Chapter 10.1, Problem 10.53P

To determine

Find the force representing the reaction at A using the virtual work method.

Find the couple representing the reaction at A using the virtual work method.

Answer to Problem 10.53P

The force representing the reaction at A is $\underset{\_}{250\text{N}(\uparrow )}$.

The couple representing the reaction at A is $\underset{\_}{450\text{N}\cdot \text{m}\left(\text{Counter}\text{clockwise}\right)}$.

Explanation of Solution

Find the force at A;

Show the free-body diagram of the continuous beam as in Figure 1.

Consider the member AB is horizontal.

The vertical displacement at point A is $\delta y_A\downarrow$.

Find the vertical displacement $\left(\delta y_A\right)$ at A with respect to vertical displacement $\left(\delta y_B\right)$ at B as follows;

$\delta y_A=\delta y_B$

Find the vertical displacement $\left(\delta y_C\right)$ at C with respect to vertical displacement $\left(\delta y_B\right)$ at B as follows;

$\begin{array}{c}\frac{\delta y_B}{2.4}=\frac{\delta y_C}{1.5}\\ \delta y_C=\frac{1.5}{2.4}\delta y_B\\ =\frac{5}{8}\delta y_B\end{array}$

Find the vertical displacement $\left(\delta y_E\right)$ at E with respect to vertical displacement $\left(\delta y_B\right)$ at B as follows;

$\begin{array}{c}\frac{\delta y_B}{2.4}=\frac{\delta y_E}{1.2}\\ \delta y_E=\frac{1.2}{2.4}\delta y_B\\ =\frac{\delta y_B}{2}\end{array}$

Find the vertical displacement $\left(\delta y_F\right)$ at F with respect to vertical displacement $\left(\delta y_B\right)$ at B as follows;

$\begin{array}{c}\frac{\delta y_E}{1.8}=\frac{\delta y_F}{1.5}\\ \delta y_F=\frac{1.5}{1.8}\delta y_E\\ =\frac{5}{6}\left(\frac{\delta y_B}{2}\right)\\ =\frac{5}{12}\delta y_B\end{array}$

Use the concept of virtual work;

$\begin{array}{l}\delta U=0;\\ -A\delta y_A+800\delta y_C-600\delta y_F=0\end{array}$

Substitute $\delta y_B$ for $\delta y_A$, $\frac{5}{8}\delta y_B$ for $\delta y_C$, and $\frac{5}{12}\delta y_B$ for $\delta y_F$.

$\begin{array}{l}-A\delta y_B+800\left(\frac{5}{8}\delta y_B\right)-600\left(\frac{5}{12}\delta y_B\right)=0\\ -A+500-250=0\\ A=250\text{N}(\uparrow )\end{array}$

Therefore, the force representing the reaction at A is $\underset{\_}{250\text{N}(\uparrow )}$.

Find the couple at A;

Show the free-body diagram of the continuous beam as in Figure 2.

Rotate the member AB in the beam through $\delta \theta$ at the point A.

Find the vertical reaction $\left(\delta y_B\right)$ at point B as follows;

$\begin{array}{l}\delta \theta =\frac{\delta y_B}{1.8}\\ \delta y_B=1.8\delta \theta \end{array}$

Find the vertical displacement $\left(\delta y_C\right)$ at C with respect to vertical displacement $\left(\delta y_B\right)$ at B as follows;

$\begin{array}{c}\frac{\delta y_B}{2.4}=\frac{\delta y_C}{1.5}\\ \delta y_C=\frac{1.5}{2.4}\delta y_B\\ =\frac{5}{8}\left(1.8\delta \theta \right)\\ =1.125\delta \theta \end{array}$

Find the vertical displacement $\left(\delta y_E\right)$ at E with respect to vertical displacement $\left(\delta y_B\right)$ at B as follows;

$\begin{array}{c}\frac{\delta y_B}{2.4}=\frac{\delta y_E}{1.2}\\ \delta y_E=\frac{1.2}{2.4}\delta y_B\\ =\frac{1}{2}\left(1.8\delta \theta \right)\\ =0.9\delta \theta \end{array}$

Find the vertical displacement $\left(\delta y_F\right)$ at F with respect to vertical displacement $\left(\delta y_B\right)$ at B as follows;

$\begin{array}{c}\frac{\delta y_E}{1.8}=\frac{\delta y_F}{1.5}\\ \delta y_F=\frac{1.5}{1.8}\delta y_E\\ =\frac{5}{6}\left(0.9\delta \theta \right)\\ =0.75\delta \theta \end{array}$

Use the concept of virtual work;

$\begin{array}{l}\delta U=0;\\ -M_A\delta \theta +800\delta y_C-600\delta y_F=0\end{array}$

Substitute $1.125\delta \theta$ for $\delta y_C$ and $0.75\delta \theta$ for $\delta y_F$.

$\begin{array}{l}-M_A\delta \theta +800\left(1.125\delta \theta \right)-600\left(0.75\delta \theta \right)=0\\ -M_A+900-450=0\\ M_A=450\text{N}\cdot \text{m}\left(\text{Anti}\text{clockwise}\right)\end{array}$

Therefore, the couple representing the reaction at A is $\underset{\_}{450\text{N}\cdot \text{m}\left(\text{Counter}\text{clockwise}\right)}$.