Example 2.3: Kinematic analysis of a quick-return mechanism#
Position analysis#
The figure shows a 4-bar quick-return mechanism. At the instant when the crank forms an angle \(\theta_2 = 20^{\circ}\) with the horizontal, calculate the relative position of the slider with respect to \(O_4\), the angle that link 4 forms with the horizontal, and the coordinates of point B.
Data:
Link lengths, in meters: \(\overline{O_2 A} = 1.0;\, \overline{O_4 B} = 3.0;\, \overline{O_2 O_4} = 3.0;\)
Position analysis: graphical method#
The objective is to construct the mechanism to scale to measure the unknowns: the angle of link 4 (\(\theta_4\)), the position of slider A along link 4 (distance \(\overline{O_4A}\)), and the coordinates of point B.
Construction Data:#
Link 1 (Frame): \(L_1 = \overline{O_2O_4} = 3.0 \text{ m}\)
Link 2 (Crank): \(L_2 = \overline{O_2A} = 1.0 \text{ m}\)
Link 4 (Slotted bar): The total length is \(L_4 = \overline{O_4B} = 3.0 \text{ m}\).
Input Angle: \(\theta_2 = 20^\circ\).
Construction Procedure:#
Establish the frame: Place the fixed pivot \(O_2\) at the origin (0,0) and the fixed pivot \(O_4\) at (3, 0).
Position the crank (Link 2): From \(O_2\), draw link 2 with its length \(L_2 = 1.0 \text{ m}\) and angle \(\theta_2 = 20^\circ\). The end is the position of pin A.
Determine the slotted bar (Link 4): Draw a straight line passing through \(O_4\) and A. This line defines the orientation of link 4.
Locate point B: Extend the line from \(O_4\) in the direction of A until it has the total length of link 4, \(L_4 = 3.0 \text{ m}\). The end of this segment is point B.
Measurement of results:
Measure the angle of the line \(O_4B\) with the horizontal to get \(\theta_4\).
Measure the distance of the segment \(\overline{O_4A}\) to get the relative position of the slider.
Read the coordinates of point B from the reference system.
Position analysis: analytical method#
Vector equations are used to find expressions for the unknowns (\(\theta_4\), \(d = \overline{O_4A}\), and coordinates of B).
Vector Loop Formulation#
We consider the loop formed by points \(O_2, O_4, A\). The vector equation is: \( \vec{R}_{O_2A} = \vec{R}_{O_2O_4} + \vec{R}_{O_4A} \) Rearranging to solve for the vector defining link 4: \( \vec{R}_{O_4A} = \vec{R}_{O_2A} - \vec{R}_{O_2O_4} \) Where:
\(\vec{R}_{O_2A} = L_2(\cos\theta_2 \hat{i} + \sin\theta_2 \hat{j})\)
\(\vec{R}_{O_2O_4} = L_1 \hat{i}\)
\(\vec{R}_{O_4A}\) is a vector whose direction gives us \(\theta_4\) and whose magnitude is the distance \(d = \overline{O_4A}\).
