Question 17.2: Displacements in a Frame A planar rectangular frame 1234 is ......
Displacements in a Frame
A planar rectangular frame 1234 is fixed at both supports l and 4 (Figure 17.12). The load on the frame consists of a horizontal force P acting at joint 2 and a moment M applied at joint 3. By the FEA, calculate the nodal displacements.
Given: P =4 kips, M =2 kips · in., L=5 ft, E=30 \times 10^6 psi, and A =5 in.² for all elements; I =120 in.^4 for elements 1 and 3 and I =60 in. ^4 for element 2.
The global coordinate axes xy are indicated in Figure 17.12. Through the use of Equation (17.24) and Table 17.2, the element stiffness matrices are obtained as
\hspace{50 pt} \begin{array}{llllll} & \hspace{11 pt}u_1 & \hspace{6 pt} \upsilon_1 & \hspace{10 pt} \theta_1 & \hspace{8 pt} u_2 & \hspace{6 pt} \upsilon_2 & \hspace{9 pt} \theta_2 \end{array} \\ [k]_1=5\left(10^5\right)\left[\begin{array}{cccccc} 0.4 & 0 & -12 & -0.4 & 0 & -12 \\ 0 & 5 & 0 & 0 & -5 & 0 \\ -12 & 0 & 480 & 12 & 0 & 240 \\ -0.4 & 0 & 12 & 0.4 & 0 & 12 \\ 0 & -5 & 0 & 0 & 5 & 0 \\ -12 & 0 & 240 & 12 & 0 & 480 \end{array}\right] \left\{\begin{array}{l} u_1 \\ \upsilon_1 \\ \theta_1 \\ u_2 \\ \upsilon_2 \\ \theta_2 \end{array}\right\}
\hspace{50 pt} \begin{array}{llllll} & \hspace{10 pt}u_2 & \hspace{5 pt} \upsilon_2 & \hspace{9 pt} \theta_2 & \hspace{6 pt} u_3 & \hspace{5 pt} \upsilon_3 & \hspace{8 pt} \theta_3 \end{array} \\ [k]_2=5\left(10^5\right)\left[\begin{array}{cccccc} 5 & 0 & 0 & -5 & 0 & 0 \\ 0 & 0.2 & 6 & 0 & -0.2 & 6 \\ 0 & 6 & 240 & 0 & -6 & 120 \\ -5 & 0 & 0 & 5 & 0 & 0 \\ 0 & -0.2 & -6 & 0 & 0.2 & -6 \\ 0 & 6 & 120 & 0 & -6 & 240 \end{array}\right] \left\{\begin{array}{l} u_2 \\ \upsilon_2 \\ \theta_2 \\ u_3 \\ \upsilon_3 \\ \theta_3 \end{array}\right\}
\hspace{73 pt} \begin{array}{llllll} u_3 & \hspace{6 pt} \upsilon_3 & \hspace{9 pt} \theta_3 & \hspace{9 pt} u_4 & \hspace{6 pt} \upsilon_4 & \hspace{8 pt} \theta_4 \end{array} \\ [k]_3=5\left(10^5\right)\left[\begin{array}{cccccc} 0.4 & 0 & -12 & -0.4 & 0 & -12 \\ 0 & 5 & 0 & 0 & -5 & 0 \\ -12 & 0 & 480 & 12 & 0 & 240 \\ -0.4 & 0 & 12 & 0.4 & 0 & 12 \\ 0 & -5 & 0 & 0 & 5 & 0 \\ -12 & 0 & 240 & 12 & 0 & 480 \end{array}\right]\left\{\begin{array}{l} u_3 \\ \upsilon _3 \\ \theta_3 \\ u_4 \\ \upsilon _4 \\ \theta_4 \end{array}\right\}
We superpose the element stiffness matrices and apply the boundary conditions:
u_1=\upsilon_1=\theta_1=0, \quad u_4=\upsilon_4=\theta_4=0
at nodes 1 and 4. This leads to the following reduced set of equations:
\left\{\begin{array}{c} 4000 \\ 0 \\ 0 \\ 0 \\ 0 \\ 2000 \end{array}\right\}= 5\left(10^5\right) \\ \hspace{28 pt} \begin{array}{llllll} u_2 & \hspace{26 pt} \upsilon _2 & \hspace{5 pt} \theta_2 & \hspace{5 pt} u_3 & \hspace{3 pt} \upsilon _3 & \hspace{6 pt} \theta_3 \end{array} \\\left[\begin{array}{cccccc} 5.4 & 0 & 12 & -5 & 0 & 0 \\ & 5.2 & 6 & 0 & -0.2 & 6 \\ & & 720 & 0 & -6 & 120 \\ & & & 5.4 & 0 & 12 \\ & & & & 52 & -6 \\ \text { Symmetric } & & & & & 720 \end{array}\right]\left\{\begin{array}{l} u_2 \\ \upsilon _2 \\ \theta_2 \\ u_3 \\ \upsilon _3 \\ \theta_3 \end{array}\right\}
Solving, the nodal deflections and rotations are
\left\{\begin{array}{l} u_2 \\ \upsilon _2 \\ \theta_2 \\ u_3 \\ \upsilon _3 \\ \theta_3 \end{array}\right\}=\left\{\begin{array}{c} 18.208 \\ 0.582 \\ -0.271 \\ 17.408 \\ -0.582 \\ -0.248 \end{array}\right\}\left(10^{-3}\right) \begin{matrix} \text{in.} \\ \text{in.} \\ \text{rad} \\ \text{in.} \\ \text{in.} \\ \text{rad} \end{matrix}
The negative sign indicates a downward displacement or clockwise rotation.
| TABLE 17.2 Data for the Frame of Figure 17.12 |
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| Element | 1 | 2 | 3 |
| \theta | 90^{\circ} | 0^{\circ} | 270^{\circ} |
| c | 0 | 1 | 0 |
| s | 1 | 0 | -1 |
| 12 I / L^2 | 0.4 | 0.2 | 0 4 |
| 6 I / L | 12 | 6 | 12 |
| E / L | 5 \times 10^5 | 5 \times 10^5 | 5 \times 10^5 |