Chapter 10.1, Problem 10.53P

Using the method of virtual work, determine separately the force and couple representing the reaction at A.

Fig. P10.53 and P10.54

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 $\underset{\_}{450\text{N}\cdot \text{m}\left(\text{Counter}\text{clockwise}\right)}$.