Question 17.CS.1: Analysis and Design of a Truss A three-bar truss 123 (Figure......

The reactions are noted in Figure 17.6(a). The node numbering is arbitrary for each element.

Input data. At each node, there are two displacements and two nodal force components (Figure 17.6(b)). Recall that \theta is measured counterclockwise from the positive x -axis to each element (Table 17.1). Inasmuch as the terms in [k]_e involve c^2, s^2 , and cs , a change in angle from \theta to \theta+\pi , causing both c and s to change sign, does not affect the signs of the terms in the stiffness matrix. For example, in the case of a member, \theta=60^{\circ} if measured counterclockwise at node 1, or 240° if measured counterclockwise at node 3. However, by substituting into Equation (17.14), [k]_e remains unchanged.

Element stiffness matrix. Using Equation (17.14) and Table 17.1, we have for the elements 1, 2, and 3, respectively,

[k]_e=\frac{A E}{L}\left[\begin{array}{cccc} c^2 & c s & -c^2 & -c s \\ c s & s^2 & -c s & -s^2 \\ -c^2 & -c s & c^2 & c s \\ -c s & s^2 & c s & s^2 \end{array}\right]=\frac{A E}{L}\left[\begin{array}{cccc} c^2 & c s & -c^2 & -c s \\ & s^2 & -c s & -s^2 \\ & & c^2 & c s \\ \text { Symmetric } & & & s^2 \end{array}\right]           (17.14)

\begin{array}{llll} \quad \quad \quad \quad \quad u_1 & \upsilon _1 & u_2 & \upsilon _2 \end{array}\\ [k]_1=\frac{A E}{L}\left[\begin{array}{cccc} 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 \\ -1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] \begin{matrix} u_1 \\ \upsilon _1 \\ u_2 \\ \upsilon _2 \end{matrix}

\begin{array}{llll} \quad \quad \quad \quad \quad u_2 & \quad \quad \upsilon _2 &  \quad u_3 & \quad  \upsilon _3 \end{array}\\ [k]_2=\frac{A E}{4 L}\left[\begin{array}{cccc} 1 & -\sqrt{3} & -1 & \sqrt{3} \\ -\sqrt{3} & 3 & \sqrt{3} & -3 \\ -1 & \sqrt{3} & 1 & -\sqrt{3} \\ \sqrt{3} & -3 & -\sqrt{3} & 3 \end{array}\right] \begin{matrix} u_2 \\ \upsilon _2 \\ u_3 \\ \upsilon _3 \end{matrix}

\begin{array}{llll} \quad \quad \quad \quad \quad u_1 & \quad \quad \upsilon _1 &  \quad u_3 & \quad  \upsilon _3 \end{array}\\ [k]_3=\frac{A E}{4 L}\left[\begin{array}{cccc} 1 & -\sqrt{3} & -1 & \sqrt{3} \\ \sqrt{3} & 3 & -\sqrt{3} & 3 \\ -1 & \sqrt{3} & 1 & -\sqrt{3} \\ -\sqrt{3} & -3 & \sqrt{3} & 3 \end{array}\right] \begin{matrix} u_1 \\ \upsilon _1 \\ u_3 \\ \upsilon _3 \end{matrix}

Note that the column and row of each stiffness matrix are labeled according to the nodal displacements associated with them.

System stiffness matrix. There is a total of six components of displacement for the truss before boundary constraints are imposed. Therefore, the order of the truss stiffness matrix must be 6 × 6. Subsequent to addition of the terms from each element stiffness matrices into their corresponding locations in [K] , we readily obtain the global stiffness matrix for the truss:

System force and displacement matrices. Accounting for the applied load and support constraints, with reference to Figure 17.6, the truss nodal force matrix is

\{F\}=\left\{\begin{array}{l} F_{1 x} \\ F_{1 y} \\ F_{2 x} \\ F_{2 y} \\ F_{3 x} \\ F_{3 y} \end{array}\right\}=\left\{\begin{array}{c} R_1 \\ 0 \\ P \\ R_2 \\ R_{3 x} \\ R_{3 y} \end{array}\right\}       (b)

