김경종
22 KiB
After multiplying the matrices in Eq. (3.5.7), we have
\underline {{C}} ^ {\prime} = \frac {E}{L} [ - C \quad - S \quad C \quad S ] \tag {3.5.8}
Example 3.4
For the bar shown in Figure 3–13, determine the axial stress. Let A = 4 \times 1 0 ^ { - 4 } ~ \mathrm { m } ^ { 2 } ; , E = 2 1 0 \mathrm { \ G P a } , and L = 2 ~ \mathrm { m } , and let the angle between x and x^ be 60
. Assume the global displacements have been previously determined to be d _ { 1 x } = 0 . 2 5 mm, d _ { 1 y } = 0 . 0 , d _ { 2 x } = 0 . 5 0 mm, and d _ { 2 y } = 0 . 7 5 mm.
text_image
y x 1 2 x̂ 60°
Figure 3–13 Bar element for stress evaluation
We can use Eq. (3.5.6) to evaluate the axial stress. Therefore, we first calculate \underline { { C ^ { \prime } } } from Eq. (3.5.8) as
\underline {{C}} ^ {\prime} = \frac {2 1 0 \times 1 0 ^ {6} \mathrm{kN} / \mathrm{m} ^ {2}}{2 \mathrm{m}} \left[ \frac {- 1}{2} \quad \frac {- \sqrt {3}}{2} \quad \frac {1}{2} \quad \frac {\sqrt {3}}{2} \right] \tag {3.5.9}
where we have used C = \cos { 6 0 ^ { \circ } } = { \frac { 1 } { 2 } } and S = \sin 6 0 ^ { \circ } = { \sqrt { 3 } } / 2 in Eq. (3.5.9). Now \underline { d } is given by
\underline {{d}} = \left\{ \begin{array}{l} d _ {1 x} \\ d _ {1 y} \\ d _ {2 x} \\ d _ {2 y} \end{array} \right\} = \left\{ \begin{array}{l} 0. 2 5 \times 1 0 ^ {- 3} \mathrm{m} \\ 0. 0 \\ 0. 5 0 \times 1 0 ^ {- 3} \mathrm{m} \\ 0. 7 5 \times 1 0 ^ {- 3} \mathrm{m} \end{array} \right\} \tag {3.5.10}
Using Eqs. (3.5.9) and (3.5.10) in Eq. (3.5.6), we obtain the bar axial stress as
\begin{array}{l} \sigma_ {x} = \frac {2 1 0 \times 1 0 ^ {6}}{2} \left[ \begin{array}{c c c c} - 1 & - \sqrt {3} & \frac {1}{2} & \frac {\sqrt {3}}{2} \end{array} \right] \left\{ \begin{array}{l} 0. 2 5 \\ 0. 0 \\ 0. 5 0 \\ 0. 7 5 \end{array} \right\} \times 1 0 ^ {- 3} \\ = 8 1. 3 2 \times 1 0 ^ {3} \mathrm{kN} / \mathrm{m} ^ {2} = 8 1. 3 2 \mathrm{MPa} \\ \end{array}
d 3.6 Solution of a Plane Truss
We will now illustrate the use of equations developed in Sections 3.4 and 3.5, along with the direct stiffness method of assembling the total stiffness matrix and equations, to solve the following plane truss example problems. A plane truss is a structure composed of bar elements that all lie in a common plane and are connected by frictionless pins. The plane truss also must have loads acting only in the common plane and all loads must be applied at the nodes or joints.
Example 3.5
For the plane truss composed of the three elements shown in Figure 3–14 subjected to a downward force of 10,000 lb applied at node 1, determine the x and y displacements at node 1 and the stresses in each element. Let E = 3 0 \times 1 0 ^ { 6 } psi and A = 2 \mathrm { i n } . ^ { 2 } for all elements. The lengths of the elements are shown in the figure.
text_image
10 ft 45° 45° 10,000 lb 10 ft 3 4 y x
Figure 3–14 Plane truss
First, we determine the global stiffness matrices for each element by using Eq. (3.4.23). This requires determination of the angle y between the global x axis and the local x^ axis for each element. In this example, the direction of the x^ axis for each element is taken in the direction from node 1 to the other node. The node numbering is arbitrary for each element. However, once the direction is chosen, the angle y is then established as positive when measured counterclockwise from positive x to x^. For element 1, the local x^ axis is directed from node 1 to node 2; therefore, \theta ^ { ( 1 ) } = 9 0 ^ { \circ } . For element 2, the local x^ axis is directed from node 1 to node 3 and \theta ^ { ( 2 ) } = 4 5 ^ { \circ } . For element 3, the local x^ axis is directed from node 1 to node 4 and \theta ^ { ( 3 ) } = 0 ^ { \circ } . It is convenient to construct Table 3–1 to aid in determining each element stiffness matrix.
