김경종
513 lines
23 KiB
Markdown
513 lines
23 KiB
Markdown
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<details>
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<summary>text_image</summary>
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ˆu^(1) = \frac{ˆd_{2x} - \hat{d}_{1x}}{L} \hat{x} + \hat{d}_{1x} \quad \hat{u}^{(2)} = \frac{\hat{d}_{3x} - \hat{d}_{2x}}{L} \hat{x} + \hat{d}_{2x} \nˆd_{1x}^{(1)} \quad ① \quad \hat{d}_{2x}^{(1)} \quad \hat{d}_{2x}^{(2)} \quad ② \quad \hat{d}_{3x}^{(2)} \n1 \quad L \quad 2 \quad L \quad 3 \quad \hat{x}
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</details>
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Figure 3–5 Interelement continuity of a two-bar structure
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common to two or more elements remain common to these elements upon deformation and thus prevent overlaps or voids between elements. For example, consider the two-bar structure shown in Figure 3–5. For the two-bar structure, the linear function for u^ [Eq. (3.1.2)] within each element will ensure that elements 1 and 2 remain connected; the displacement at node 2 for element 1 will ethe displacement at the same node 2 for element 2; that is, $\mathbf { \hat { \boldsymbol { d } } } _ { 2 x } ^ { ( 1 ) } = \hat { \boldsymbol { d } } _ { 2 x } ^ { ( 2 ) }$ d2x . This rule was also illustrated by Eq. (2.3.3). The linear function is then called a conforming, or compatible, function for the bar element because it ensures the satisfaction both of continuity between adjacent elements and of continuity within the element.
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In general, the symbol $C ^ { m }$ is used to describe the continuity of a piecewise field (such as axial displacement), where the superscript m indicates the degree of derivative that is interelement continuous. A field is then $C ^ { 0 }$ continuous if the function itself is interelement continuous. For instance, for the field variable being the axial displacement illustrated in Figure 3–5, the displacement is continuous across the common node 2. Hence the displacement field is said to be $C ^ { 0 }$ continuous. Bar elements, plane elements (see Chapter $7 ) _ { : }$ , and solid elements (Chapter 11) are $C ^ { 0 }$ elements in that they enforce displacement continuity across the common boundaries.
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If the function has both its field variable and its first derivative continuous across the common boundary, then the field variable is said to be $C ^ { 1 }$ continuous. We will later see that the beam and plate elements are $C ^ { 1 }$ continuous. That is, they enforce both displacement and slope continuity across common boundaries.
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4. The approximation function should allow for rigid-body displacement and for a state of constant strain within the element. The onedimensional displacement function [Eq. (3.1.1)] satisfies these criteria because the $a _ { 1 }$ term allows for rigid-body motion (constant motion of the body without straining) and the $a _ { 2 } \hat { x }$ term allows for constant strain because $\varepsilon _ { x } = d \hat { u } / d \hat { x } = a _ { 2 }$ is a constant. (This state of constant strain in the element can, in fact, occur if elements are chosen small enough.) The simple polynomial Eq. (3.1.1) satisfying this fourth guideline is then said to be complete for the bar element.
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<details>
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<summary>line</summary>
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| Number of elements | Displacement |
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| ------------------ | ------------ |
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| 0 | 0 |
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| 2.5 | ~0.5 |
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| 5 | ~1.0 |
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| 10 | ~1.5 |
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| 15 | ~2.0 |
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</details>
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Figure 3–6 Convergence to the exact solution for displacement as the number of elements of a finite element solution is increased
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This idea of completeness also means in general that the lowerorder term cannot be omitted in favor of the higher-order term. For the simple linear function, this means $a _ { 1 }$ cannot be omitted while keeping $a _ { 2 } { \hat { x } } .$ . Completeness of a function is a necessary condition for convergence to the exact answer, for instance, for displacements and stresses (Figure 3–6) (see Reference [3]). Figure 3–6 illustrates monotonic convergence toward an exact solution for displacement as the number of elements in a finite element solution is increased. Monotonic convergence is then the process in which successive approximation solutions (finite element solutions) approach the exact solution consistently without changing sign or direction.
