김경종
21 KiB
The global stiffness matrices for each element are obtained as follows:
Element 1
For element 1, we have the local x^ axis coincident with the global x axis. Therefore, we obtain
C = \frac {x _ {2} - x _ {1}}{L ^ {(1)}} = \frac {3}{3} = 1 \quad S = \frac {z _ {2} - z _ {1}}{L ^ {(1)}} = \frac {3 - 3}{3} = 0
Other expressions needed to evaluate the stiffness matrix are
\begin{array}{l} \frac {1 2 E I}{L ^ {3}} = \frac {1 2 (2 1 0 \times 1 0 ^ {6} \mathrm{kN} / \mathrm{m} ^ {2}) (1 6 . 6 \times 1 0 ^ {- 5} \mathrm{m} ^ {4})}{(3 \mathrm{m}) ^ {3}} = 1. 5 5 \times 1 0 ^ {4} \\ \frac {6 E I}{L ^ {2}} = \frac {6 (2 1 0 \times 1 0 ^ {6}) (1 6 . 6 \times 1 0 ^ {- 5})}{(3) ^ {2}} = 2. 3 2 \times 1 0 ^ {4} \tag {5.4.41} \\ \end{array}
\frac {G J}{L} = \frac {(8 4 \times 1 0 ^ {6}) (4 . 6 \times 1 0 ^ {- 5})}{3} = 1. 2 8 \times 1 0 ^ {3}
\frac {4 E I}{L} = \frac {4 (2 1 0 \times 1 0 ^ {6}) (1 6 . 6 \times 1 0 ^ {- 5})}{3} = 4. 6 5 \times 1 0 ^ {4}
Considering the boundary condition Eqs. (5.4.40), using the results of Eqs. (5.4.41) in Eq. (5.4.17) for \underline { { \hat { k } } } _ { G } ^ { \mathrm { ~ ~ } } and Eq. (5.4.18) for \underline { { T } } _ { G } , and then applying Eq. (5.4.19), we obtain the reduced part of the global stiffness matrix associated only with the degrees of freedom at node 2 as
\underline {{k}} ^ {(1)} = \left[ \begin{array}{c c c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right] \left[ \begin{array}{c c c} 1. 5 5 & 0 & - 2. 3 2 \\ 0 & 0. 1 2 8 & 0 \\ - 2. 3 2 & 0 & 4. 6 5 \end{array} \right] (1 0 ^ {4}) \left[ \begin{array}{c c c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right]
Since the local axes associated with element 1 are parallel to the global axes, we observe that \underline { { T } } _ { G } is merely the identity matrix; therefore, \underline { { k } } _ { G } = \underline { { \hat { k } } } _ { G } . Performing the matrix multiplications, we obtain
\underline {{k}} ^ {(1)} = \left[ \begin{array}{c c c} 1. 5 5 & 0 & - 2. 3 2 \\ 0 & 0. 1 2 8 & 0 \\ - 2. 3 2 & 0 & 4. 6 5 \end{array} \right] (1 0 ^ {4}) \frac {\mathrm{kN}}{\mathrm{m}} \tag {5.4.42}
Element 2
For element 2, we assume the local x^ axis to be directed from node 2 to node 3 for the formulation of \underline { { k } } . . Therefore,
C = \frac {x _ {3} - x _ {2}}{L ^ {(2)}} = \frac {0 - 0}{3} = 0 \quad S = \frac {z _ {3} - z _ {2}}{L ^ {(2)}} = \frac {0 - 3}{3} = - 1 \tag {5.4.43}
Other expressions used in Eq. (5.4.17) are identical to those obtained in Eqs. (5.4.41) for element 1. Evaluating Eq. (5.4.19) for the global stiffness matrix, we obtain