There are a total of eight nodal components of displacement, or degrees of freedom, for the truss before boundary constraints are imposed. Thus the order of the
Table 3–1 Data for the truss of Figure 3–14
| Element | $\theta^{\circ}$ | $C$ | $S$ | $C^{2}$ | $S^{2}$ | $CS$ |
| 1 | $90^{\circ}$ | 0 | 1 | 0 | 1 | 0 |
| 2 | $45^{\circ}$ | $\sqrt{2}/2$ | $\sqrt{2}/2$ | $\frac{1}{2}$ | $\frac{1}{2}$ | $\frac{1}{2}$ |
| 3 | $0^{\circ}$ | 1 | 0 | 1 | 0 | 0 |
total stiffness matrix must be 8 \times 8 . . We could then expand the \underline { { k } } matrix for each element to the order 8 \times 8 by adding rows and columns of zeros as explained in the first part of Section 2.4. Alternatively, we could label the rows and columns of each element stiffness matrix according to the displacement components associated with it as explained in the latter part of Section 2.4. Using this latter approach, we construct the total stiffness matrix \underline { { K } } simply by adding terms from the individual element stiffness matrices into their corresponding locations in \underline { { K } } . . This approach will be used here and throughout this text.
For element 1, using Eq. (3.4.23), along with Table 3–1 for the direction cosines, we obtain
\underline {{k}} ^ {(1)} = \frac {(3 0 \times 1 0 ^ {6}) (2)}{1 2 0} \left[ \begin{array}{c c c c} d _ {1 x} & d _ {1 y} & d _ {2 x} & d _ {2 y} \\ 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & - 1 \\ 0 & 0 & 0 & 0 \\ 0 & - 1 & 0 & 1 \end{array} \right] \tag {3.6.1}
Similarly, for element 2, we have
\underline {{k}} ^ {(2)} = \frac {(3 0 \times 1 0 ^ {6}) (2)}{1 2 0 \times \sqrt {2}} \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 & - 0. 5 & 0. 5 & 0. 5 \\ - 0. 5 & - 0. 5 & 0. 5 & 0. 5 \end{array} \right] \tag {3.6.2}
and for element 3, we have
\underline {{{{k}}}} ^ {(3)} = \frac {(3 0 \times 1 0 ^ {6}) (2)}{1 2 0} \left[ \begin{array}{c c c c} d _ {1 x} & d _ {1 y} & d _ {4 x} & d _ {4 y} \\ 1 & 0 & - 1 & 0 \\ 0 & 0 & 0 & 0 \\ - 1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right] \tag {3.6.3}
The common factor of 3 0 \times 1 0 ^ { 6 } \times 2 / 1 2 0 ( = 5 0 0 , 0 0 0 ) can be taken from each of \operatorname { E q s } . (3.6.1)–(3.6.3). After adding terms from the individual element stiffness matrices into
their corresponding locations in \underline { { K } } , we obtain the total stiffness matrix as
\underline {{{K}}} = (5 0 0, 0 0 0) \left[ \begin{array}{c c c c c c c c} d _ {1 x} & d _ {1 y} & d _ {2 x} & d _ {2 y} & d _ {3 x} & d _ {3 y} & d _ {4 x} & d _ {4 y} \\ 1. 3 5 4 & 0. 3 5 4 & 0 & 0 & - 0. 3 5 4 & - 0. 3 5 4 & - 1 & 0 \\ 0. 3 5 4 & 1. 3 5 4 & 0 & - 1 & - 0. 3 5 4 & - 0. 3 5 4 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & - 1 & 0 & 1 & 0 & 0 & 0 & 0 \\ - 0. 3 5 4 & - 0. 3 5 4 & 0 & 0 & 0. 3 5 4 & 0. 3 5 4 & 0 & 0 \\ - 0. 3 5 4 & - 0. 3 5 4 & 0 & 0 & 0. 3 5 4 & 0. 3 5 4 & 0 & 0 \\ - 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{array} \right] \tag {3.6.4}
