Chapter 10, Problem 10.21P

To determine

The couple $C_o$ required to hold the mechanism in the position shown.

Answer to Problem 10.21P

The couple $C_o$ required to hold the mechanism in the position shown is $103\text{N}\cdot \text{m}$.

Explanation of Solution

Given:

The mass of each uniform bar is $18\text{ kg/m}$.

Concept used:

Apply the Principle of virtual work by providing a small angular virtual displacement of $\delta \theta$ to bar AB.

By applying the virtual angular displacement to bar AB then bar AB will rotate, Bar BC will translate and bar CD will rotate

The virtual kinematic diagram after the angular displacement of $\delta \theta$ is shown below:

Write the expression to calculate linear displacement of bar AB.

  $\delta y_{AB}=l_{AB}\delta \theta$ ...... (1)

Here, $\delta y_{AB}$ is the linear displacement of end of bar AB, $l_{AB}$ is the length of bar AB and $\delta \theta$ is the virtual angular displacement.

Write the expression to calculate linear displacement of bar CD.

  $\delta y_{CD}=l_{CD}\delta \theta$ ...... (2)

Here, $\delta y_{CD}$ is the linear displacement of end of bar CD, $l_{CD}$ is the length of bar CD and $\delta \theta$ is the virtual angular displacement.

The weight force is the only force applied on the system thus, but similar triangle the displacement of center of bars AB and CD is $0.5\delta y_{AB}$ and $0.5\delta y_{CD}$.

Write the expression for the weight of bar.

  $W=\dot{m}gl$ ...... (3)

Here, $W$ is the weight of bar, $\dot{m}$ is the mass per unit length, $g$ is acceleration due to gravity and $l$ is the length of bar.

Write the expression for virtual work done on system.

  $\delta U=C_0\delta \theta -W_{AB}\left(0.5\delta y_{AB}\right)-W_{BC}\delta y_{BC}-W_{CD}\left(0.5\delta y_{CD}\right)$ ...... (4)

Here, $\delta U$ is the total virtual work done on system, $C_0$ is the initial couple, $W_{AB}$ is the weight of bar AB, $W_{BC}$ is the weight of bar BC, $W_{CD}$ is the weight of bar CD and $\delta y_{BC}$ is the displacement of bar BC.

Calculation:

Substitute $480\text{mm}$ for $l_{AB}$ in equation (1).

  $\begin{array}{c}\delta y_{AB}=\left(480\text{ mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)\delta \theta \\ =0.48\delta \theta \end{array}$

Displacement of bar BC will be equal to displacement of end B as there is no constraint therefore, $\delta y_{BC}$ is $0.48\delta \theta$.

Substitute $480\text{mm}$ for $l_{CD}$ in equation (2).

  $\begin{array}{c}\delta y_{CD}=\left(480\text{ mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)\delta \theta \\ =0.48\delta \theta \end{array}$

Substitute $W_{AB}$ for $W$, $18\text{ kg/m}$ for $\dot{m}$, $9.81{\text{ m/s}}^2$ for $g$ and $480\text{ mm}$ for $l$ in equation (3).

  $\begin{array}{c}{W}_{AB}=18\left(9.81\right)\left(480\text{ mm}\left(\frac{1\text{ m}}{1000\text{mm}}\right)\right)\\ =84.76\text{N}\end{array}$

Substitute $W_{BC}$ for $W$, $18\text{ kg/m}$ for $\dot{m}$, $9.81{\text{ m/s}}^2$ for $g$ and $600\text{ mm}$ for $l$ in equation (3).

  $\begin{array}{c}{W}_{BC}=18\left(9.81\right)\left(600\text{ mm}\left(\frac{1\text{ m}}{1000\text{mm}}\right)\right)\\ =105.95\text{N}\end{array}$

Substitute $W_{CD}$ for $W$, $18\text{ kg/m}$ for $\dot{m}$, $9.81{\text{ m/s}}^2$ for $g$ and $750\text{ mm}$ for $l$ in equation (3).

  $\begin{array}{c}{W}_{CD}=18\left(9.81\right)\left(750\text{ mm}\left(\frac{1\text{ m}}{1000\text{mm}}\right)\right)\\ =132.44\text{N}\end{array}$

Substitute $0$ for $\delta U$, $84.76\text{ N}$ for $W_{AB}$, $105.95\text{N}$ for $W_{BC}$, $132.44\text{N}$ for $W_{CD}$, $0.48\delta \theta$ for $\delta y_{AB}$, $0.48\delta \theta$ for $\delta y_{BC}$ and $0.48\delta \theta$ for $\delta y_{CD}$ in equation (4).

  $\begin{array}{c}0=C_0\delta \theta -84.76\left(0.5\right)\left(0.48\right)\delta \theta -105.95\left(0.48\right)\delta \theta -132.44\left(0.5\right)\left(0.48\right)\delta \theta \\ 0=\left(C_0-102.98\right)\delta \theta \\ C_0\approx 103\text{ N}\cdot \text{m}\end{array}$

The couple $C_o$ required to hold the mechanism in the position shown is $103\text{N}\cdot \text{m}$.

Conclusion:

Thus, the couple $C_o$ required to hold the mechanism in the position shown is $103\text{N}\cdot \text{m}$.