\underline {{k}} ^ {(2)} = \left[ \begin{array}{c c c} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & - 1 & 0 \end{array} \right] \left[ \begin{array}{c c c} 1. 5 5 & 0 & 2. 3 2 \\ 0 & 0. 1 2 8 & 0 \\ 2. 3 2 & 0 & 4. 6 5 \end{array} \right] (1 0 ^ {4}) \left[ \begin{array}{c c c} 1 & 0 & 0 \\ 0 & 0 & - 1 \\ 0 & 1 & 0 \end{array} \right]
where the reduced part of \underline { { k } } is now associated with node 2 for element 2. Again performing the matrix multiplications, we have
\underline {{k}} ^ {(2)} = \left[ \begin{array}{l l l} 1. 5 5 & 2. 3 2 & 0 \\ 2. 3 2 & 4. 6 5 & 0 \\ 0 & 0 & 0. 1 2 8 \end{array} \right] (1 0 ^ {4}) \frac {\mathrm{kN}}{\mathrm{m}} \tag {5.4.44}
Superimposing the global stiffness matrices from Eqs. (5.4.42) and (5.4.44), we obtain the total global stiffness matrix (with boundary conditions applied) as
\underline {{{K}}} _ {G} = \left[ \begin{array}{c c c} 3. 1 0 & 2. 3 2 & - 2. 3 2 \\ 2. 3 2 & 4. 7 8 & 0 \\ - 2. 3 2 & 0 & 4. 7 8 \end{array} \right] (1 0 ^ {4}) \frac {\mathrm{kN}}{\mathrm{m}} \tag {5.4.45}
The grid matrix equation becomes
\left\{ \begin{array}{l} F _ {2 y} = - 2 2 \\ M _ {2 x} = 0 \\ M _ {2 z} = 0 \end{array} \right\} = \left[ \begin{array}{c c c} 3. 1 0 & 2. 3 2 & - 2. 3 2 \\ 2. 3 2 & 4. 7 8 & 0 \\ - 2. 3 2 & 0 & 4. 7 8 \end{array} \right] \left\{ \begin{array}{l} d _ {2 y} \\ \phi_ {2 x} \\ \phi_ {2 z} \end{array} \right\} (1 0 ^ {4}) \tag {5.4.46}
Solving for the displacement and the rotations in Eq. (5.4.46), we obtain
d _ {2 y} = - 0. 2 5 9 \times 1 0 ^ {- 2} \mathrm{m}
\phi_ {2 x} = 0. 1 2 6 \times 1 0 ^ {- 2} \mathrm{rad} \tag {5.4.47}
\phi_ {2 z} = - 0. 1 2 6 \times 1 0 ^ {- 2} \mathrm{rad}
We determine the local element forces by applying the local equation \underline { { \hat { f } } } = \underline { { \hat { k } } } _ { G } \underline { { T } } _ { G } \underline { { d } } for each element as follows:
Element 1
Using Eq. (5.4.17) for \underline { { \hat { k } } } _ { G } . , Eq. (5.4.18) for \underline { { T } } _ { G } , and Eqs. (5.4.47), we obtain
\underline {{T}} _ {G} \underline {{d}} = \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 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end{array} \right] \left\{ \begin{array}{c} 0 \\ 0 \\ 0 \\ - 0. 2 5 9 \times 1 0 ^ {- 2} \\ 0. 1 2 6 \times 1 0 ^ {- 2} \\ - 0. 1 2 6 \times 1 0 ^ {- 2} \end{array} \right\}
Multiplying the matrices, we have
\underline {{T}} _ {G} \underline {{d}} = \left\{ \begin{array}{c} 0 \\ 0 \\ 0 \\ - 0. 2 5 9 \times 1 0 ^ {- 2} \\ 0. 1 2 6 \times 1 0 ^ {- 2} \\ - 0. 1 2 6 \times 1 0 ^ {- 2} \end{array} \right\} \tag {5.4.48}
Using Eqs. (5.4.17), (5.4.41), and (5.4.48), we obtain the local element forces as