Solving for \(\theta_4\) and \(d\)#
Decompose into components: \( (x_A - x_{O_4}) \hat{i} + (y_A - y_{O_4}) \hat{j} = (L_2\cos\theta_2 - L_1) \hat{i} + (L_2\sin\theta_2) \hat{j} \)
Calculate \(\theta_4\): The angle of link 4 is the angle of the vector \(\vec{R}_{O_4A}\). \( \theta_4 = \arctan\left(\frac{L_2\sin\theta_2}{L_2\cos\theta_2 - L_1}\right) \)
Calculate \(d\) (slider position): The distance is the magnitude of the vector \(\vec{R}_{O_4A}\). \( d = |\vec{R}_{O_4A}| = \sqrt{(L_2\cos\theta_2 - L_1)^2 + (L_2\sin\theta_2)^2} \)
Calculation of the Coordinates of Point B#
Point B is at the end of link 4. Its position vector is: \( \vec{R}_B = \vec{R}_{O_4} + L_4 (\cos\theta_4 \hat{i} + \sin\theta_4 \hat{j}) \)
\(x_B = x_{O_4} + L_4\cos\theta_4\)
\(y_B = y_{O_4} + L_4\sin\theta_4\)
Numerical Calculation for the Example#
Initial Data:
Lengths: \(L_1=3.0 \text{ m}, L_2=1.0 \text{ m}, L_4=3.0 \text{ m}\)
Input angle: \(\theta_2 = 20^\circ\)
Calculation of \(\theta_4\) and \(d\): First, we calculate the components of the vector \(\vec{R}_{O_4A}\):
x-component: \(L_2\cos\theta_2 - L_1 = 1.0\cos(20^\circ) - 3.0 = 0.9397 - 3.0 = -2.0603 \text{ m}\)
y-component: \(L_2\sin\theta_2 = 1.0\sin(20^\circ) = 0.3420 \text{ m}\)
Now, we calculate the angle and magnitude: \( \theta_4 = \arctan\left(\frac{0.3420}{-2.0603}\right) = 170.56^\circ \) \( d = \sqrt{(-2.0603)^2 + (0.3420)^2} = \sqrt{4.2448 + 0.1170} = 2.0885 \text{ m} \)
Calculation of Coordinates of B: We use \(\theta_4 = 170.56^\circ\) and \(L_4 = 3.0 \text{ m}\).
\(x_B = 3.0 + 3.0\cos(170.56^\circ) = 3.0 + 3.0(-0.9865) = 3.0 - 2.9595 = 0.0405 \text{ m}\)
\(y_B = 0 + 3.0\sin(170.56^\circ) = 3.0(0.1640) = 0.4920 \text{ m}\)
Final Position Results:
Slider position A: \(\mathbf{d = \overline{O_4A} = 2.089 \text{ m}}\)
Angle of link 4: \(\mathbf{\theta_4 = 170.56^\circ}\)
Coordinates of B: (0.0405, 0.4920) m
Velocity analysis#
The slider represents the expansion movement of a cylinder (e.g., a truck’s hydraulic cylinder) at a constant velocity of \(0.1\) m/s. At the instant when the crank forms an angle \(\theta_2 = 20^{\circ}\) with the horizontal, calculate the velocity of point B and the angular velocities of links 2 and 4.
Data:
Link lengths, in meters: \(\overline{O_2 A} = 1.0;\, \overline{O_4 B} = 3.0;\, \overline{O_2 O_4} = 3.0\). Angle of link 4 with the horizontal, from the position analysis: \(\theta_4 = 170.56^{\circ}\). Position of slider A relative to \(O_4\): \(d = \overline{O_4 A} = 2.089\) m.
Resolution#
The mechanism has one degree of freedom, which corresponds to the relative velocity of the slider with respect to link 4, \(\vec{v}_{A3/A4} = \vec{v}_{A2/A4}\). We use this known data to calculate the angular velocities of links 2 and 4 by applying the relative velocity equation, knowing that at the analyzed instant, points \(A_2\), \(A_3\), and \(A_4\) coincide. It’s worth remembering that \(A_2\) and \(A_3\), being connected by a revolute joint, have the same absolute velocity, which must be perpendicular to link 2. Meanwhile, 3 and 4 are joined by a prismatic joint, and between \(A_3\) and \(A_4\) there is a relative velocity in the direction of link 4, which is precisely the data provided. With this, we have:
\(\vec{v}_{A_2} = \vec{v}_{A_4} + \vec{v}_{A_2/A_4}\)
where each term can be expressed as:
\(\vec{v}_{A_2} = \vec{\omega}_2 \times \vec{r}_{O2A}\)
clear, clc
syms w2k real
rO2A = [cosd(20) sind(20) 0];
w2 = [0 0 w2k];
vA2 = cross(w2, rO2A);
\(\vec{v}_{A_4} = \vec{\omega}_4 \times \vec{r}_{O4A}\)
syms w4k real
rO4A = 2.089 * [cosd(170.56) sind(170.56) 0];
w4 = [0 0 w4k];
vA4 = cross(w4, rO4A);
\(\vec{v}_{A_2/A_4} = 0.1(\cos 170.56^{\circ} \hat{i} + \sin 170.56^{\circ} \hat{j})\)
vA2_A4 = 0.1 * [cosd(170.56) sind(170.56) 0];
Now we can solve the relative velocity equation:
eq1 = vA2 == vA4 + vA2_A4;
sol_v1 = solve(eq1, w2k, w4k);
w2k = double(sol_v1.w2k);
w4k = double(sol_v1.w4k);
fprintf('w2k=%0.3f w4k=%0.3f\n', w2k, w4k)
w2k=0.203 w4k=-0.085
Finally, we calculate the velocity of point B now that we know \(\vec{\omega}_4\):
rO4B = 3 * [cosd(170.56) sind(170.56) 0];
vB = subs(cross(w4, rO4B));
fprintf('|vB|=%0.3f\n', norm([vB(1) vB(2)]))
|vB|=0.254
Velocity Cinema#
The velocity cinema for this example is not provided, but it can be created using the same principles demonstrated in previous examples.