The global K matrix, Eq. (3.6.4), relates the global forces to the global displacements. We thus write the total structure stiffness equations, accounting for the applied force at node 1 and the boundary constraints at nodes 2–4 as follows:
\left\{ \begin{array}{c} 0 \\ - 1 0, 0 0 0 \\ F _ {2 x} \\ F _ {2 y} \\ F _ {3 x} \\ F _ {3 y} \\ F _ {4 x} \\ F _ {4 y} \end{array} \right\} = (5 0 0, 0 0 0) \left[ \begin{array}{c c c c c c c c} 1. 3 5 4 & 0. 3 5 4 & 0 & 0 & - 0. 3 5 4 & - 0. 3 5 4 & - 1 & 0 \\ 0. 3 5 4 & 1. 3 5 4 & 0 & - 1 & - 0. 3 5 4 & - 0. 3 5 4 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & - 1 & 0 & 1 & 0 & 0 & 0 & 0 \\ - 0. 3 5 4 & - 0. 3 5 4 & 0 & 0 & 0. 3 5 4 & 0. 3 5 4 & 0 & 0 \\ - 0. 3 5 4 & - 0. 3 5 4 & 0 & 0 & 0. 3 5 4 & 0. 3 5 4 & 0 & 0 \\ - 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{array} \right]
\times \left\{ \begin{array}{c} d _ {1 x} \\ d _ {1 y} \\ d _ {2 x} = 0 \\ d _ {2 y} = 0 \\ d _ {3 x} = 0 \\ d _ {3 y} = 0 \\ d _ {4 x} = 0 \\ d _ {4 y} = 0 \end{array} \right\} \tag {3.6.5}
We could now use the partitioning scheme described in the first part of Section 2.5 to obtain the equations used to determine unknown displacements d _ { 1 x } and d _ { 1 y } —that is, partition the first two equations from the third through the eighth in Eq. (3.6.5). Alternatively, we could eliminate rows and columns in the total stiffness matrix corresponding to zero displacements as previously described in the latter part of Section 2.5. Here we will use the latter approach; that is, we eliminate rows and column 3–8 in Eq. (3.6.5) because those rows and columns correspond to zero displacements.
(Remember, this direct approach must be modified for nonhomogeneous boundary conditions as was indicated in Section 2.5.) We then obtain
\left\{ \begin{array}{c} 0 \\ - 1 0, 0 0 0 \end{array} \right\} = (5 0 0, 0 0 0) \left[ \begin{array}{l l} 1. 3 5 4 & 0. 3 5 4 \\ 0. 3 5 4 & 1. 3 5 4 \end{array} \right] \left\{ \begin{array}{l} d _ {1 x} \\ d _ {1 y} \end{array} \right\} \tag {3.6.6}
Equation (3.6.6) can now be solved for the displacements by multiplying both sides of the matrix equation by the inverse of the 2 2 stiffness matrix or by solving the two equations simultaneously. Using either procedure for solution yields the displacements
d _ {1 x} = 0. 4 1 4 \times 1 0 ^ {- 2} \text { in. } \quad d _ {1 y} = - 1. 5 9 \times 1 0 ^ {- 2} \text { in. }
The minus sign in the d _ { 1 y } result indicates that the displacement component in the y direction at node 1 is in the direction opposite that of the positive y direction based on the assumed global coordinates, that is, a downward displacement occurs at node 1.