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The idea that the interpolation (approximation) function must allow for a rigidbody displacement means that the function must be capable of yielding a constant value $( \operatorname { s a y } , a _ { 1 } )$ , because such a value can, in fact, occur. Therefore, we must consider the case
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$$
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\hat {u} = a _ {1} \tag {3.2.1}
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$$
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For $\hat { u } = a _ { 1 }$ requires nodal displacements $\hat { d } _ { 1 x } = \hat { d } _ { 2 x }$ to obtain a rigid-body displacement. Therefore
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$$
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a _ {1} = \hat {d} _ {1 x} = \hat {d} _ {2 x} \tag {3.2.2}
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$$
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Using Eq. (3.2.2) in Eq. (3.1.3), we have
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$$
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\hat {u} = N _ {1} \hat {d} _ {1 x} + N _ {2} \hat {d} _ {2 x} = (N _ {1} + N _ {2}) a _ {1} \tag {3.2.3}
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$$
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From Eqs. (3.2.1) and (3.2.3), we then have
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$$
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\hat {u} = a _ {1} = (N _ {1} + N _ {2}) a _ {1} \tag {3.2.4}
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$$
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Therefore, by Eq. (3.2.4), we obtain
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$$
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N _ {1} + N _ {2} = 1 \tag {3.2.5}
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$$
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Thus Eq. (3.2.5) shows that the displacement interpolation functions must add to unity at every point within the element so that u^ will yield a constant value when a rigid-body displacement occurs.
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# 3.3 Transformation of Vectors in Two Dimensions
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In many problems it is convenient to introduce both local and global (or reference) coordinates. Local coordinates are always chosen to represent the individual element conveniently. Global coordinates are chosen to be convenient for the whole structure.
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Given the nodal displacement of an element, represented by the vector d in Figure 3–7, we want to relate the components of this vector in one coordinate system to components in another. For general purposes, we will assume in this section that d is not coincident with either the local or the global axis. In this case, we want to relate global displacement components to local ones. In so doing, we will develop a transformation matrix that will subsequently be used to develop the global stiffness matrix for a bar element. We define the angle y to be positive when measured counterclockwise from x to x^. We can express vector displacement d in both global and local coordinates by
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$$
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\mathbf {d} = d _ {x} \mathbf {i} + d _ {y} \mathbf {j} = \hat {d} _ {x} \hat {\mathbf {i}} + \hat {d} _ {y} \hat {\mathbf {j}} \tag {3.3.1}
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$$
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where i and j are unit vectors in the x and y global directions and ^i and ^j are unit vectors in the x^ and y^ local directions. We will now relate i and j to ^i and ^j through use of Figure 3–8.
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<details>
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<summary>text_image</summary>
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y
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j
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d
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x̂
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î
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</details>
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Figure 3–7 General displacement vector d
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<details>
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<summary>text_image</summary>
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y
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y
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b'
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j
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ĵ
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a'
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</details>
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Figure 3–8 Relationship between local and global unit vectors
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Using Figure 3–8 and vector addition, we obtain
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$$
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\mathbf {a} + \mathbf {b} = \mathbf {i} \tag {3.3.2}
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$$
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Also, from the law of cosines,
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$$
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| \mathbf {a} | = | \mathbf {i} | \cos \theta \tag {3.3.3}
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$$
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and because i is, by definition, a unit vector, its magnitude is given by
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$$
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| \mathbf {i} | = 1 \tag {3.3.4}
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$$
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Therefore, we obtain $| \mathbf { a } | = 1 \cos \theta$ ð3:3:5Þ
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Similarly, $| \mathbf { b } | = 1 \sin \theta$ ð3:3:6Þ