\left\{ \begin{array}{l} \hat {f} _ {1 y} \\ \hat {m} _ {1 x} \\ \hat {m} _ {1 z} \\ \hat {f} _ {2 y} \\ \hat {m} _ {2 x} \\ \hat {m} _ {2 z} \end{array} \right\} = (1 0 ^ {4}) \left[ \begin{array}{c c c c c c} 1. 5 5 & 0 & 2. 3 2 & - 1. 5 5 & 0 & 2. 3 2 \\ & 0. 1 2 8 & 0 & 0 & - 0. 1 2 8 & 0 \\ & & 4. 6 5 & - 2. 3 2 & 0 & 2. 3 3 \\ & & & 1. 5 5 & 0 & - 2. 3 2 \\ & & & & 0. 1 2 8 & 0 \\ \text {Symmetry} & & & & & 4. 6 5 \end{array} \right] \left\{ \begin{array}{c} 0 \\ 0 \\ 0 \\ - 0. 2 5 9 \times 1 0 ^ {- 2} \\ 0. 1 2 6 \times 1 0 ^ {- 2} \\ - 0. 1 2 6 \times 1 0 ^ {- 2} \end{array} \right\} \tag {5.4.49}
Multiplying the matrices in Eq. (5.4.49), we obtain
\hat {f} _ {1 y} = 1 1. 0 \mathrm{kN} \quad \hat {m} _ {1 x} = - 1. 5 0 \mathrm{kN} \cdot \mathrm{m} \quad \hat {m} _ {1 z} = 3 1. 0 \mathrm{kN} \cdot \mathrm{m} \tag {5.4.50}
\hat {f} _ {2 y} = - 1 1. 0 \mathrm{kN} \quad \hat {m} _ {2 x} = 1. 5 0 \mathrm{kN} \cdot \mathrm{m} \quad \hat {m} _ {2 z} = 1. 5 0 \mathrm{kN} \cdot \mathrm{m}
Element 2
We can obtain the local element forces for element 2 in a similar manner. Because the procedure is the same as that used to obtain the element 1 local forces, we will not show the details but will only list the final results:
\hat {f} _ {2 y} = - 1 1. 0 \mathrm{kN} \quad \hat {m} _ {2 x} = 1. 5 0 \mathrm{kN} \cdot \mathrm{m} \quad \hat {m} _ {2 z} = - 1. 5 0 \mathrm{kN} \cdot \mathrm{m} \tag {5.4.51}
\hat {f} _ {3 y} = 1 1. 0 \mathrm{kN} \quad \hat {m} _ {3 x} = - 1. 5 0 \mathrm{kN} \cdot \mathrm{m} \quad \hat {m} _ {3 z} = - 3 1. 0 \mathrm{kN} \cdot \mathrm{m}
Free-body diagrams showing the local element forces are shown in Figure 5–22.
5.5 Beam Element Arbitrarily Oriented in Space
In this section, we develop the stiffness matrix for the beam element arbitrarily oriented in space, or three dimensions. This element can then be used to analyze frames in three-dimensional space.
First we consider bending about two axes, as shown in Figure 5–23.
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31.0 kN · m 1.50 kN · m 1.50 kN · m 31.0 kN · m 11 kN 11 kN 11 kN 3 1.50 kN · m 3 11 kN 11 kN 11 kN 11 kN 11 kN 11 kN 11 kN 11 kN 11 kN 11 kN 11 kN 11 kN 11 kN 11 kN 11 kN 11 kN 11 kN 11 kN 11 kN 11 kN 11 k
Figure 5–22 Free-body diagram of each element of Figure 5–21
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m̂₁y, φ̂₁y m̂₁z, φ̂₁z 1 L m̂₂y, φ̂₂y m̂₂z, φ̂₂z 2 f̂₂z, d̂₂z f̂₁y, d̂₁y x̂
Figure 5–23 Bending about two axes \hat { y } and \hat { \boldsymbol { z } }
We establish the following sign convention for the axes. Now we choose positive \hat { x } from node 1 to 2. Then \hat { y } is the principal axis for which the moment of inertia is minimum, I _ { y } . By the right-hand rule we establish { \hat { z } } , and the maximum moment of inertia is I _ { z } .
