Question 17.1: Displacements and Forces in a Statically Indeterminate Beam ......
Displacements and Forces in a Statically Indeterminate Beam
A propped cantilevered beam of flexural rigidity EI is subjected to end load P as shown in Figure 17.9(a). Using the FEM, find:
a. The nodal displacements.
b. The nodal forces and moments.
We discretize the beam into elements with nodes 1, 2, and 3, as shown in Figure 17.9(a). By Equation (17.19),
\left\{\begin{array}{l} \bar{F}_{1 y} \\ \bar{M}_1 \\ \bar{F}_{2 y} \\ \bar{M}_2 \end{array}\right\}_e=\frac{E I}{L^3} \left[\begin{array}{cccc} 12 & 6 L & -12 & 6 L \\ & 4 L^2 & -6 L & 2 L^2 \\ & & 12 & -6 L \\ \text { Symmetric } & & & 4 L^2 \end{array}\right] \left\{\begin{array}{c} \overline{ \upsilon }_1 \\ \bar{\theta}_1 \\ \overline{ \upsilon }_2 \\ \bar{\theta}_2 \end{array}\right\} _e (17.19a)
\{\bar{F}\}_e=[\bar{k}]_e\{\delta\}_e (17.19b)
\begin{array}{llll} & && & & & &\upsilon _1 &&& \theta_1 && \upsilon _2 && \theta_2 \end{array} \\ [k]_1=\frac{E I}{L^3} \left[\begin{array}{cccc} 12 & 6 L & -12 & 6 L \\ & 4 L^2 & -6 L & 2 L^2 \\ & & 12 & -6 L \\ \text { Symmetric } & & & 4 L^2 \end{array}\right] \begin{aligned} & \upsilon _1 \\ & \theta_1 \\ & \upsilon _2 \\ & \theta_2 \end{aligned}
\begin{array}{llll} & && & & & &\upsilon _2 &&& \theta_2 && \upsilon _3 && \theta_3 \end{array} \\ [k]_2=\frac{E I}{L^3} \left[\begin{array}{cccc} 12 & 6 L & -12 & 6 L \\ & 4 L^2 & -6 L & 2 L^2 \\ & & 12 & -6 L \\ \text { Symmetric } & & & 4 L^2 \end{array}\right] \begin{aligned} & \upsilon _2 \\ & \theta_2 \\ & \upsilon _3 \\ & \theta_3 \end{aligned}
a. The global stiffness matrix of the beam can now be assembled: [K]=[k]_1+[k]_2 . The governing equations for the beam are then
\left\{\begin{array}{l} F_{1 y} \\ M_1 \\ F_{2 y} \\ M_2 \\ F_{3 y} \\ M_3 \end{array}\right\}=\frac{E I}{L^3} \left[\begin{array}{cccccc} 12 & 6 L & -12 & 6 L & 0 & 0 \\ & 4 L^2 & -6 L & 2 L^2 & 0 & 0 \\ & & 24 & 0 & -12 & 6 L \\ & & & 8 L^2 & -6 L & 2 L^2 \\ & & & & 12 & -6 L \\ \text { Symmetric } & & & & & 4 L^2 \end{array}\right] \left\{\begin{array}{l} \upsilon _1 \\ \theta_1 \\ \upsilon _2 \\ \theta_2 \\ \upsilon _3 \\ \theta_3 \end{array}\right\} (17.20a)
or
\{F\}=[K]\{\delta\} (17.20b)
The boundary conditions are \upsilon_2=0, \theta_3=0, \text { and } \upsilon_3=0 . Partitioning the first, second, and fourth of these equations associated with the unknown displacements,
\left\{\begin{array}{c} -P \\ 0 \\ 0 \end{array}\right\}=\frac{E I}{L^3}\left[\begin{array}{ccc} 12 & 6 L & 6 L \\ 6 L & 4 L^2 & 2 L^2 \\ 6 L & 2 L^2 & 8 L^2 \end{array}\right]\left\{\begin{array}{l} \upsilon_1 \\ \theta_1 \\ \theta_2 \end{array}\right\}
Solving for nodal displacements, we obtain
\upsilon_1=-\frac{7 P L^3}{12 E I}, \quad \theta_1=\frac{3 P L^2}{4 E I}, \quad \theta_2=\frac{P L^2}{4 E I}
b. Introducing these equations into Equation (17.20a), after multiplying, the nodal forces and moments are found to be
F_{1 y}=-P \quad M_1=0, \quad F_{2 y}=\frac{5}{2} P,
M_2=0 \quad F_{3 y}=-\frac{3}{2} P, \quad M_3=\frac{1}{2} P L
Note that M_1 and M_2 are 0, since no reactive moments are present on the beam at nodes 1 and 2.
Comments: In general, it is necessary to determine the local nodal forces and moments associated with each element to analyze the entire structure. For the case under consideration, it may readily be observed from a free-body diagram of element 1 that \left( M _2\right)_1=-P L . Hence, we obtain the shear and moment diagrams for the beam as shown in Figures 17.9(b) and (c), respectively.