김경종
363 lines
22 KiB
Markdown
363 lines
22 KiB
Markdown
<!-- source-page: 231 -->
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4.47 Derive the equations for the beam element on an elastic foundation (Figure P4–47) using the principle of minimum potential energy. Here $k _ { f }$ is the subgrade spring constant per unit length. The potential energy of the beam is
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$$
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\pi_ {p} = \int_ {0} ^ {L} \frac {1}{2} E I \left(v ^ {\prime \prime}\right) ^ {2} d x + \int_ {o} ^ {L} \frac {k _ {f} v ^ {2}}{2} d x - \int_ {0} ^ {L} w v d x
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$$
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4.48 Derive the equations for the beam element on an elastic foundation (see Figure P4–47) using Galerkin’s method. The basic differential equation for the beam on an elastic foundation is
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$$
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(E I v ^ {\prime \prime}) ^ {\prime \prime} = - w + k _ {f} v
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$$
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4.49–76 Solve problems 4.5–4.11, 4.19–4.36, and 4.40–4.42 using a suitable computer program.
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4.77 For the beam shown, use a computer program to determine the deflection at the mid-span using four beam elements, making the shear area zero and then making the shear area equal 5/6 times the cross-sectional area (b times h). Then make the beam have decreasing spans of 200 mm, 100 mm, and 50 mm with zero shear area and then 5/6 times the cross-sectional area. Compare the answers. Based on your program answers, can you conclude whether your program includes the effects of transverse shear deformation?
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<details>
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<summary>text_image</summary>
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50,000 N
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200 mm
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400 mm
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h = 50 mm
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b = 25 mm
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</details>
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Figure P4–77
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4.78 For the beam shown in Figure P4–77, use a longhand solution to solve the problem. Compare answers using the beam stiffness matrix, Eq. (4.1.14), without transverse shear deformation effects and then Eq. (4.1.15o), which includes the transverse shear effects.
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<!-- source-page: 232 -->
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# Introduction
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Many structures, such as buildings (Figure 5–1) and bridges, are composed of frames and/or grids. This chapter develops the equations and methods for solution of plane frames and grids.
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First, we will develop the stiffness matrix for a beam element arbitrarily oriented in a plane. We will then include the axial nodal displacement degree of freedom in the local beam element stiffness matrix. Then we will combine these results to develop the stiffness matrix, including axial deformation effects, for an arbitrarily oriented beam element, thus making it possible to analyze plane frames. Specific examples of plane frame analysis follow. We will then consider frames with inclined or skewed supports.
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Next, we will develop the grid element stiffness matrix. We will present the solution of a grid deck system to illustrate the application of the grid equations. We will then develop the stiffness matrix for a beam element arbitrarily oriented in space. We will also consider the concept of substructure analysis.
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# 5.1 Two-Dimensional Arbitrarily Oriented Beam Element
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We can derive the stiffness matrix for an arbitrarily oriented beam element, as shown in Figure 5–2, in a manner similar to that used for the bar element in Chapter 3. The local axes x^ and y^ are located along the beam element and transverse to the beam element, respectively, and the global axes x and y are located to be convenient for the total structure.