Bending in xˆ-zˆ Plane
First consider bending in the \hat { x } { = } \hat { z } plane due to \hat { m } _ { y } . Then clockwise rotation \hat { \phi } _ { \nu } is in the same sense as before for single bending. The stiffness matrix due to bending in the x^-^z plane is then
\hat {\underline {{k}}} _ {y} = \frac {E I _ {y}}{L ^ {4}} \left[ \begin{array}{c c c c} 1 2 L & - 6 L ^ {2} & - 1 2 L & - 6 L ^ {2} \\ & 4 L ^ {3} & 6 L ^ {2} & 2 L ^ {3} \\ & & 1 2 L & 6 L ^ {2} \\ \text { Symmetry } & & & 4 L ^ {3} \end{array} \right] \tag {5.5.1}
where I _ { y } is the moment of inertia of the cross section about the principal axis \hat { y } , the weak axis; that is, I _ { y } < I _ { z } .
Bending in the \hat { z } { \boldsymbol { - } } \hat { y } Plane
Now we consider bending in the \hat { x } { - } \hat { y } plane due to \hat { m } _ { z } . Now positive rotation \hat { \phi } _ { z } is counterclockwise instead of clockwise. Therefore, some signs change in the stiffness
matrix for bending in the x^-y^ plane. The resulting stiffness matrix is
\hat {\underline {{k}}} _ {z} = \frac {E I _ {z}}{L ^ {4}} \left[ \begin{array}{c c c c} 1 2 L & 6 L ^ {2} & - 1 2 L & 6 L ^ {2} \\ & 4 L ^ {3} & - 6 L ^ {2} & 2 L ^ {3} \\ & & 1 2 L & - 6 L ^ {2} \\ \text { Symmetry } & & & 4 L ^ {3} \end{array} \right] \tag {5.5.2}
Direct superposition of Eqs. (5.5.1) and (5.5.2) with the axial stiffness matrix Eq. (3.1.14) and the torsional stiffness matrix Eq. (5.4.14) yields the element stiffness matrix for the beam or frame element in three-dimensional space as
\hat {k} = \left[ \begin{array}{c c c c c c c c c c c c} \hat {d} _ {1 x} & \hat {d} _ {1 y} & \hat {d} _ {1 z} & \hat {\phi} _ {1 x} & \hat {\phi} _ {1 y} & \hat {\phi} _ {1 z} & \hat {d} _ {2 x} & \hat {d} _ {2 y} & \hat {d} _ {2 z} & \hat {\phi} _ {2 x} & \hat {\phi} _ {2 y} & \hat {\phi} _ {2 z} \\ \frac {A E}{L} & 0 & 0 & 0 & 0 & 0 & - \frac {A E}{L} & 0 & 0 & 0 & 0 & 0 \\ 0 & \frac {1 2 E I _ {z}}{L ^ {3}} & 0 & 0 & 0 & \frac {6 E I _ {z}}{L ^ {2}} & 0 & - \frac {1 2 E I _ {z}}{L ^ {3}} & 0 & 0 & 0 & \frac {6 E I _ {z}}{L ^ {2}} \\ 0 & 0 & \frac {1 2 E I _ {y}}{L ^ {3}} & 0 & - \frac {6 E I _ {y}}{L ^ {2}} & 0 & 0 & 0 & - \frac {1 2 E I _ {y}}{L ^ {3}} & 0 & - \frac {6 E I _ {y}}{L ^ {2}} & 0 \\ 0 & 0 & 0 & \frac {G J}{L} & 0 & 0 & 0 & 0 & 0 & - \frac {G J}{L} & 0 & 0 \\ 0 & 0 & - \frac {6 E I _ {y}}{L ^ {2}} & 0 & \frac {4 E I _ {y}}{L} & 0 & 0 & 0 & \frac {6 E I _ {y}}{L ^ {2}} & 0 & \frac {2 E I _ {y}}{L} & 0 \\ 0 & \frac {6 E I _ {z}}{L ^ {2}} & 0 & 0 & 0 & \frac {4 E I _ {z}}{L} & 0 & - \frac {6 E I _ {z}}{L ^ {2}} & 0 & 0 & 0 & \frac {2 E I _ {z}}{L} \\ \hline - \frac {A E}{L} & 0 & 0 & 0 & 0 & 0 & \frac {A E}{L} & 0 & 0 & 0 & 0 & 0 \\ 0 & - \frac {1 2 E I _ {z}}{L ^ {3}} & 0 & 0 & 0 & - \frac {6 E I _ {z}}{L ^ {2}} & 0 & \frac {1 2 E I _ {z}}{L ^ {3}} & 0 & 0 & 0 & - \frac {6 E I _ {z}}{L ^ {2}} \\ 0 & 0 & - \frac {1 2 E I _ {y}}{L ^ {3}} & 0 & \frac {6 E I _ {y}}{L ^ {2}} & 0 & 0 & 0 & \frac {1 2 E I _ {y}}{L ^ {3}} & 0 & \frac {6 E I _ {y}}{L ^ {2}} & 0 \\ 0 & 0 & 0 & - \frac {G J}{L} & 0 & 0 & 0 & 0 & 0 & \frac {G J}{L} & 0 & 0 \\ 0 & 0 & - \frac {6 E I _ {y}}{L ^ {2}} & 0 & \frac {2 E I _ {y}}{L} & 0 & 0 & 0 & \frac {6 E I _ {y}}{L ^ {2}} & 0 & \frac {4 E I _ {y}}{L} & 0 \\ 0 & \frac {6 E I _ {z}}{L ^ {2}} & 0 & 0 & 0 & \frac {2 E I _ {z}}{L} & 0 & - \frac {6 E I _ {z}}{L ^ {2}} & 0 & 0 & 0 & \frac {4 E I _ {z}}{L} \end{array} \right] \tag {5.5.3}