Using Eq. (3.5.6) and Table 3–1, we determine the stresses in each element as follows:
\sigma^ {(1)} = \frac {3 0 \times 1 0 ^ {6}}{1 2 0} \left[ \begin{array}{l l l l} 0 & - 1 & 0 & 1 \end{array} \right] \left\{ \begin{array}{l} d _ {1 x} = 0. 4 1 4 \times 1 0 ^ {- 2} \\ d _ {1 y} = - 1. 5 9 \times 1 0 ^ {- 2} \\ d _ {2 x} = 0 \\ d _ {2 y} = 0 \end{array} \right\} = 3 9 6 5 \mathrm{psi}
\begin{array}{l} \sigma^ {(2)} = \frac {3 0 \times 1 0 ^ {6}}{1 2 0 \sqrt {2}} \left[ \begin{array}{c c c c} - \sqrt {2} & - \sqrt {2} & \sqrt {2} & \sqrt {2} \\ \hline 2 & 2 & 2 & 2 \end{array} \right] \left\{ \begin{array}{l} d _ {1 x} = 0. 4 1 4 \times 1 0 ^ {- 2} \\ d _ {1 y} = - 1. 5 9 \times 1 0 ^ {- 2} \\ d _ {3 x} = 0 \\ d _ {3 y} = 0 \end{array} \right\} \\ = 1 4 7 1 \mathrm{psi} \\ \end{array}
\sigma^ {(3)} = \frac {3 0 \times 1 0 ^ {6}}{1 2 0} \left[ - 1 \quad 0 \quad 1 \quad 0 \right] \left\{ \begin{array}{l} d _ {1 x} = 0. 4 1 4 \times 1 0 ^ {- 2} \\ d _ {1 y} = - 1. 5 9 \times 1 0 ^ {- 2} \\ d _ {4 x} = 0 \\ d _ {4 y} = 0 \end{array} \right\} = - 1 0 3 5 \text { psi }
We now verify our results by examining force equilibrium at node 1; that is, summing forces in the global x and y directions, we obtain
\sum F _ {x} = 0 \quad (1 4 7 1 \mathrm{psi}) (2 \mathrm{in} ^ {2}) \frac {\sqrt {2}}{2} - (1 0 3 5 \mathrm{psi}) (2 \mathrm{in} ^ {2}) = 0
\sum F _ {y} = 0 \quad (3 9 6 5 \mathrm{psi}) (2 \mathrm{in} ^ {2}) + (1 4 7 1 \mathrm{psi}) (2 \mathrm{in} ^ {2}) \frac {\sqrt {2}}{2} - 1 0, 0 0 0 = 0
Example 3.6
For the two-bar truss shown in Figure 3–15, determine the displacement in the y direction of node 1 and the axial force in each element. A force of P = 1 0 0 0 kN is applied at node 1 in the positive y direction while node 1 settles an amount \delta = 5 0 mm in the negative x direction. Let E ¼ 210 GPa and A = 6 . 0 0 \times 1 0 ^ { - 4 } \mathrm { m } ^ { 2 } for each element. The lengths of the elements are shown in the figure.
text_image
2 3 m ① θ(1) x 1 P = 1000 kN δ = 50 mm 3 ② y 4 m
Figure 3–15 Two-bar truss
We begin by using Eq. (3.4.23) to determine each element stiffness matrix.
Element 1
\cos \theta^ {(1)} = \frac {3}{5} = 0. 6 0 \quad \sin \theta^ {(1)} = \frac {4}{5} = 0. 8 0
\underline {{{k}}} ^ {(1)} = \frac {(6 . 0 \times 1 0 ^ {- 4} \mathrm{m} ^ {2}) (2 1 0 \times 1 0 ^ {6} \mathrm{kN} / \mathrm{m} ^ {2})}{5 \mathrm{m}} \left[ \begin{array}{c c c c} 0. 3 6 & 0. 4 8 & - 0. 3 6 & - 0. 4 8 \\ & 0. 6 4 & - 0. 4 8 & - 0. 6 4 \\ & & 0. 3 6 & 0. 4 8 \\ \text { Symmetry } & & & 0. 6 4 \end{array} \right] \tag {3.6.7}
Simplifying Eq. (3.6.7), we obtain