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Recall that we can relate local displacements to global displacements by using Eq. (3.3.16), repeated here for convenience as
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$$
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\left\{ \begin{array}{l} \hat {d} _ {x} \\ \hat {d} _ {y} \end{array} \right\} = \left[ \begin{array}{c c} C & S \\ - S & C \end{array} \right] \left\{ \begin{array}{l} d _ {x} \\ d _ {y} \end{array} \right\} \tag {5.1.1}
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$$
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<!-- source-page: 233 -->
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<details>
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<summary>natural_image</summary>
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Exterior view of a large stadium under construction with scaffolding and cranes, surrounded by open land and trees (no signage or text visible)
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</details>
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Figure 5–1 The Arizona Cardinal Football Stadium under construction—a rigid building frame (Courtesy Ed Yack)
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<details>
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<summary>text_image</summary>
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y
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\hat{d}_{2y}
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\hat{y}
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\hat{d}_{1y}
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\hat{\phi}_{2}
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2
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x
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\theta
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L
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\hat{\phi}_{1}
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1
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x
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</details>
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Figure 5–2 Arbitrarily oriented beam element
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Using the second equation of Eqs. (5.1.1) for the beam element, we relate local nodal degrees of freedom to global degrees of freedom by
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$$
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\left\{ \begin{array}{c} \hat {d} _ {1 y} \\ \hat {\phi} _ {1} \\ \hat {d} _ {2 y} \\ \hat {\phi} _ {2} \end{array} \right\} = \left[ \begin{array}{c c c c c c} - S & C & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & - S & C & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end{array} \right] \left\{ \begin{array}{c} d _ {1 x} \\ d _ {1 y} \\ \phi_ {1} \\ d _ {2 x} \\ d _ {2 y} \\ \phi_ {2} \end{array} \right\} \tag {5.1.2}
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$$
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where, for a beam element, we define
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$$
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\underline {{{T}}} = \left[ \begin{array}{c c c c c c} - S & C & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & - S & C & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end{array} \right] \tag {5.1.3}
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$$
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<!-- source-page: 234 -->
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as the transformation matrix. The axial effects are not yet included. Equation (5.1.2) indicates that rotation is invariant with respect to either coordinate system. For example, $\hat { \phi } _ { 1 } = \phi _ { 1 }$ , and moment $\hat { m } _ { 1 } = m _ { 1 }$ can be considered to be a vector pointing normal to the $\hat { x } { - } \hat { y }$ plane or to the $x { - } y$ plane by the usual right-hand rule. From either viewpoint, the moment is in the $\hat { z } = z$ direction. Therefore, moment is unaffected as the element changes orientation in the x-y plane.
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Substituting Eq. (5.1.3) for $\underline { T }$ and Eq. (4.1.14) for $\underline { { \hat { k } } }$ into Eq. (3.4.22), ${ \underline { { k } } } = { \underline { { T } } } ^ { T } { \underline { { \hat { k } } } } { \underline { { T } } }$ , we obtain the global element stiffness matrix as
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$$
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\underline {{{{k}}}} = \frac {E I}{L ^ {3}} \left[ \begin{array}{c c c c c c} d _ {1 x} & d _ {1 y} & \phi_ {1} & d _ {2 x} & d _ {2 y} & \phi_ {2} \\ 1 2 S ^ {2} & - 1 2 S C & - 6 L S & - 1 2 S ^ {2} & 1 2 S C & - 6 L S \\ & 1 2 C ^ {2} & 6 L C & 1 2 S C & - 1 2 C ^ {2} & 6 L C \\ & & 4 L ^ {2} & 6 L S & - 6 L C & 2 L ^ {2} \\ & & & 1 2 S ^ {2} & - 1 2 S C & 6 L S \\ & & & & 1 2 C ^ {2} & - 6 L C \\ \text {Symmetry} & & & & & 4 L ^ {2} \end{array} \right] \tag {5.1.4}
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$$
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where, again, $C = \cos \theta$ and $S = \sin \theta .$ It is not necessary here to expand T given by Eq. (5.1.3) to make it a square matrix to be able to use Eq. (3.4.22). Because Eq. (3.4.22) is a generally applicable equation, the matrices used must merely be of the correct order for matrix multiplication (see Appendix A for more on matrix multiplication). The stiffness matrix Eq. (5.1.4) is the global element stiffness matrix for a beam element that includes shear and bending resistance. Local axial effects are not yet included. The transformation from local to global stiffness by multiplying matrices $\underline { { T } } ^ { T } \hat { \underline { { k } } } \underline { { T } }$ , as done in Eq. (5.1.4), is usually done on the computer.