Acceleration analysis#
The slider moves at the same constant velocity of \(0.1\) m/s. At the instant when the crank forms an angle \(\theta_2 = 20^{\circ}\) with the horizontal, calculate the acceleration of point B and the angular accelerations of links 2 and 4.
Data:
Link lengths, in meters: \(\overline{O_2 A} = 1.0;\, \overline{O_4 B} = 3.0;\, \overline{O_2 O_4} = 3.0\). Angle of link 4 with the horizontal (from position analysis): \(\theta_4 = 170.56^{\circ}\) and position of slider A relative to \(O_4\): \(d = \overline{O_4 A} = 2.089\) m. Angular velocities (from velocity analysis): \(\vec{\omega}_2 = 0.203\hat{k}\) rad/s; \(\vec{\omega}_4 = -0.085\hat{k}\) rad/s.
Resolution#
The procedure for calculating the mechanism’s accelerations is similar to that for velocities, taking into account that this time the Coriolis acceleration is involved:
\(\vec{a}_{A_2} = \vec{a}_{A_4} + \vec{a}_{A_2/A_4} + \vec{a}_{Cor}\)
where each term can be expressed as:
\(\vec{a}_{A_2} = \vec{\omega}_2 \times (\vec{\omega}_2 \times \vec{r}_{O2A}) + \vec{\alpha}_2 \times \vec{r}_{O2A}\)
syms af2k real
af2 = [0 0 af2k];
aA2 = cross(w2, cross(w2, rO2A)) + cross(af2, rO2A);
\(\vec{a}_{A_4} = \vec{\omega}_4 \times (\vec{\omega}_4 \times \vec{r}_{O4A}) + \vec{\alpha}_4 \times \vec{r}_{O4A}\)
syms af4k real
af4 = [0 0 af4k];
aA4 = cross(w4, cross(w4, rO4A)) + cross(af4, rO4A);
\(\vec{a}_{A_2/A_4} = 0\) (constant expansion velocity)
aA2_A4 = [0 0 0];
\(\vec{a}_{Cor} = 2\vec{\omega}_4 \times \vec{v}_{A2/A4}\)
aCor = 2 * cross(w4, vA2_A4);
Now we can solve the relative acceleration equation:
eq2 = aA2 == aA4 + aA2_A4 + aCor;
sol_a1 = solve(eq2, af2k, af4k);
af2k = double(subs(sol_a1.af2k));
af4k = double(subs(sol_a1.af4k));
fprintf('af2k=%0.3f af4k=%0.3f\n', af2k, af4k)
af2k=-0.104 af4k=0.061
Finally, we calculate the acceleration of point B now that we know \(\vec{\alpha}_4\):
aB = double(subs(cross(w4, (cross(w4, rO4B))) + cross(af4, rO4B)));
fprintf('aBi=%0.3f aBj=%0.3f\n', aB(1), aB(2))
aBi=-0.009 aBj=-0.185
fprintf('|aB|=%0.3f\n', norm([aB(1) aB(2)]))
|aB|=0.185
Acceleration Cinema#
The acceleration cinema for this example is not provided, but it can be created using the same principles demonstrated in previous examples.