Problem

# The two-cylinder engine is designed so that the pistons are connected to the crankshaft BE...

The two-cylinder engine is designed so that the pistons are connected to the crankshaft BE using a master rod ABC and articulated rod AD. If the crankshaft is rotating at ω = 30 rad/s, determine the velocities of the pistons C and D at the instant shown.

#### Step-by-Step Solution

Solution 1

Draw the kinematic diagram:

Calculate the velocity of point $$B$$.

$$v_{B}=30 \mathbf{k} \times(0.05 \mathbf{j})$$

Find the position vector of point $$C$$ with respect to point $$B$$.

$$r_{C / B}=\left[\left(0.25 \times \cos 60^{\circ}\right) \mathbf{i}+\left(0.25 \times \sin 60^{\circ}\right) \mathbf{j}\right]$$

Write the velocity of point $$C$$ in vector form.

$$v_{C}=-\left(v_{C} \times \cos 45^{\circ}\right) \mathbf{i}-\left(v_{C} \times \sin 45^{\circ}\right) \mathbf{j}$$

According to general planer motion for the link $$B C$$ :

$$v_{C}=v_{B}+\left(\omega_{B C} \times r_{C / B}\right) \ldots \ldots .(1)$$

For the link $$B C$$ :

Substitute, $$-\left(v_{C} \times \cos 45^{\circ}\right) \mathbf{i}-\left(v_{C} \times \sin 45^{\circ}\right) \mathbf{j}$$ for $$v_{C}, 30 \mathbf{k} \times(0.05 \mathbf{j})$$ for $$v_{B}, \omega_{B C} \mathbf{k}$$ for $$\omega_{B C}$$,

$$\left[\left(0.25 \times \cos 60^{\circ}\right) \mathbf{i}+\left(0.25 \times \sin 60^{\circ}\right) \mathbf{j}\right]$$ for $$r_{C / B}$$ in the equation (1):

$$-\left(v_{C} \times \cos 45^{\circ} \mathbf{i}\right)-\left(v_{C} \times \sin 45^{\circ}\right) \mathbf{j}=30 \mathbf{k} \times(0.05 \mathbf{j})+\left(\omega_{B C} \mathbf{k}\right)\left[0.25 \times \cos 60^{\circ} \mathbf{i}+0.25 \times \sin 60^{\circ} \mathbf{j}\right]$$

$$-\left(v_{C} \times \cos 45^{\circ}\right) \mathbf{i}-\left(v_{C} \times \sin 45^{\circ}\right) \mathbf{j}=30 \mathbf{k} \times(0.05 \mathbf{j})+\left(\omega_{B C} \mathbf{k}\right)[0.125 \mathbf{i}+0.217 \mathbf{j}]$$

$$-\left(v_{C} \times \cos 45^{\circ}\right) \mathbf{i}-\left(v_{C} \times \sin 45^{\circ}\right) \mathbf{j}=-1.5 \mathbf{i}+\left(\omega_{B C}\right) 0.125 \mathbf{j}-\left(\omega_{B C}\right) 0.217 \mathbf{i}$$

Equate i components:

$$-v_{C} \times \cos 45^{\circ}=-1.5-\left(\omega_{B C}\right) 0.217 \ldots \ldots .(2)$$

Equate jcomponents:

$$-v_{C} \times \sin 45^{\circ}=\left(\omega_{B C}\right) 0.125$$...... (3)

$$\omega_{B C}=-5.66 \times v_{C}$$

Substitute, $$-\left(5.66 \times v_{C}\right)$$ for $$\omega_{B C}$$ in the equation (2):

$$-v_{C} \times 0.707=-1.5-\left(-5.66 \times v_{C}\right) 0.217$$

$$-0.707 v_{C}=-1.5+1.23 v_{C}$$

$$1.93522 v_{C}=1.5$$

$$v_{C}=0.775 \mathrm{~m} / \mathrm{s}$$

Therefore, the velocity of the piston $$C$$ at the instant shown is $$0.775 \mathrm{~m} / \mathrm{s}$$

Substitute, this value $$0.775 \mathrm{~m} / \mathrm{s}$$ for $$v_{C}$$ in the equation (3):

$$\omega_{B C}=-5.66 \times 0.775$$

$$=-4.39 \mathrm{rad} / \mathrm{s}$$

As $$A B C$$ is a single member:

$$\omega_{\mathrm{AB}}=\omega_{\mathrm{BC}}$$

Calculate the velocity of point $$A$$ with respect to pint $$B$$.