The transformation from local to global axis system is accomplished as follows:
\underline {{k}} = \underline {{T}} ^ {T} \hat {\underline {{k}}} \underline {{T}} \tag {5.5.4}
where \underline { { \hat { k } } } is given by Eq. (5.5.3) and T is given by
\underline {{T}} = \left[ \begin{array}{c c c c} \underline {{\lambda}} _ {3 \times 3} & & & \\ & \underline {{\lambda}} _ {3 \times 3} & & \\ & & \underline {{\lambda}} _ {3 \times 3} & \\ & & & \underline {{\lambda}} _ {3 \times 3} \end{array} \right] \tag {5.5.5}
text_image
y θŷ θx̂ θx̂ x θx̂ θx̂ z ŷ x̂ θx̂
Figure 5–24 Direction cosines associated with the x axis
text_image
y x̂ ŷ x z × x̂ = ŷ z
Figure 5–25 Illustration showing how local \hat{y} axis is determined
where
\underline {{{\lambda}}} = \left[ \begin{array}{c c c} C _ {x \hat {x}} & C _ {y \hat {x}} & C _ {z \hat {x}} \\ C _ {x \hat {y}} & C _ {y \hat {y}} & C _ {z \hat {y}} \\ C _ {x \hat {z}} & C _ {y \hat {z}} & C _ {z \hat {z}} \end{array} \right] \tag {5.5.6}
Here C_{y\hat{x}} and C_{x\hat{y}} are not necessarily equal. The direction cosines are shown in part in Figure 5–24.
Remember that direction cosines of the \hat{x} axis member are
\hat {x} = \cos \theta_ {x \hat {x}} \bar {i} + \cos \theta_ {y \hat {x}} \bar {j} + \cos \theta_ {z \hat {x}} \bar {k} \tag {5.5.7}
where
\cos \theta_ {x \hat {x}} = \frac {x _ {2} - x _ {1}}{L} = l
\cos \theta_ {y \hat {x}} = \frac {y _ {2} - y _ {1}}{L} = m \tag {5.5.8}
\cos \theta_ {z \hat {x}} = \frac {z _ {2} - z _ {1}}{L} = n
The \hat{y} axis is selected to be perpendicular to the \hat{x} and z axes in such a way that the cross product of global z with \hat{x} results in the \hat{y} axis, as shown in Figure 5–25. Therefore,
z \times \hat {x} = \hat {y} = \frac {1}{D} \left| \begin{array}{c c c} \bar {i} & \bar {j} & \overline {{k}} \\ 0 & 0 & 1 \\ l & m & n \end{array} \right| \tag {5.5.9}
\hat {y} = - \frac {m}{D} \bar {i} + \frac {l}{D} \bar {j} \tag {5.5.10}
and D = (l^{2} + m^{2})^{1/2}
The \hat{z} axis will be determined by the orthogonality condition \hat{z} = \hat{x} \times \hat{y} as follows:
\hat {z} = \hat {x} \times \hat {y} = \frac {1}{D} \left| \begin{array}{c c c} \bar {i} & \bar {j} & \overline {{k}} \\ l & m & n \\ - m & l & 0 \end{array} \right| \tag {5.5.11}
or
\hat {z} = - \frac {l n}{D} \bar {i} - \frac {m n}{D} \bar {j} + D \bar {k} \tag {5.5.12}
Combining Eqs. (5.5.7), (5.5.10), and (5.5.12), the 3 \times 3 transformation matrix becomes
\underline {{{\lambda}}} _ {3 \times 3} = \left[ \begin{array}{c c c} l & m & n \\ - \frac {m}{D} & \frac {l}{D} & 0 \\ - \frac {l n}{D} & - \frac {m n}{D} & D \end{array} \right] \tag {5.5.13}
This vector \underline { { \lambda } } rotates a vector from the local coordinate system into the global one. This is the \underline { { \lambda } } used in the \underline { T } matrix. In summary, we have
