Chapter 10, Problem 10.31P

To determine

The couple $C_0$ required to hold the mechanism in equilibrium.

Answer to Problem 10.31P

The couple $C_0$ required to hold the mechanism in equilibrium is $9.468\text{N}\cdot \text{m}$.

Explanation of Solution

Given:

For equilibrium the value of $\theta$ is $25°$.

The value of force $Q$ is $\text{200 N}$.

Concept used:

To determine the couple $C_0$, apply a virtual angular displacement at point A.

Draw the kinematic diagram for the mechanism as shown below:

  

The instantaneous center of the system is at point D after theapplication of virtual angular displacement at point A the bar AB rotates about point A and bar BC will rotate about point D.

From above figure, by help of geometry angle $\alpha$ can be calculated as follows:

  $\begin{array}{c}\frac{\sin \alpha }{50}=\frac{\sin \theta }{40}\\ \sin \alpha =\frac{50\theta }{40}\end{array}$

Simplify further for $\alpha$.

  $\alpha ={\sin }^{-1}\left(1.25\theta \right)$ ...... (1)

Here, $\alpha$ is the angle made by link BC with horizontal and $\theta$ is the angle made by link AB with horizontal.

Write the expression for the distance between points A and C.

  $d_{AC}=50\cos \theta +40\cos \alpha$......(2)

Here, $d_{AC}$ is the distance between the points A and C.

Write the expression for the distance between point C and D.

  $d_{CD}=d_{AC}\tan \theta$......(3)

Here, $d_{CD}$ is the distance between the points C and D.

Write the expression for the distance between point A and D.

  $d_{AD}=\frac{d_{AC}}{\cos \theta }$......(4)

Here, $d_{AD}$ is the distance between the points A and D.

Write the expression for the distance between point B and D.

  $d_{BD}=d_{AD}-d_{AB}$......(5)

Here, $d_{BD}$ is the distance between the points B and D.

Write the expression for displacement of point B fro bar AB.

  $\delta r_B=50\delta \theta$ ...... (6)

Here, $\delta r_B$ is the displacement of point B and $\delta \theta$ is the virtual angular displacement at point A.

Write the expression for displacement of point B for bar BC.

  $\delta r_B=d_{BD}\delta {\theta }_{BC}$ ...... (7)

Here, $\delta {\theta }_{BC}$ is the angular displacement for bar BC about point D.

Write the expression for displacement of point C for bar BC.

  $\delta r_C=d_{CD}\delta {\theta }_{BC}$ ...... (8)

Here, $\delta r_C$ is the displacement of point C.

Write the expression for net virtual work done on the system.

  $\delta U=C_0\delta \theta -Q\delta r_C$ ...... (9)

Here, $\delta U$ is the net virtual work, $C_0$ is the applied couple and $Q$ is the applied force.

Calculation:

Substitute $25°$ for $\theta$ in equation (1).

  $\begin{array}{c}\alpha ={\sin }^{-1}\left(1.25\left(25°\right)\right)\\ \alpha =31.89°\end{array}$

Substitute $25°$ for $\theta$ and $31.89°$ for $\alpha$ in equation (2).

  $\begin{array}{c}{d}_{AC}=50\cos \left(25°\right)+40\cos \left(31.89°\right)\\ =79.28\text{ mm}\end{array}$

Substitute $25°$ for $\theta$ and $79.28\text{mm}$ for $d_{AC}$ in equation (3).

  $\begin{array}{c}{d}_{CD}=79.28\text{mm}\left(\tan 25°\right)\\ =36.97\text{mm}\end{array}$

Substitute $25°$ for $\theta$ and $79.28\text{mm}$ for $d_{AC}$ in equation (4).

  $\begin{array}{c}{d}_{AD}=\frac{79.28\text{ mm}}{\cos 25°}\\ =87.48\text{mm}\end{array}$

Substitute $87.48\text{mm}$ for $d_{AD}$ and $50\text{mm}$ for $d_{AB}$ in equation (5).

  $\begin{array}{c}{d}_{BD}=87.48-50\\ =37.48\text{mm}\end{array}$

Substitute $50\delta \theta$ for $\delta r_B$ and $37.48\text{mm}$ for $d_{BD}$ in equation (7).

  $\begin{array}{c}50\delta \theta =\left(37.48\right)\delta {\theta }_{BC}\\ \delta {\theta }_{BC}=1.334\delta \theta \end{array}$

Substitute $1.334\delta \theta$ for $\delta {\theta }_{BC}$ and $36.97\text{}\text{mm}$ for $d_{CD}$ in equation (8).

  $\begin{array}{c}\delta r_C=36.97\left(1.334\delta \theta \right)\\ =49.32\delta \theta \end{array}$

Substitute $200\text{ N}$ for $Q$, $0$ for $\delta U$ and $49.32\delta \theta$ for $\delta r_C$ in equation (9).

  $\begin{array}{c}0=C_0\delta \theta -200\left(49.32\delta \theta \right)\\ 0=\left(C_0-9864\right)\delta \theta \\ C_0=9468\text{N}\cdot \text{mm}\left(\frac{1\text{ m}}{1000\text{ mm}}\right)\\ C_0=9.468\text{}\text{N}\cdot \text{m}\end{array}$

The couple $C_0$ required to hold the mechanism in equilibrium is $9.468\text{N}\cdot \text{m}$.

Conclusion:

Thus, the couple $C_0$ required to hold the mechanism in equilibrium is $9.468\text{N}\cdot \text{m}$.