$$\mathbf{v}_{\mathrm{A}}=\mathbf{v}_{\mathrm{B}}+\omega_{\mathrm{AB}} \times \mathbf{r}_{\mathrm{A} / \mathrm{B}}$$

Substitute $$30 \mathbf{k} \times(0.05 \mathbf{j})$$ for $$v_{B},-4.39 \mathbf{k} \mathrm{rad} / \mathrm{s}$$ for $$\omega_{\mathrm{AB}}$$ and

$$(-0.05 \mathrm{~m}) \cos 45^{\circ} \mathbf{i}+0.05 \sin 45^{\circ} \mathbf{j}$$ for $$\mathbf{r}_{\mathbf{A} / \mathrm{B}}$$

$$\mathbf{v}_{\mathbf{A}}=\mathbf{v}_{\mathbf{B}}+\omega_{\mathbf{A B}} \times \mathbf{r}_{\mathbf{A} / \mathrm{B}}$$

$$=30 \mathbf{k} \times 0.05 \mathbf{j}+(-4.39 \mathbf{k}) \times\left(-0.05 \cos 45^{\circ} \mathbf{i}+0.05 \sin 45^{\circ} \mathbf{j}\right)$$

$$=-1.5 \mathbf{i}+0.155 \mathbf{j}+0.1552 \mathbf{i}$$

$$=-1.344 \mathbf{i}+0.155 \mathbf{j} \mathrm{m} / \mathrm{s}$$

According to general planer motion for the link $$A D:$$ $$v_{D}=v_{A}+\omega_{A D} \times r_{D / A} \ldots \ldots . .(4)$$

Write the position vector of point $$D$$ with respect to point $$A$$.

$$\mathbf{r}_{\mathbf{D} / \mathbf{A}}=-0.25 \cos 45^{\circ} \mathbf{i}+0.25 \sin 45^{\circ} \mathbf{j}$$

Substitute all these values in the equation (4):

$$\left(v_{D} \times \cos 45^{\circ}\right) \mathbf{i}-\left(v_{D} \times \sin 45^{\circ}\right) \mathbf{j}=(-1.344 \mathbf{i}+0.155 \mathbf{j})+\omega_{A D} \mathbf{k} \times\left(-0.25 \cos 45^{\circ} \mathbf{i}+0.25 \sin 45^{\circ} \mathbf{j}\right)$$

$$\left(v_{D} \times 0.707\right) \mathbf{i}-\left(v_{D} \times 0.707\right) \mathbf{j}=(-1.344 \mathbf{i}+0.155 \mathbf{j})+\left(-\omega_{A D} \times 0.176 \mathbf{j}\right)+\left(-\omega_{A D} \times 0.1767\right) \mathbf{i}$$

$$\left(v_{D} \times 0.707\right) \mathbf{i}-\left(v_{D} \times 0.707\right) \mathbf{j}=\left(-1.344+\left(-\omega_{A D} \times 0.1767\right)\right) \mathbf{i}+\left(0.155+\left(-\omega_{A D} \times 0.176\right)\right) \mathbf{j}$$

Equate i components:

$$\left(v_{D} \times 0.707\right)=\left(-1.344+\left(-\omega_{A D} \times 0.1767\right)\right) \ldots \ldots .(5)$$

Equate $$\mathbf{j}$$ components:

$$-\left(v_{D} \times 0.707\right)=\left(0.155+\left(-\omega_{A D} \times 0.176\right)\right)$$

$$\omega_{A D}=\frac{\left(v_{D} \times 0.707\right)+0.155}{0.176}$$

Substitute, $$\frac{\left(v_{D} \times 0.707\right)+0.155}{0.176}$$ for $$\omega_{A D}$$ in the equation (5):

$$\left(v_{D} \times 0.707\right)=\left(-1.344+\left(-\omega_{A D} \times 0.1767\right)\right)$$

$$\left(v_{D} \times 0.707\right)=\left(-1.344+\left(-\left(\frac{\left(v_{D} \times 0.707\right)+0.155}{0.176}\right) \times 0.1767\right)\right)$$

$$\left(v_{D} \times 0.707\right)=\left(-1.344-\left(v_{D} \times 0.707+0.155\right)\right)$$

$$0.707 v_{D}+0.707 v_{D}=-1.499$$

$$v_{D}=-1.06 \mathrm{~m} / \mathrm{s}$$

Negative sign indicates that the direction will be opposite of the assumed direction.

Therefore, velocity of the piston $$D$$ at the instant shown $$v_{D}$$ is $$1.06 \mathrm{~m} / \mathrm{s}$$.