\begin{array}{l} \cos \theta_ {x \hat {y}} = - \frac {m}{D} \\ \cos \theta_ {y \hat {y}} = \frac {l}{D} \\ \cos \theta_ {z \hat {y}} = 0 \\ \cos \theta_ {x \hat {z}} = - \frac {\ln}{D} \tag {5.5.14} \\ \cos \theta_ {y \hat {z}} = - \frac {m n}{D} \\ \cos \theta_ {z \hat {z}} = D \\ \end{array}
Two exceptions arise when local and global axes have special orientations with respect to each other. If the local x^ axis coincides with the global z axis, then the member is parallel to the global z axis and the \hat { y } axis becomes uncertain, as shown in Figure 5–26(a). In this case the local y^ axis is selected as the global y axis. Then, for
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y ŷ ẑ x z
(a) x in same direction as z
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y x z ŷ x̂ ẑ
(b)x in opposite direction of z
Figure 5–26 Special cases of transformation matrices
the positive x^ axis in the same direction as the global z , \underline { { \lambda } } becomes
\underline {{\lambda}} = \left[ \begin{array}{c c c} 0 & 0 & 1 \\ 0 & 1 & 0 \\ - 1 & 0 & 0 \end{array} \right] \tag {5.5.15}
For the positive x^ axis opposite the global z [Figure 5–26(b)], l becomes
\underline {{{\lambda}}} = \left[ \begin{array}{c c c} 0 & 0 & - 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{array} \right] \tag {5.5.16}
Example 5.7
Determine the direction cosines and the rotation matrix of the local \hat { x } , \hat { y } , \hat { z } axes in reference to the global x , y , z axes for the beam element oriented in space with end nodal coordinates of 1 ( 0 , 0 , 0 ) and 2 \left( 3 , 4 , 1 2 \right) , as shown in Figure 5–27.
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y 1 (0, 0, 0) x ŷ ẑ x̂ 2 (3, 4, 12) 4 12 3 z
Figure 5–27 Beam element oriented in space
First we determine the length of the element as
L = \sqrt {3 ^ {2} + 4 ^ {2} + 1 2 ^ {2}} = 1 3
Now using Eq. (5.5.8), we obtain the direction cosines of the \hat{x} axis as follows:
l _ {x} = \frac {x _ {2} - x _ {1}}{L} = \frac {3 - 0}{1 3} = \frac {3}{1 3}
m _ {x} = \frac {y _ {2} - y _ {1}}{L} = \frac {4 - 0}{1 3} = \frac {4}{1 3} \tag {5.5.17}
n _ {x} = \frac {z _ {2} - z _ {1}}{L} = \frac {1 2 - 0}{1 3} = \frac {1 2}{1 3}
By Eq. (5.5.10) or (5.5.14), we obtain the direction cosines of the \hat{y} axis as follows:
D = (l ^ {2} + m ^ {2}) ^ {1 / 2} = \left[ \left(\frac {3}{1 3}\right) ^ {2} + \left(\frac {4}{1 3}\right) ^ {2} \right] ^ {1 / 2} = \frac {5}{1 3} \tag {5.5.18}
Define the direction cosines of the \hat{y} axis as l_{y} , m_{y} , and n_{y} , where
l _ {y} = - \frac {m}{D} = - \frac {4}{5}
m _ {y} = \frac {l}{D} = \frac {3}{5} \tag {5.5.19}
n _ {y} = 0
For the \hat{z} axis, define the direction cosines as l_{z}, m_{z}, n_{z} and again use Eq. (5.5.12) or (5.5.14) as follows:
l _ {z} = - \frac {\ln}{D} = \frac {\left(- \frac {3}{1 3}\right) \left(\frac {1 2}{1 3}\right)}{\frac {5}{1 3}} = - \frac {3 6}{6 5}
m _ {z} = - \frac {m n}{D} = \frac {\left(- \frac {4}{1 3}\right) \left(\frac {1 2}{1 3}\right)}{\frac {5}{1 3}} = - \frac {4 8}{6 5} \tag {5.5.20}
n _ {z} = D = \frac {5}{1 3}
Now check that l^{2} + m^{2} + n^{2} = 1 .