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Now a is in the $\hat { \bf i }$ direction and b is in the $- \hat { \bf j }$ direction. Therefore,
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$$
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\mathbf {a} = | \mathbf {a} | \hat {\mathbf {i}} = (\cos \theta) \hat {\mathbf {i}} \tag {3.3.7}
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$$
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and $\mathbf { b } = | \mathbf { b } | ( - \hat { \mathbf { j } } ) = ( \sin \theta ) ( - \hat { \mathbf { j } } )$ ð3:3:8Þ
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Using Eqs. (3.3.7) and (3.3.8) in Eq. (3.3.2) yields
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$$
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\mathbf {i} = \cos \theta \hat {\mathbf {i}} - \sin \theta \hat {\mathbf {j}} \tag {3.3.9}
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$$
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Similarly, from Figure 3–8, we obtain
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$$
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\mathbf {a} ^ {\prime} + \mathbf {b} ^ {\prime} = \mathbf {j} \tag {3.3.10}
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$$
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$$
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\mathbf {a} ^ {\prime} = \cos \theta \hat {\mathbf {j}} \tag {3.3.11}
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$$
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$$
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\mathbf {b} ^ {\prime} = \sin \theta \hat {\mathbf {i}} \tag {3.3.12}
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$$
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Using Eqs. (3.3.11) and (3.3.12) in Eq. (3.3.10), we have
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$$
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\mathbf {j} = \sin \theta \hat {\mathbf {i}} + \cos \theta \hat {\mathbf {j}} \tag {3.3.13}
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$$
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Now, using Eqs. (3.3.9) and (3.3.13) in Eq. (3.3.1), we have
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$$
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d _ {x} (\cos \theta \hat {\mathbf {i}} - \sin \theta \hat {\mathbf {j}}) + d _ {y} (\sin \theta \hat {\mathbf {i}} + \cos \theta \hat {\mathbf {j}}) = \hat {d} _ {x} \hat {\mathbf {i}} + \hat {d} _ {y} \hat {\mathbf {j}} \tag {3.3.14}
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$$
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Combining like coefficients of ^i and $\hat { \bf j }$ in Eq. (3.3.14), we obtain
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$$
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d _ {x} \cos \theta + d _ {y} \sin \theta = \hat {d} _ {x}
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$$
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ð3:3:15Þ and $- d _ { x } \sin \theta + d _ { y } \cos \theta = \hat { d } _ { y }$
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In matrix form, Eqs. (3.3.15) are written as
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$$
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\left\{ \begin{array}{l} \hat {d} _ {x} \\ \hat {d} _ {y} \end{array} \right\} = \left[ \begin{array}{c c} C & S \\ - S & C \end{array} \right] \left\{ \begin{array}{l} d _ {x} \\ d _ {y} \end{array} \right] \tag {3.3.16}
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$$
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where $C = \cos \theta$ and $S = \sin \theta$ .
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Equation (3.3.16) relates the global displacement $\underline { d }$ to the local displacement $\hat { \underline { d } } .$ The matrix
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$$
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\left[ \begin{array}{c c} C & S \\ - S & C \end{array} \right] \tag {3.3.17}
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$$
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<details>
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<summary>text_image</summary>
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y
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x̂
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dy
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dx̂
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θ
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dx
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</details>
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Figure 3–9 Relationship between local and global displacements
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is called the transformation (or rotation) matrix. For an additional description of this matrix, see Appendix A. It will be used in Section 3.4 to develop the global stiffness matrix for an arbitrarily oriented bar element and to transform global nodal displacements and forces to local ones.
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Now, for the case of $\hat { d } _ { y } = 0 \mathrm { { ; } }$ , we have, from Eq. (3.3.1),
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$$
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d _ {x} \mathbf {i} + d _ {y} \mathbf {j} = \hat {d} _ {x} \hat {\mathbf {i}} \tag {3.3.18}
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$$
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Figure 3–9 shows $\hat { d } _ { x }$ expressed in terms of global x and y components. Using trigonometry and Figure 3–9, we then obtain the magnitude of $\hat { d } _ { x }$ as
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$$
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\hat {d} _ {x} = C d _ {x} + S d _ {y} \tag {3.3.19}
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$$
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Equation (3.3.19) is equivalent to equation 1 of Eq. (3.3.16).
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# Example 3.2
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The global nodal displacements at node 2 have been determined to be $d _ { 2 x } = 0 . 1$ in. and $d _ { 2 y } = 0 . 2$ in. for the bar element shown in Figure 3–10. Determine the local x^ displacement at node 2.
