {x} \hat {u} _ {s} d S - \sum_ {i = 1} ^ {M} \hat {f} _ {i x} \hat {d} _ {i x} \tag {3.10.6}
|
||
$$
|
||
|
||
where the first, second, and third terms on the right side of Eq. (3.10.6) represent the potential energy of (1) body forces $\hat { X } _ { b } .$ , typically from the self-weight of the bar (in units of force per unit volume) moving through displacement function u^, (2) surface loading or traction ${ \hat { T } } _ { x } ,$ typically from distributed loading acting along the surface of the element (in units of force per unit surface area) moving through displacements $\hat { u } _ { s }$ , where $\hat { u } _ { s }$ are the displacements occurring over surface $S _ { 1 }$ , and (3) nodal concentrated forces $\hat { f } _ { i x }$ moving through nodal displacements $\hat { d } _ { i x }$ . The forces $\hat { X } _ { b } , \hat { T } _ { x }$ , and $\hat { f } _ { i x }$ are considered to act in the local x^ direction of the bar as shown in Figure 3–27. In Eqs. (3.10.5) and (3.10.6), V is the volume of the body and $S _ { 1 }$ is the part of the surface S on which surface loading acts. For a bar element with two nodes and one degree of freedom per node, M ¼ 2.
|
||
|
||
We are now ready to describe the finite element formulation of the bar element equations by using the principle of minimum potential energy.
|
||
|
||
The finite element process seeks a minimum in the potential energy within the constraint of an assumed displacement pattern within each element. The greater the number of degrees of freedom associated with the element (usually meaning increasing the number of nodes), the more closely will the solution approximate the true one and ensure complete equilibrium (provided the true displacement can, in the limit, be approximated). An approximate finite element solution found by using the stiffness method will always provide an approximate value of potential energy greater than or equal to the correct one. This method also results in a structure behavior that is predicted to be physically stiffer than, or at best to have the same stiffness as, the actual one. This is explained by the fact that the structure model is allowed to displace only into shapes defined by the terms of the assumed displacement field within each element of the structure. The correct shape is usually only approximated by the assumed field, although the correct shape can be the same as the assumed field. The assumed field effectively constrains the structure from deforming in its natural manner. This constraint effect stiffens the predicted behavior of the structure.
|
||
|
||
Apply the following steps when using the principle of minimum potential energy to derive the finite element equations.
|
||
|
||
1. Formulate an expression for the total potential energy.
|
||
2. Assume the displacement pattern to vary with a finite set of undetermined parameters (here these are the nodal displacements $d _ { i x } )$ which are substituted into the expression for total potential energy.
|
||
3. Obtain a set of simultaneous equations minimizing the total potential energy with respect to these nodal parameters. These resulting equations represent the element equations.
|
||
|
||
The resulting equations are the approximate (or possibly exact) equilibrium equations whose solution for the nodal parameters seeks to minimize the potential energy when back-substituted into the potential energy expression. The preceding
|
||
|
||
<!-- source-page: 130 -->
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
S₁
|
||
T̂ₓ
|
||
f̂₁ₓ
|
||
L
|
||
X̂ₕ
|
||
f̂₂ₓ
|
||
x̂
|
||
</details>
|
||
|
||
Figure 3–27 General forces acting on a one-dimensional bar
|
||
|
||
three steps will now be followed to derive the bar element equations and stiffness matrix.
|
||
|
||
Consider the bar element of length L, with constant cross-sectional area A, shown in Figure 3–27. Using Eqs. (3.10.5) and (3.10.6), we find that the total potential energy, Eq. (3.10.1), becomes
|
||
|
||
$$
|
||
\pi_ {p} = \frac {A}{2} \int_ {0} ^ {L} \sigma_ {x} \varepsilon_ {x} d \hat {x} - \hat {f} _ {1 x} \hat {d} _ {1 x} - \hat {f} _ {2 x} \hat {d} _ {2 x} - \iint_ {S _ {1}} \hat {u} _ {s} \hat {T} _ {x} d S - \iint_ {V} \hat {u} \hat {X} _ {b} d V \tag {3.10.7}
|
||
$$
|
||
|
||
because A is a constant and variables $\sigma _ { x }$ and $\varepsilon _ { x }$ at most vary with x^.
|
||
|
||
From Eqs. (3.1.3) and (3.1.4), we have the axial displacement function expressed in terms of the shape functions and nodal displacements by
|
||
|
||
$$
|
||
\hat {u} = [ N ] \{\hat {d} \} \quad \hat {u} _ {s} = [ N _ {S} ] \{\hat {d} \} \tag {3.10.8}
|
||
$$
|
||
|
||
where
|
||
|
||
$$
|
||
[ N ] = \left[ 1 - \frac {\hat {x}}{L} \quad \frac {\hat {x}}{L} \right] \tag {3.10.9}
|
||
$$
|
||
|
||
$[ N _ { S } ]$ is the shape function matrix evaluated over the surface that the distributed surface traction acts and
|
||
|
||
$$
|
||
\{\hat {d} \} = \left\{ \begin{array}{l} \hat {d} _ {1 x} \\ \hat {d} _ {2 x} \end{array} \right\} \tag {3.10.10}
|
||
$$
|
||
|
||
Then, using the strain/displacement relationship $\varepsilon _ { x } = d \hat { u } / d \hat { x }$ , we can write the axial strain as
|
||
|
||
$$
|
||
\{\varepsilon_ {x} \} = \left[ - \frac {1}{L} \quad \frac {1}{L} \right] \{\hat {d} \} \tag {3.10.11}
|
||
$$
|
||
|
||
or $\{ \varepsilon _ { x } \} = [ B ] \{ \hat { d } \}$ ð3:10:12Þ
|