\underline {{k}} ^ {(1)} = (2 5, 2 0 0) \left[ \begin{array}{c c c c} d _ {1 x} & d _ {1 y} & d _ {2 x} & d _ {2 y} \\ 0. 3 6 & 0. 4 8 & - 0. 3 6 & - 0. 4 8 \\ & 0. 6 4 & - 0. 4 8 & - 0. 6 4 \\ & & 0. 3 6 & 0. 4 8 \\ \text { Symmetry } & & & 0. 6 4 \end{array} \right] \tag {3.6.8}
Element 2
\cos \theta^ {(2)} = 0. 0 \quad \sin \theta^ {(2)} = 1. 0
\underline {{k}} ^ {(2)} = \frac {(6 . 0 \times 1 0 ^ {- 4}) (2 1 0 \times 1 0 ^ {6})}{4} \left[ \begin{array}{c c c c} 0 & 0 & 0 & 0 \\ & 1 & 0 & - 1 \\ & & 0 & 0 \\ \text { Symmetry } & & & 1 \end{array} \right] \tag {3.6.9}
\underline {{k}} ^ {(2)} = (2 5, 2 0 0) \left[ \begin{array}{c c c c} d _ {1 x} & d _ {1 y} & d _ {3 x} & d _ {3 y} \\ 0 & 0 & 0 & 0 \\ & 1. 2 5 & 0 & - 1. 2 5 \\ & & 0 & 0 \\ \text { Symmetry } & & & 1. 2 5 \end{array} \right] \tag {3.6.10}
where, for computational simplicity, Eq. (3.6.10) is written with the same factor (25,200) in front of the matrix as Eq. (3.6.8). Superimposing the element stiffness matrices, Eqs. (3.6.8) and (3.6.10), we obtain the global K matrix and relate the global forces to global displacements by
\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\} = (2 5, 2 0 0) \left[ \begin{array}{c c c c c c} 0. 3 6 & 0. 4 8 & - 0. 3 6 & - 0. 4 8 & 0 & 0 \\ & 1. 8 9 & - 0. 4 8 & - 0. 6 4 & 0 & - 1. 2 5 \\ & & 0. 3 6 & 0. 4 8 & 0 & 0 \\ & & & 0. 6 4 & 0 & 0 \\ & & & & 0 & 0 \\ \text { Symmetry } & & & & & 1. 2 5 \end{array} \right] \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\} \tag {3.6.11}
We can again partition equations with known displacements and then simultaneously solve those associated with unknown displacements. To do this partitioning, we consider the boundary conditions given by
d _ {1 x} = \delta \quad d _ {2 x} = 0 \quad d _ {2 y} = 0 \quad d _ {3 x} = 0 \quad d _ {3 y} = 0 \tag {3.6.12}
Therefore, using Eqs. (3.6.12), we partition equation 2 from equations 1, 3, 4, 5, and 6 of Eq. (3.6.11) and are left with
P = 2 5, 2 0 0 (0. 4 8 \delta + 1. 8 9 d _ {1 y}) \tag {3.6.13}
where F _ { 1 y } = P and d _ { 1 x } = \delta were substituted into Eq. (3.6.13). Expressing Eq. (3.6.13) in terms of P and d allows these two influences on d _ { \mathrm { l } y } to be clearly separated. Solving Eq. (3.6.13) for d _ { \mathrm { l } y } , we have
d _ {1 y} = 0. 0 0 0 0 2 1 P - 0. 2 5 4 \delta \tag {3.6.14}
Now, substituting the numerical values P ¼ 1000 kN and \delta = - 0 . 0 5 m into Eq. (3.6.14), we obtain
d _ {1 y} = 0. 0 3 3 7 \mathrm{m} \tag {3.6.15}
where the positive value indicates horizontal displacement to the left.
The local element forces are obtained by using Eq. (3.4.11). We then have the following.