\text { For } \hat {x}: \frac {3 ^ {2} + 4 ^ {2} + 1 2 ^ {2}}{1 3 ^ {2}} = 1
\text { For } \hat {y}: \frac {(- 4) ^ {2} + 3 ^ {2}}{5 ^ {2}} = 1 \tag {5.5.21}
\text { For } \hat {z}: \left(- \frac {3 6}{6 5}\right) ^ {2} + \left(- \frac {4 8}{6 5}\right) ^ {2} + \left(\frac {2 5}{6 5}\right) ^ {2} = 1
By Eq. (5.5.13), the rotation matrix is
\underline {{{\lambda}}} _ {3 \times 3} = \left[ \begin{array}{c c c} \frac {3}{1 3} & \frac {4}{1 3} & \frac {1 2}{1 3} \\ - \frac {4}{5} & \frac {3}{5} & 0 \\ - \frac {3 6}{6 5} & - \frac {4 8}{6 5} & \frac {5}{1 3} \end{array} \right] \tag {5.5.22}
Based on the resulting direction cosines from Eqs. (5.5.17), (5.5.19), and (5.5.20), the local axes are also shown in Figure 5–27.
Example 5.8
Determine the displacements and rotations at the free node (node 1) and the element local forces and moments for the space frame shown in Figure 5–28. Also verify equilibrium at node 1. Let E ¼ 30;000 ksi, G = 1 0 , 0 0 0 \ \mathrm { k s i } , J = 5 0 \ \mathrm { i n } . { } ^ { 4 } , I _ { y } = 1 0 0 \ \mathrm { i n } . { } ^ { 4 } ; I _ { z } = 1 0 0 \mathrm { i n . } ^ { 4 } , \mathrm { A } = 1 0 \mathrm { i n . } ^ { 2 } , and L ¼ 100 in. for all three beam elements.
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y x Mx = -1000 k-in. 2 x̂ L = 100 in. ① Fy = -50 k L = 100 in. ② I ③ L = 100 in. x̂ 4 Joint 1 Plan
Figure 5–28 Space frame for analysis
Use Eq. (5.5.4) to obtain the global stiffness matrix for each element. This requires us to first use Eq. (5.5.3) to obtain each local stiffness matrix, Eq. (5.5.5) to obtain the transformation matrix for each element, and Eqs. (5.5.6) and (5.5.14) to obtain the direction cosine matrix for each element.
Element 1
We establish the local x^ axis to go from node 2 to node 1 as shown in Figure 5–28. Therefore, using Eq. (5.5.8), we obtain the direction cosines of the x^ axis as follows:
l = 1, \quad m = 0, \quad n = 0 \tag {5.5.23}
Also, D = ( l ^ { 2 } + m ^ { 2 } ) ^ { 1 / 2 } = 1 .
Using Eqs. (5.5.10) and (5.5.14), we obtain the direction cosines of the ^y axis as follows:
l _ {y} = - \frac {m}{D} = 0 \quad m _ {y} = \frac {l}{D} = 1 \quad n _ {y} = 0 \tag {5.5.24}