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<details>
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<summary>text_image</summary>
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y
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2
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x̂
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60°
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1
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</details>
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Figure 3–10 Bar element
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Using Eq. (3.3.19), we obtain
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$$
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\hat {d} _ {2 x} = (\cos 6 0 ^ {\circ}) (0. 1) + (\sin 6 0 ^ {\circ}) (0. 2) = 0. 2 2 3 \text { in. }
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$$
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# d 3.4 Global Stiffness Matrix
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We will now use the transformation relationship Eq. (3.3.16) to obtain the global stiffness matrix for a bar element. We need the global stiffness matrix of each element to assemble the total global stiffness matrix of the structure. We have shown in Eq. (3.1.13) that for a bar element in the local coordinate system,
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$$
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\left\{ \begin{array}{l} \hat {f} _ {1 x} \\ \hat {f} _ {2 x} \end{array} \right\} = \frac {A E}{L} \left[ \begin{array}{c c} 1 & - 1 \\ - 1 & 1 \end{array} \right] \left\{ \begin{array}{l} \hat {d} _ {1 x} \\ \hat {d} _ {2 x} \end{array} \right\} \tag {3.4.1}
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$$
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or $\underline { { \hat { f } } } = \underline { { \hat { k } } } \underline { { \hat { d } } }$ ð3:4:2Þ
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We now want to relate the global element nodal forces $\underline { { f } }$ to the global nodal displacements d for a bar element arbitrarily oriented with respect to the global axes as was shown in Figure 3–2. This relationship will yield the global stiffness matrix $\underline { { k } }$ of the element. That is, we want to find a matrix $\underline { { k } }$ such that
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$$
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\left\{ \begin{array}{l} f _ {1 x} \\ f _ {1 y} \\ f _ {2 x} \\ f _ {2 y} \end{array} \right\} = \underline {{k}} \left\{ \begin{array}{l} d _ {1 x} \\ d _ {1 y} \\ d _ {2 x} \\ d _ {2 y} \end{array} \right\} \tag {3.4.3}
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$$
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or, in simplified matrix form, Eq. (3.4.3) becomes
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$$
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\underline {{f}} = \underline {{k}} \underline {{d}} \tag {3.4.4}
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$$
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We observe from Eq. (3.4.3) that a total of four components of force and four of displacement arise when global coordinates are used. However, a total of two components of force and two of displacement appear for the local-coordinate representation of a spring or a bar, as shown by Eq. (3.4.1). By using relationships between local and global force components and between local and global displacement components, we will be able to obtain the global stiffness matrix. We know from transformation relationship Eq. (3.3.15) that
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$$
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\begin{array}{l} \hat {d} _ {1 x} = d _ {1 x} \cos \theta + d _ {1 y} \sin \theta \\ \hat {\hat {d}} _ {2 x} = d _ {2 x} \cos \theta + d _ {2 y} \sin \theta \end{array} \tag {3.4.5}
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$$
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$$
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\hat {d} _ {2 x} = d _ {2 x} \cos \theta + d _ {2 y} \sin \theta
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$$
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In matrix form, Eqs. (3.4.5) can be written as
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$$
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\left\{ \begin{array}{l} \hat {d} _ {1 x} \\ \hat {d} _ {2 x} \end{array} \right\} = \left[ \begin{array}{c c c c} C & S & 0 & 0 \\ 0 & 0 & C & S \end{array} \right] \left\{ \begin{array}{l} d _ {1 x} \\ d _ {1 y} \\ d _ {2 x} \\ d _ {2 y} \end{array} \right\} \tag {3.4.6}