Element 1
\left\{ \begin{array}{l} \hat {f} _ {1 x} \\ \hat {f} _ {2 x} \end{array} \right\} = (2 5, 2 0 0) \left[ \begin{array}{c c} 1 & - 1 \\ - 1 & 1 \end{array} \right] \left[ \begin{array}{c c c c} 0. 6 0 & 0. 8 0 & 0 & 0 \\ 0 & 0 & 0. 6 0 & 0. 8 0 \end{array} \right] \left\{ \begin{array}{l} d _ {1 x} = - 0. 0 5 \\ d _ {1 y} = 0. 0 3 3 7 \\ d _ {2 x} = 0 \\ d _ {2 y} = 0 \end{array} \right\} \tag {3.6.16}
Performing the matrix triple product in Eq. (3.6.16) yields
\hat {f} _ {1 x} = - 7 6. 6 \mathrm{kN} \quad \hat {f} _ {2 x} = 7 6. 6 \mathrm{kN} \tag {3.6.17}
Element 2
\left\{ \begin{array}{l} \hat {f} _ {1 x} \\ \hat {f} _ {3 x} \end{array} \right\} = (3 1, 5 0 0) \left[ \begin{array}{c c} 1 & - 1 \\ - 1 & 1 \end{array} \right] \left[ \begin{array}{c c c c} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right] \left\{ \begin{array}{l} d _ {1 x} = - 0. 0 5 \\ d _ {1 y} = 0. 0 3 3 7 \\ d _ {3 x} = 0 \\ d _ {3 y} = 0 \end{array} \right\} \tag {3.6.18}
Performing the matrix triple product in Eq. (3.6.18), we obtain
\hat {f} _ {1 x} = 1 0 6 1 \mathrm{kN} \quad \hat {f} _ {3 x} = - 1 0 6 1 \mathrm{kN} \tag {3.6.19}
Verification of the computations by checking that equilibrium is satisfied at node 1 is left to your discretion.
Example 3.7
To illustrate how we can combine spring and bar elements in one structure, we now solve the two-bar truss supported by a spring shown in Figure 3–16. Both bars have E = 2 1 0 ~ \mathrm { G P a } and A = 5 . 0 \times 1 0 ^ { - 4 } \dot { \mathrm { m } } ^ { 2 } . Bar one has a length of 5 m and bar two a length of 10 m. The spring stiffness is k = 2 0 0 0 \mathrm { k N } / \mathrm { m } .
text_image
25 kN 5 m ① 45° 10 m ② 3 ③ k = 2000 kN/m 4
Figure 3–16 Two-bar truss with spring support
We begin by using Eq. (3.4.23) to determine each element stiffness matrix.
Element 1
\theta^ {(1)} = 1 3 5 ^ {\circ}, \quad \cos \theta^ {(1)} = - \sqrt {2} / 2, \quad \sin \theta^ {(1)} = \sqrt {2} / 2
\underline {{k}} ^ {(1)} = \frac {(5 . 0 \times 1 0 ^ {- 4} \mathrm{m} ^ {2}) (2 1 0 \times 1 0 ^ {6} \mathrm{kN} / \mathrm{m} ^ {2})}{5 \mathrm{m}} \left[ \begin{array}{r r r r} 0. 5 & - 0. 5 & - 0. 5 & 0. 5 \\ - 0. 5 & 0. 5 & 0. 5 & - 0. 5 \\ - 0. 5 & 0. 5 & 0. 5 & - 0. 5 \\ 0. 5 & - 0. 5 & - 0. 5 & 0. 5 \end{array} \right] \tag {3.6.20}
Simplifying Eq. (3.6.20), we obtain
\underline {{k}} ^ {(1)} = 1 0 5 \times 1 0 ^ {5} \left[ \begin{array}{c c c c} 1 & - 1 & - 1 & 1 \\ - 1 & 1 & 1 & - 1 \\ - 1 & 1 & 1 & - 1 \\ 1 & - 1 & - 1 & 1 \end{array} \right] \tag {3.6.21}
Element 2
\theta^ {(2)} = 1 8 0 ^ {\circ}, \quad \cos \theta^ {(2)} = - 1. 0, \quad \sin \theta^ {(2)} = 0
\underline {{k}} ^ {(2)} = \frac {(5 \times 1 0 ^ {- 4} \mathrm{m} ^ {2}) (2 1 0 \times 1 0 ^ {6} \mathrm{kN} / \mathrm{m} ^ {2})}{1 0 \mathrm{m}} \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] \tag {3.6.22}
Simplifying Eq. (3.6.22), we obtain
\underline {{k}} ^ {(2)} = 1 0 5 \times 1 0 ^ {5} \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] \tag {3.6.23}
Element 3
\theta^ {(3)} = 2 7 0 ^ {\circ}, \quad \cos \theta^ {(3)} = 0, \quad \sin \theta^ {(3)} = 1. 0