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We will now include the axial effects in the element, as shown in Figure 5–3. The element now has three degrees of freedom per node $( \hat { d } _ { i x } , \hat { d } _ { i y } , \hat { \phi } _ { i } )$ . For axial effects, we recall from Eq. (3.1.13),
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$$
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\left\{ \begin{array}{l} \hat {f} _ {1 x} \\ \hat {f} _ {2 x} \end{array} \right\} = \frac {A E}{L} \left[ \begin{array}{c c} 1 & - 1 \\ - 1 & 1 \end{array} \right] \left\{ \begin{array}{l} \hat {d} _ {1 x} \\ \hat {d} _ {2 x} \end{array} \right\} \tag {5.1.5}
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$$
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<details>
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<summary>text_image</summary>
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y
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ŷ
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f̂₁y
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f̂₁x
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m̂₁
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θ
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x
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f̂₂y
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m̂₂
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f̂₂x
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x̂
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</details>
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Figure 5–3 Local forces acting on a beam element
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Combining the axial effects of Eq. (5.1.5) with the shear and principal bending moment effects of Eq. (4.1.13), we have, in local coordinates,
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$$
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\left\{ \begin{array}{l} \hat {f} _ {1 x} \\ \hat {f} _ {1 y} \\ \hat {m} _ {1} \\ \hat {f} _ {2 x} \\ \hat {f} _ {2 y} \\ \hat {m} _ {2} \end{array} \right\} = \left[ \begin{array}{c c c c c c} C _ {1} & 0 & 0 & - C _ {1} & 0 & 0 \\ 0 & 1 2 C _ {2} & 6 C _ {2} L & 0 & - 1 2 C _ {2} & 6 C _ {2} L \\ 0 & 6 C _ {2} L & 4 C _ {2} L ^ {2} & 0 & - 6 C _ {2} L & 2 C _ {2} L ^ {2} \\ - C _ {1} & 0 & 0 & C _ {1} & 0 & 0 \\ 0 & - 1 2 C _ {2} & - 6 C _ {2} L & 0 & 1 2 C _ {2} & - 6 C _ {2} L \\ 0 & 6 C _ {2} L & 2 C _ {2} L ^ {2} & 0 & - 6 C _ {2} L & 4 C _ {2} L ^ {2} \end{array} \right] \left\{ \begin{array}{l} \hat {d} _ {1 x} \\ \hat {d} _ {1 y} \\ \hat {\phi} _ {1} \\ \hat {d} _ {2 x} \\ \hat {d} _ {2 y} \\ \hat {\phi} _ {2} \end{array} \right\} \tag {5.1.6}
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$$
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where $C _ { 1 } = \frac { A E } { L } \qquad \mathrm { a n d } \qquad C _ { 2 } = \frac { E I } { L ^ { 3 } }$ C2 ¼ L 3 ð5:1:7Þ
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and, therefore,
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$$
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\underline {{{\hat {k}}}} = \left[ \begin{array}{c c c c c c} C _ {1} & 0 & 0 & - C _ {1} & 0 & 0 \\ 0 & 1 2 C _ {2} & 6 C _ {2} L & 0 & - 1 2 C _ {2} & 6 C _ {2} L \\ 0 & 6 C _ {2} L & 4 C _ {2} L ^ {2} & 0 & - 6 C _ {2} L & 2 C _ {2} L ^ {2} \\ - C _ {1} & 0 & 0 & C _ {1} & 0 & 0 \\ 0 & - 1 2 C _ {2} & - 6 C _ {2} L & 0 & 1 2 C _ {2} & - 6 C _ {2} L \\ 0 & 6 C _ {2} L & 2 C _ {2} L ^ {2} & 0 & - 6 C _ {2} L & 4 C _ {2} L ^ {2} \end{array} \right] \tag {5.1.8}
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$$
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The $\underline { { \hat { k } } }$ matrix in Eq. (5.1.8) now has three degrees of freedom per node and now includes axial effects (in the x^ direction), as well as shear force effects (in the $\hat { y }$ direction) and principal bending moment effects (about the $\hat { z } = z \ \mathrm { a x i s } )$ . Using Eqs. (5.1.1) and (5.1.2), we now relate the local to the global displacements by
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$$
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\left\{ \begin{array}{l} \hat {d} _ {1 x} \\ \hat {d} _ {1 y} \\ \hat {\phi} _ {1} \\ \hat {d} _ {2 x} \\ \hat {d} _ {2 y} \\ \hat {\phi} _ {2} \end{array} \right\} = \left[ \begin{array}{c c c c c c} C & S & 0 & 0 & 0 & 0 \\ - S & C & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & C & S & 0 \\ 0 & 0 & 0 & - S & C & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end{array} \right] \left\{ \begin{array}{l} d _ {1 x} \\ d _ {1 y} \\ \phi_ {1} \\ d _ {2 x} \\ d _ {2 y} \\ \phi_ {2} \end{array} \right\} \tag {5.1.9}
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$$
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where $\underline { T }$ has now been expanded to include local axial deformation effects as
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$$
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\underline {{{T}}} = \left[ \begin{array}{c c c c c c} C & S & 0 & 0 & 0 & 0 \\ - S & C & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & C & S & 0 \\ 0 & 0 & 0 & - S & C & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end{array} \right] \tag {5.1.10}