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$$
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or as $\underline { { \hat { d } } } = \underline { { T } } ^ { * } \underline { { d } }$ ð3:4:7Þ
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where $\underline { { { T } } } ^ { * } = \left[ \begin{array} { l l l l } { { C } } & { { S } } & { { 0 } } & { { 0 } } \\ { { 0 } } & { { 0 } } & { { C } } & { { S } } \end{array} \right]$ ð3:4:8Þ
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Similarly, because forces transform in the same manner as displacements, we have
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$$
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\left\{ \begin{array}{l} \hat {f} _ {1 x} \\ \hat {f} _ {2 x} \end{array} \right\} = \left[ \begin{array}{c c c c} C & S & 0 & 0 \\ 0 & 0 & C & S \end{array} \right] \left\{ \begin{array}{l} f _ {1 x} \\ f _ {1 y} \\ f _ {2 x} \\ f _ {2 y} \end{array} \right\} \tag {3.4.9}
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$$
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Using Eq. (3.4.8), we can write Eq. (3.4.9) as
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$$
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\underline {{\hat {f}}} = \underline {{T}} ^ {*} \underline {{f}} \tag {3.4.10}
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$$
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Now, substituting Eq. (3.4.7) into Eq. (3.4.2), we obtain
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$$
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\underline {{\hat {f}}} = \underline {{\hat {k}}} \underline {{T}} ^ {*} \underline {{d}} \tag {3.4.11}
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$$
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and using Eq. (3.4.10) in Eq. (3.4.11) yields
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$$
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\underline {{T}} ^ {*} \underline {{f}} = \hat {\underline {{k}}} \underline {{T}} ^ {*} \underline {{d}} \tag {3.4.12}
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$$
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However, to write the final expression relating global nodal forces to global nodal displacements for an element, we must invert $\underline { { T } } ^ { * }$ in Eq. (3.4.12). This is not immediately possible because $\underline { { T } } ^ { * }$ is not a square matrix. Therefore, we must expand $\underline { { \hat { d } } } , \hat { f } ,$ and $\underline { { \hat { k } } }$ to the order that is consistent with the use of global coordinates even though $\hat { f } _ { 1 y }$ and $\hat { f } _ { 2 y }$ are zero. Using Eq. (3.3.16) for each nodal displacement, we thus obtain
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$$
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\left\{ \begin{array}{l} \hat {d} _ {1 x} \\ \hat {d} _ {1 y} \\ \hat {d} _ {2 x} \\ \hat {d} _ {2 y} \end{array} \right\} = \left[ \begin{array}{c c c c} C & S & 0 & 0 \\ - S & C & 0 & 0 \\ 0 & 0 & C & S \\ 0 & 0 & - S & C \end{array} \right] \left\{ \begin{array}{l} d _ {1 x} \\ d _ {1 y} \\ d _ {2 x} \\ d _ {2 y} \end{array} \right\} \tag {3.4.13}
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$$
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or $\hat { \underline { d } } = { \underline { T } } { \underline { d } }$ ð3:4:14Þ
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where $\underline { { T } } = \left[ \begin{array} { r r r r } { C } & { S } & { 0 } & { 0 } \\ { - S } & { C } & { 0 } & { 0 } \\ { 0 } & { 0 } & { C } & { S } \\ { 0 } & { 0 } & { - S } & { C } \end{array} \right]$ ð3:4:15Þ
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Similarly, we can write
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$$
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\underline {{\hat {f}}} = \underline {{T}} \underline {{f}} \tag {3.4.16}
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$$
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because forces are like displacements—both are vectors. Also, $\underline { { \hat { k } } }$ must be expanded to a $4 \times 4$ matrix. Therefore, Eq. (3.4.1) in expanded form becomes
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$$
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\left\{ \begin{array}{l} \hat {f} _ {1 x} \\ \hat {f} _ {1 y} \\ \hat {f} _ {2 x} \\ \hat {f} _ {2 y} \end{array} \right\} = \frac {A E}{L} \left[ \begin{array}{c c c c} 1 & 0 & - 1 & 0 \\ 0 & 0 & 0 & 0 \\ - 1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right] \left\{ \begin{array}{l} \hat {d} _ {1 x} \\ \hat {d} _ {1 y} \\ \hat {d} _ {2 x} \\ \hat {d} _ {2 y} \end{array} \right\} \tag {3.4.17}
