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 \(1.06 \mathrm{~m} / \mathrm{s}\).

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