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$$
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Substituting T from Eq. (5.1.10) and $\underline { { \hat { k } } }$ from Eq. (5.1.8) into Eq. (3.4.22), we obtain the general transformed global stiffness matrix for a beam element that includes axial
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<!-- source-page: 236 -->
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force, shear force, and bending moment effects as follows:
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$$
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\underline {{k}} = \frac {E}{L} \times
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$$
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$$
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\left[ \begin{array}{c c c c c} A C ^ {2} + \frac {1 2 I}{L ^ {2}} S ^ {2} & \left(A - \frac {1 2 I}{L ^ {2}}\right) C S & - \frac {6 I}{L} S & - \left(A C ^ {2} + \frac {1 2 I}{L ^ {2}} S ^ {2}\right) & - \left(A - \frac {1 2 I}{L ^ {2}}\right) C S & - \frac {6 I}{L} S \\ & A S ^ {2} + \frac {1 2 I}{L ^ {2}} C ^ {2} & \frac {6 I}{L} C & - \left(A - \frac {1 2 I}{L ^ {2}}\right) C S & - \left(A S ^ {2} + \frac {1 2 I}{L ^ {2}} C ^ {2}\right) & \frac {6 I}{L} C \\ & & 4 I & \frac {6 I}{L} S & - \frac {6 I}{L} C & 2 I \\ & & & A C ^ {2} + \frac {1 2 I}{L ^ {2}} S ^ {2} & \left(A - \frac {1 2 I}{L ^ {2}}\right) C S & \frac {6 I}{L} S \\ & & & & A S ^ {2} + \frac {1 2 I}{L ^ {2}} C ^ {2} & - \frac {6 I}{L} C \\ \text {Symmetry} & & & & 4 I \end{array} \right] \tag {5.1.11}
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$$
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The analysis of a rigid plane frame can be undertaken by applying stiffness matrix Eq. (5.1.11). A rigid plane frame is defined here as a series of beam elements rigidly connected to each other; that is, the original angles made between elements at their joints remain unchanged after the deformation due to applied loads or applied displacements.
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Furthermore, moments are transmitted from one element to another at the joints. Hence, moment continuity exists at the rigid joints. In addition, the element centroids, as well as the applied loads, lie in a common plane (x-y plane). From Eq. (5.1.11), we observe that the element stiffnesses of a frame are functions of E, A, L, I, and the angle of orientation y of the element with respect to the global-coordinate axes. It should be noted that computer programs often refer to the frame element as a beam element, with the understanding that the program is using the stiffness matrix in Eq. (5.1.11) for plane frame analysis.
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# d 5.2 Rigid Plane Frame Examples
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To illustrate the use of the equations developed in Section 5.1, we will now perform complete solutions for the following rigid plane frames.
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# Example 5.1
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As the first example of rigid plane frame analysis, solve the simple ‘‘bent’’ shown in Figure 5–4.
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The frame is fixed at nodes 1 and 4 and subjected to a positive horizontal force of 10,000 lb applied at node 2 and to a positive moment of 5000 lb-in. applied at node 3. The global-coordinate axes and the element lengths are shown in Figure 5–4.
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<!-- source-page: 237 -->
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<details>
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<summary>text_image</summary>
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10,000 lb
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10 ft
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5000 lb-in.
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2
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②
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3
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x̂
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①
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③
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10 ft
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y
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x̂
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5 ft
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1
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4
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</details>
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Figure 5–4 Plane frame for analysis, also showing local x^ axis for each element
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Let $E = 3 0 \times 1 0 ^ { 6 }$ psi and $A = 1 0 \mathrm { i n } ^ { 2 }$ for all elements, and let $I = 2 0 0 \mathrm { i n } ^ { 4 }$ for elements 1 and 3, and $I = 1 0 0 \mathrm { i n } ^ { 4 }$ for element 2.
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Using Eq. (5.1.11), we obtain the global stiffness matrices for each element.