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$$
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<!-- source-page: 98 -->
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In Eq. (3.4.17), because $\hat { f } _ { 1 y }$ and $\hat { f } _ { 2 \nu }$ are zero, rows of zeros corresponding to the row numbers $\hat { f } _ { 1 y }$ and $\hat { f } _ { 2 y }$ appear in $\underline { { \hat { k } } } .$ . Now, using Eqs. (3.4.14) and (3.4.16) in Eq. (3.4.2), we obtain
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||
$$
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\underline {{T}} \underline {{f}} = \hat {\underline {{k}}} \underline {{T}} \underline {{d}} \tag {3.4.18}
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||
$$
|
||
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Equation (3.4.18) is Eq. (3.4.12) expanded. Premultiplying both sides of Eq. (3.4.18) by $\underline { { T } } ^ { - 1 }$ , we have
|
||
|
||
$$
|
||
\underline {{f}} = \underline {{T}} ^ {- 1} \underline {{\hat {k}}} \underline {{T}} \underline {{d}} \tag {3.4.19}
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$$
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where $\underline { { T } } ^ { - 1 }$ is the inverse of $\underline { { T } } .$ . However, it can be shown (see Problem 3.28) that
|
||
|
||
$$
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\underline {{T}} ^ {- 1} = \underline {{T}} ^ {T} \tag {3.4.20}
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$$
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where $\underline { { T } } ^ { T }$ is the transpose of $\underline { { T } } .$ . The property of square matrices such as $\underline { T }$ given by Eq. (3.4.20) defines $\underline { T }$ to be an orthogonal matrix. For more about orthogonal matrices, see Appendix A. The transformation matrix $\underline { T }$ between rectangular coordinate frames is orthogonal. This property of $\underline { T }$ is used throughout this text. Substituting Eq. (3.4.20) into Eq. (3.4.19), we obtain
|
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|
||
$$
|
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\underline {{f}} = \underline {{T}} ^ {T} \hat {\underline {{k}}} \underline {{T}} \underline {{d}} \tag {3.4.21}
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||
$$
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Equating Eqs. (3.4.4) and (3.4.21), we obtain the global stiffness matrix for an element as
|
||
|
||
$$
|
||
\underline {{k}} = \underline {{T}} ^ {T} \hat {\underline {{k}}} \underline {{T}} \tag {3.4.22}
|
||
$$
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||
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Substituting Eq. (3.4.15) for $\underline { T }$ and the expanded form of $\underline { { \hat { k } } }$ given in Eq. (3.4.17) into Eq. (3.4.22), we obtain $\underline { { k } }$ given in explicit form by
|
||
|
||
$$
|
||
\underline {{k}} = \frac {A E}{L} \left[ \begin{array}{c c c c} C ^ {2} & C S & - C ^ {2} & - C S \\ & S ^ {2} & - C S & - S ^ {2} \\ & & C ^ {2} & C S \\ \text { Symmetry } & & & S ^ {2} \end{array} \right] \tag {3.4.23}
|
||
$$
|
||
|
||
Now, because the trial displacement function Eq. (3.1.1) was assumed piecewisecontinuous element by element, the stiffness matrix for each element can be summed by using the direct stiffness method to obtain
|
||
|
||
$$
|
||
\sum_ {e = 1} ^ {N} \underline {{k}} ^ {(e)} = \underline {{K}} \tag {3.4.24}
|
||
$$
|
||
|
||
where $\underline { { K } }$ is the total stiffness matrix and N is the total number of elements. Similarly, each element global nodal force matrix can be summed such that
|
||
|
||
$$
|
||
\sum_ {e = 1} ^ {N} \underline {{f}} ^ {(e)} = \underline {{F}} \tag {3.4.25}
|
||
$$
|
||
|
||
K now relates the global nodal forces $\underline { { F } }$ to the global nodal displacements $\underline { d }$ for the whole structure by
|
||
|
||
$$
|
||
\underline {{F}} = \underline {{K}} \underline {{d}} \tag {3.4.26}
|
||
$$
|
||
|
||
<!-- source-page: 99 -->
|
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|
||
For the bar element shown in Figure 3–11, evaluate the global stiffness matrix with respect to the x-y coordinate system. Let the bar’s cross-sectional area equal 2 in.2, length equal 60 in., and modulus of elasticity equal $3 0 \times 1 0 ^ { 6 }$ psi. The angle the bar makes with the x axis is 30
|
||
.