Using Eq. (3.4.23) but replacing AE/L with the spring constant k , we obtain the stiffness matrix of the spring as
\underline {{k}} ^ {(3)} = 2 0 \times 1 0 ^ {5} \left[ \begin{array}{c c c c} 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & - 1 \\ 0 & 0 & 0 & 0 \\ 0 & - 1 & 0 & 1 \end{array} \right] \tag {3.6.24}
Applying the boundary conditions, we have
d _ {2 x} = d _ {2 y} = d _ {3 x} = d _ {3 y} = d _ {4 x} = d _ {4 y} = 0 \tag {3.6.25}
Using the boundary conditions in Eq. (3.6.25), the reduced assembled global equations are given by:
\left\{ \begin{array}{l} F _ {1 x} = 0 \\ F _ {1 y} = - 2 5 \mathrm{kN} \end{array} \right\} = 1 0 ^ {5} \left[ \begin{array}{c c} 2 1 0 & - 1 0 5 \\ - 1 0 5 & 1 2 5 \end{array} \right] \left\{ \begin{array}{l} d _ {1 x} \\ d _ {1 y} \end{array} \right\} \tag {3.6.26}
Solving Eq. (3.6.26) for the global displacements, we obtain
d _ {1 x} = - 1. 7 2 4 \times 1 0 ^ {- 3} \mathrm{m} \quad d _ {1 y} = - 3. 4 4 8 \times 1 0 ^ {- 3} \mathrm{m} \tag {3.6.27}
We can obtain the stresses in the bar elements by using Eq. (3.5.6) as
\sigma^ {(1)} = \frac {2 1 0 \times 1 0 ^ {3} \mathrm{MN} / \mathrm{m} ^ {2}}{5 \mathrm{m}} [ 0. 7 0 7 \quad - 0. 7 0 7 \quad - 0. 7 0 7 \quad 0. 7 0 7 ] \left\{ \begin{array}{c} - 1. 7 2 4 \times 1 0 ^ {- 3} \\ - 3. 4 4 8 \times 1 0 ^ {- 3} \\ 0 \\ 0 \end{array} \right\}
Simplifying, we obtain
\sigma^ {(1)} = 5 1. 2 \mathrm{MPa} (T)
Similarly, we obtain the stress in element two as
\sigma^ {(2)} = \frac {2 1 0 \times 1 0 ^ {3} \mathrm{MN} / \mathrm{m} ^ {2}}{1 0 \mathrm{m}} [ 1. 0 \quad 0 \quad - 1. 0 \quad 0 ] \left\{ \begin{array}{c} - 1. 7 2 4 \times 1 0 ^ {- 3} \\ - 3. 4 4 8 \times 1 0 ^ {- 3} \\ 0 \\ 0 \end{array} \right\}
Simplifying, we obtain
\sigma^ {(2)} = - 3 6. 2 \mathrm{MPa(C)}
3.7 Transformation Matrix and Stiffness Matrix for a Bar in Three-Dimensional Space
We will now derive the transformation matrix necessary to obtain the general stiffness matrix of a bar element arbitrarily oriented in three-dimensional space as shown in Figure 3–17. Let the coordinates of node 1 be taken as x _ { 1 } , y _ { 1 } , and z _ { 1 } , and let those of node 2 be taken as x _ { 2 } , y _ { 2 } ; and z2. Also, let \theta _ { x } , \theta _ { y } , and \theta _ { z } be the angles measured from the global x , y , , and z axes, respectively, to the local \hat { x } axis. Here x^ is directed along the element from node 1 to node 2. We must now determine \underline { { T } } ^ { * } such that \hat { \underline { d } } = \bar { \underline { T } } ^ { * } \underline { d } . We begin the derivation of \underline { { T } } ^ { * } by considering the vector \hat { \mathbf { d } } = \mathbf { d } expressed in three dimensions as
\hat {d} _ {x} \hat {\mathbf {i}} + \hat {d} _ {y} \hat {\mathbf {j}} + \hat {d} _ {z} \hat {\mathbf {k}} = d _ {x} \mathbf {i} + d _ {y} \mathbf {j} + d _ {z} \mathbf {k} \tag {3.7.1}
where \hat { \bf i } , \hat { \bf j } , and k^ are unit vectors associated with the local { \hat { x } } , { \hat { y } } , , and z^ axes, respectively, and i, j, and k are unit vectors associated with the global x , y , , and z axes. Taking the