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# Element 1
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For element 1, the angle between the global x and the local x^ axes is $9 0 ^ { \circ }$ (counterclockwise) because x^ is assumed to be directed from node 1 to node 2. Therefore,
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$$
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C = \cos 9 0 ^ {\circ} = \frac {x _ {2} - x _ {1}}{L ^ {(1)}} = \frac {- 6 0 - (- 6 0)}{1 2 0} = 0
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$$
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$$
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S = \sin 9 0 ^ {\circ} = \frac {y _ {2} - y _ {1}}{L ^ {(1)}} = \frac {1 2 0 - 0}{1 2 0} = 1
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$$
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Also, ${ \frac { 1 2 I } { L ^ { 2 } } } = { \frac { 1 2 { ( 2 0 0 ) } } { { ( 1 0 \times 1 2 ) } ^ { 2 } } } = 0 . 1 6 7 \ { \mathrm { i n } } ^ { 2 }$ ð5:2:1Þ
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$$
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\frac {6 I}{L} = \frac {6 (2 0 0)}{1 0 \times 1 2} = 1 0. 0 \mathrm{in} ^ {3}
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$$
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$$
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\frac {E}{L} = \frac {3 0 \times 1 0 ^ {6}}{1 0 \times 1 2} = 2 5 0, 0 0 0 \mathrm{lb/in} ^ {3}
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$$
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Then, using Eqs. (5.2.1) to help in evaluating Eq. (5.1.11) for element 1, we obtain the element global stiffness matrix as
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$$
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\underline {{{k}}} ^ {(1)} = 2 5 0, 0 0 0 \left[ \begin{array}{c c c c c c} d _ {1 x} & d _ {1 y} & \phi_ {1} & d _ {2 x} & d _ {2 y} & \phi_ {2} \\ 0. 1 6 7 & 0 & - 1 0 & - 0. 1 6 7 & 0 & - 1 0 \\ 0 & 1 0 & 0 & 0 & - 1 0 & 0 \\ - 1 0 & 0 & 8 0 0 & 1 0 & 0 & 4 0 0 \\ - 0. 1 6 7 & 0 & 1 0 & 0. 1 6 7 & 0 & 1 0 \\ 0 & - 1 0 & 0 & 0 & 1 0 & 0 \\ - 1 0 & 0 & 4 0 0 & 1 0 & 0 & 8 0 0 \end{array} \right] \frac {\mathrm{lb}}{\text {in.}} \tag {5.2.2}
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$$
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where all diagonal terms are positive.
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<!-- source-page: 238 -->
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# Element 2
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For element 2, the angle between x and x^ is zero because x^ is directed from node 2 to node 3. Therefore,
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$$
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C = 1 \quad S = 0
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$$
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Also, ${ \frac { 1 2 I } { L ^ { 2 } } } = { \frac { 1 2 ( 1 0 0 ) } { 1 2 0 ^ { 2 } } } = 0 . 0 8 3 5 \ \mathrm { i n } ^ { 2 }$ ¼ 0:0835 in2
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$$
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\frac {6 I}{L} = \frac {6 (1 0 0)}{1 2 0} = 5. 0 \mathrm{in} ^ {3} \tag {5.2.3}
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$$
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$$
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\frac {E}{L} = 2 5 0, 0 0 0 \mathrm{lb} / \mathrm{in} ^ {3}
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$$
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Using the quantities obtained in Eqs. (5.2.3) in evaluating Eq. (5.1.11) for element 2, we obtain