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
y
|
||
x
|
||
30°
|
||
x̂
|
||
</details>
|
||
|
||
Figure 3–11 Bar element for stiffness matrix evaluation
|
||
|
||
To evaluate the global stiffness matrix $\underline { { k } }$ for a bar, we use Eq. (3.4.23) with angle y defined to be positive when measured counterclockwise from x to x^. Therefore,
|
||
|
||
$$
|
||
\theta = 3 0 ^ {\circ} \quad C = \cos 3 0 ^ {\circ} = \frac {\sqrt {3}}{2} \quad S = \sin 3 0 ^ {\circ} = \frac {1}{2}
|
||
$$
|
||
|
||
$$
|
||
\underline {{k}} = \frac {(2) (3 0 \times 1 0 ^ {6})}{6 0} \left[ \begin{array}{c c c c} \frac {3}{4} & \frac {\sqrt {3}}{4} & \frac {- 3}{4} & \frac {- \sqrt {3}}{4} \\ & \frac {1}{4} & \frac {- \sqrt {3}}{4} & \frac {- 1}{4} \\ & & \frac {3}{4} & \frac {\sqrt {3}}{4} \\ & & & \frac {1}{4} \\ \text {Symmetry} & & & \end{array} \right] \frac {\mathrm{lb}}{\text {in.}} \tag {3.4.27}
|
||
$$
|
||
|
||
Simplifying Eq. (3.4.27), we have
|
||
|
||
$$
|
||
\underline {{k}} = 1 0 ^ {6} \left[ \begin{array}{c c c c} 0. 7 5 & 0. 4 3 3 & - 0. 7 5 & - 0. 4 3 3 \\ & 0. 2 5 & - 0. 4 3 3 & - 0. 2 5 \\ & & 0. 7 5 & 0. 4 3 3 \\ \text { Symmetry } & & & 0. 2 5 \end{array} \right] \frac {\mathrm{lb}}{\text { in. }} \tag {3.4.28}
|
||
$$
|
||
|
||
<!-- source-page: 100 -->
|
||
|
||
# 3.5 Computation of Stress for a Bar in the x-y Plane
|
||
|
||
We will now consider the determination of the stress in a bar element. For a bar, the local forces are related to the local displacements by Eq. (3.1.13) or Eq. (3.4.17). This equation is repeated here for convenience.
|
||
|
||
$$
|
||
\left\{ \begin{array}{l} \hat {f} _ {1 x} \\ \hat {f} _ {2 x} \end{array} \right\} = \frac {A E}{L} \left[ \begin{array}{c c} 1 & - 1 \\ - 1 & 1 \end{array} \right] \left\{ \begin{array}{l} \hat {d} _ {1 x} \\ \hat {d} _ {2 x} \end{array} \right\} \tag {3.5.1}
|
||
$$
|
||
|
||
The usual definition of axial tensile stress is axial force divided by cross-sectional area. Therefore, axial stress is
|
||
|
||
$$
|
||
\sigma = \frac {\hat {f} _ {2 x}}{A} \tag {3.5.2}
|
||
$$
|
||
|
||
where $\hat { f } _ { 2 x }$ is used because it is the axial force that pulls on the bar as shown in Figure 3–12. By Eq. (3.5.1),
|
||
|
||
$$
|
||
\hat {f} _ {2 x} = \frac {A E}{L} [ - 1 \quad 1 ] \left\{ \begin{array}{l} \hat {d} _ {1 x} \\ \hat {d} _ {2 x} \end{array} \right\} \tag {3.5.3}
|
||
$$
|
||
|
||
Therefore, combining Eqs. (3.5.2) and (3.5.3) yields
|
||
|
||
$$
|
||
\underline {{\sigma}} = \frac {E}{L} [ - 1 \quad 1 ] \underline {{\hat {d}}} \tag {3.5.4}
|
||
$$
|
||
|
||
Now, using Eq. (3.4.7), we obtain
|
||
|
||
$$
|
||
\underline {{\sigma}} = \frac {E}{L} [ - 1 \quad 1 ] \underline {{T}} ^ {*} \underline {{d}} \tag {3.5.5}
|
||
$$
|
||
|
||
Equation (3.5.5) can be expressed in simpler form as
|
||
|
||
$$
|
||
\underline {{\sigma}} = \underline {{C}} ^ {\prime} \underline {{d}} \tag {3.5.6}
|
||
$$
|
||
|
||
where, when we use Eq. (3.4.8),
|
||
|
||
$$
|
||
\underline {{{C}}} ^ {\prime} = \frac {E}{L} \left[ - 1 \quad 1 \right] \left[ \begin{array}{c c c c} C & S & 0 & 0 \\ 0 & 0 & C & S \end{array} \right] \tag {3.5.7}
|
||
$$
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
y
|
||
2
|
||
f̂₂ₓ
|
||
L
|
||
1
|
||
f̂₁ₓ
|
||
x
|
||
</details>
|
||
|
||
Figure 3–12 Basic bar element with positive nodal forces
|