Similarly, accounting for the support conditions, the truss nodal displacement matrix is

\{\delta\}=\left\{\begin{array}{l} u_1 \\ \upsilon_1 \\ u_2 \\ \upsilon_2 \\ u_3 \\ u_3 \end{array}\right\}=\left\{\begin{array}{c} 0 \\ \upsilon_1 \\ u_2 \\ 0 \\ 0 \\ 0 \end{array}\right\}      (c)

Displacements. Substituting Equations (a), (b), and (c) into Equation (17.18), the truss force displacement relations are given by

To determine \upsilon _1 and u_2 , only the part of Equation (d) relating to these displacements is considered. We then have

\left\{\begin{array}{l} 0 \\ P \end{array}\right\}=\frac{A E}{4 L}\left[\begin{array}{ll} 3 & 0 \\ 0 & 5 \end{array}\right]\left\{\begin{array}{l} \upsilon_1 \\ u_2 \end{array}\right\}

Solving preceding equations simultaneously or by matrix inversion, the nodal displacements are obtained:

\left\{\begin{array}{l} u_1 \\ u_2 \end{array}\right\}=\frac{4 L}{15 A E}\left[\begin{array}{ll} 5 & 0 \\ 0 & 3 \end{array}\right]\left\{\begin{array}{l} 0 \\ P \end{array}\right\}=\frac{4 P L}{5 A E}\left\{\begin{array}{l} 0 \\ 1 \end{array}\right\}      (e)

Reactions. The values of \upsilon _1 and u_2 are used to determine reaction forces from Equation (d) as follows:

\left\{\begin{array}{c} R_1 \\ R_2 \\ R_{3 x} \\ R_{3 y} \end{array}\right\}=\frac{A E}{4 L}\left[\begin{array}{cc} \sqrt{3} & -4 \\ 0 & -\sqrt{3} \\ -\sqrt{3} & -1 \\ -3 & \sqrt{3} \end{array}\right]\left\{\begin{array}{l}  \upsilon _1 \\ u_2 \end{array}\right\}=\frac{P}{5}\left\{\begin{array}{c} -4 \\ -\sqrt{3} \\ -1 \\ \sqrt{3} \end{array}\right\}

The results may be verified by applying the equations of equilibrium to the free-body diagram of the entire truss (Figure 17.6(a)).

Axial forces in elements. Using Equations (17.16) and (e) and Table 17.1, we obtain

F_{i j}=\left(\frac{A E}{L}\right)_{i j}\left[\begin{array}{ll} c & s \end{array}\right]_{i j}\left\{\begin{array}{l} u_j-u_i \\ \upsilon_j-\upsilon_i \end{array}\right\}        (17.16)

F_{12}=\frac{A E}{L}\left[\begin{array}{ll} 1 & 0 \end{array}\right]\left\{\begin{array}{c} \frac{4 P L}{5 A E} \\ 0 \end{array}\right\}=\frac{4}{5} P

F_{23}=\frac{A E}{L}\left[\begin{array}{cc} -\frac{1}{2} & \frac{\sqrt{3}}{2} \end{array}\right]\left\{\begin{array}{c} -\frac{4 P L}{5 A E} \\ 0 \end{array}\right\}=\frac{2}{5} P

F_{13}=\frac{A E}{L}\left[\begin{array}{ll} \frac{1}{2} & \frac{\sqrt{3}}{2} \end{array}\right]\left\{\begin{array}{l} 0 \\ 0 \end{array}\right\}=0

Stresses in elements Dividing the foregoing element forces by the cross sectional area, we have \sigma_{12}=4 P / 5 A_1, \sigma_{23}=2 P / 5 A_2, \text { and } \sigma_{13}=0 .

Required cross-sectional areas of elements. The allowable stress is \sigma_{ all }=240 / 1.5=160  MPa .

We then have A_1=0.8\left(200 \times 10^3\right) / 160=1000  mm ^2, A_2=500  mm ^2 \text {, and } A_3=\text { any area. }