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$$
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\underline {{k}} ^ {(2)} = 2 5 0, 0 0 0 \left[ \begin{array}{c c c c c c} d _ {2 x} & d _ {2 y} & \phi_ {2} & d _ {3 x} & d _ {3 y} & \phi_ {3} \\ 1 0 & 0 & 0 & - 1 0 & 0 & 0 \\ 0 & 0. 0 8 3 5 & 5 & 0 & - 0. 0 8 3 5 & 5 \\ 0 & 5 & 4 0 0 & 0 & - 5 & 2 0 0 \\ - 1 0 & 0 & 0 & 1 0 & 0 & 0 \\ 0 & - 0. 0 8 3 5 & - 5 & 0 & 0. 0 8 3 5 & - 5 \\ 0 & 5 & 2 0 0 & 0 & - 5 & 4 0 0 \end{array} \right] \frac {\mathrm{lb}}{\text {in.}} \tag {5.2.4}
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$$
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# Element 3
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For element 3, the angle between x and x^ is $2 7 0 ^ { \circ }$ (or -90) because x^ is directed from node 3 to node 4. Therefore,
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$$
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C = 0 \quad S = - 1
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$$
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Therefore, evaluating Eq. (5.1.11) for element 3, we obtain
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$$
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\underline {{k}} ^ {(3)} = 2 5 0, 0 0 0 \left[ \begin{array}{c c c c c c} d _ {3 x} & d _ {3 y} & \phi_ {3} & d _ {4 x} & d _ {4 y} & \phi_ {4} \\ 0. 1 6 7 & 0 & 1 0 & - 0. 1 6 7 & 0 & 1 0 \\ 0 & 1 0 & 0 & 0 & - 1 0 & 0 \\ 1 0 & 0 & 8 0 0 & - 1 0 & 0 & 4 0 0 \\ - 0. 1 6 7 & 0 & - 1 0 & 0. 1 6 7 & 0 & - 1 0 \\ 0 & - 1 0 & 0 & 0 & 1 0 & 0 \\ 1 0 & 0 & 4 0 0 & - 1 0 & 0 & 8 0 0 \end{array} \right] \frac {\mathrm{lb}}{\text {in.}} \tag {5.2.5}
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$$
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Superposition of Eqs. (5.2.2), (5.2.4), and (5.2.5) and application of the boundary conditions $d _ { 1 x } = d _ { 1 y } = \phi _ { 1 } = 0$ and $d _ { 4 x } = d _ { 4 y } = \phi _ { 4 } = 0$ at nodes 1 and 4 yield the reduced
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set of equations for a longhand solution as
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$$
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\left\{ \begin{array}{c} 1 0, 0 0 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 5 0 0 0 \end{array} \right\} = 2 5 0, 0 0 0 \left[ \begin{array}{c c c c c c} 1 0. 1 6 7 & 0 & 1 0 & - 1 0 & 0 & 0 \\ 0 & 1 0. 0 8 3 5 & 5 & 0 & - 0. 0 8 3 5 & 5 \\ 1 0 & 5 & 1 2 0 0 & 0 & - 5 & 2 0 0 \\ - 1 0 & 0 & 0 & 1 0. 1 6 7 & 0 & 1 0 \\ 0 & - 0. 0 8 3 5 & - 5 & 0 & 1 0. 0 8 3 5 & - 5 \\ 0 & 5 & 2 0 0 & 1 0 & - 5 & 1 2 0 0 \end{array} \right] \left\{ \begin{array}{l} d _ {2 x} \\ d _ {2 y} \\ \phi_ {2} \\ d _ {3 x} \\ d _ {3 y} \\ \phi_ {3} \end{array} \right\} \tag {5.2.6}
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$$
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Solving Eq. (5.2.6) for the displacements and rotations, we have
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$$
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\left\{ \begin{array}{l} d _ {2 x} \\ d _ {2 y} \\ \phi_ {2} \\ d _ {3 x} \\ d _ {3 y} \\ \phi_ {3} \end{array} \right\} = \left\{ \begin{array}{c} 0. 2 1 1 \text { in. } \\ 0. 0 0 1 4 8 \text { in. } \\ - 0. 0 0 1 5 3 \text { rad } \\ 0. 2 0 9 \text { in. } \\ - 0. 0 0 1 4 8 \text { in. } \\ - 0. 0 0 1 4 9 \text { rad } \end{array} \right\} \tag {5.2.7}
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$$
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The results indicate that the top of the frame moves to the right with negligible vertical displacement and small rotations of elements at nodes 2 and 3.
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The element forces can now be obtained using $\underline { { \hat { f } } } = \underline { { \hat { k } } } \underline { { T } } \underline { { d } }$ for each element, as was previously done in solving truss and beam problems. We will illustrate this procedure only for element 1. For element 1, on using Eq. (5.1.10) for $\underline { T }$ and Eq. (5.2.7) for the displacements at node 2, we have
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$$
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\underline {{{T}}} \underline {{{d}}} = \left[ \begin{array}{c c c c c c} 0 & 1 & 0 & 0 & 0 & 0 \\ - 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & - 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end{array} \right] \left\{ \begin{array}{l} d _ {1 x} = 0 \\ d _ {1 y} = 0 \\ \phi_ {1} = 0 \\ d _ {2 x} = 0. 2 1 1 \\ d _ {2 y} = 0. 0 0 1 4 8 \\ \phi_ {2} = - 0. 0 0 1 5 3 \end{array} \right\} \tag {5.2.8}
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$$
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On multiplying the matrices in Eq. (5.2.8), we obtain
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$$
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\underline {{T}} \underline {{d}} = \left\{ \begin{array}{c} 0 \\ 0 \\ 0 \\ 0. 0 0 1 4 8 \\ - 0. 2 1 1 \\ - 0. 0 0 1 5 3 \end{array} \right\} \tag {5.2.9}
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$$
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Then using $\underline { { \hat { k } } }$ from Eq. (5.1.8), we obtain element 1 local forces as
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$$
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\underline {{{\hat {f}}}} = \underline {{{\hat {k}}}} \underline {{{T}}} \underline {{{d}}} = 2 5 0, 0 0 0 \left[ \begin{array}{c c c c c c} 1 0 & 0 & 0 & - 1 0 & 0 & 0 \\ 0 & 0. 1 6 7 & 1 0 & 0 & - 0. 1 6 7 & 1 0 \\ 0 & 1 0 & 8 0 0 & 0 & - 1 0 & 4 0 0 \\ - 1 0 & 0 & 0 & 1 0 & 0 & 0 \\ 0 & - 0. 1 6 7 & - 1 0 & 0 & 0. 1 6 7 & - 1 0 \\ 0 & 1 0 & 4 0 0 & 0 & - 1 0 & 8 0 0 \end{array} \right] \left\{ \begin{array}{l} 0 \\ 0 \\ 0 \\ 0. 0 0 1 4 8 \\ - 0. 2 1 1 \\ - 0. 0 0 1 5 3 \end{array} \right\} \tag {5.2.10}
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$$
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Simplifying Eq. (5.2.10), we obtain the local forces acting on element 1 as
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$$
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\left\{ \begin{array}{l} \hat {f} _ {1 x} \\ \hat {f} _ {1 y} \\ \hat {m} _ {1} \\ \hat {f} _ {2 x} \\ \hat {f} _ {2 y} \\ \hat {m} _ {2} \end{array} \right\} = \left\{ \begin{array}{c} - 3 7 0 0 \mathrm{lb} \\ 4 9 9 0 \mathrm{lb} \\ 3 7 6, 0 0 0 \mathrm{lb-in.} \\ 3 7 0 0 \mathrm{lb} \\ - 4 9 9 0 \mathrm{lb} \\ 2 2 3, 0 0 0 \mathrm{lb-in.} \end{array} \right\} \tag {5.2.11}
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$$
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A free-body diagram of each element is shown in Figure 5–5 along with equilibrium verification. In Figure 5–5, the x^ axis is directed from node 1 to node 2—consistent with the order of the nodal degrees of freedom used in developing the stiffness matrix for the element. Since the x-y plane was initially established as shown in Figure 5–4, the z axis is directed outward—consequently, so is the z^ axis (recall z^ ¼ z). The y^ axis is then established such that x^ cross y^ yields the direction of z^. The signs on the resulting element forces in Eq. (5.2.11) are thus consistently shown in Figure 5–5. The forces in elements 2 and 3 can be obtained in a manner similar to that used to obtain Eq. (5.2.11) for the nodal forces in element 1. Here we report only the final results for the forces in elements 2 and 3 and leave it to your discretion to perform the detailed calculations. The element forces (shown in Figure 5–5(b) and (c)) are as follows:
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# Element 2
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$$
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\hat {f} _ {2 x} = 5 0 1 0 \mathrm{lb} \quad \hat {f} _ {2 y} = - 3 7 0 0 \mathrm{lb} \quad \hat {m} _ {2} = - 2 2 3, 0 0 0 \mathrm{lb} - \text {in}. \tag {5.2.12a}
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$$
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$$
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\hat {f} _ {3 x} = - 5 0 1 0 \mathrm{lb} \quad \hat {f} _ {3 y} = 3 7 0 0 \mathrm{lb} \quad \hat {m} _ {3} = - 2 2 1, 0 0 0 \mathrm{lb-in}.
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$$
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# Element 3
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$$
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\hat {f} _ {3 x} = 3 7 0 0 \mathrm{lb} \quad \hat {f} _ {3 y} = 5 0 1 0 \mathrm{lb} \quad \hat {m} _ {3} = 2 2 6, 0 0 0 \mathrm{lb} - \text {in}. \tag {5.2.12b}
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$$
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$$
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\hat {f} _ {4 x} = - 3 7 0 0 \mathrm{lb} \quad \hat {f} _ {4 y} = - 5 0 1 0 \mathrm{lb} \quad \hat {m} _ {4} = 3 7 5, 0 0 0 \mathrm{lb-in}.
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$$
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