PREFACE

In the study of mechanics of machines, students and lecturers often encounter challenges in bridging theoretical knowledge with practical applications. While textbooks provide comprehensive derivations and explanations of fundamental principles, there is a need for a resource that focuses on problem-solving through computational methods. This book, Worked Examples in Mechanics of Machines using MATLAB, is designed to fill that gap by offering a structured approach to solving kinematic and dynamic problems using MATLAB.

The primary aim of this book is to serve as a practical companion for tutorials and seminars in mechanical engineering courses. It provides a collection of carefully selected worked examples that demonstrate how MATLAB can be used to analyse mechanisms, velocity and acceleration, inertia forces, energy storage, and power transmission. By following the structured format of problem statement → MATLAB solution → interpretation of results, students can develop a deeper understanding of how theoretical equations are applied in real-world engineering problems.

This book is particularly useful for:

Students, who can use it as a self-study guide to practice computational problem-solving techniques.

Lecturers, who can integrate the examples into tutorials, assignments, and seminar discussions.

Engineering professionals and researchers, who wish to apply MATLAB in machine dynamics analysis.

Each chapter covers a key topic in mechanics of machines, from velocity and acceleration analysis of linkages to flywheels, governors, cams, and gear trains. The focus is on providing step-by-step MATLAB solutions that reinforce conceptual understanding and computational skills. The book does not replace standard textbooks but complements them by offering hands-on exercises that allow students to experiment with different parameters and observe their effects on system behaviour.

To enhance learning, students and lecturers are encouraged to:

Modify the MATLAB scripts to test different scenarios and deepen their understanding.

Discuss the engineering significance of the results, particularly in design and optimization contexts.

Apply the computational techniques to real-world mechanical systems and case studies.

We hope that this book will prove to be a valuable resource in advancing both teaching and learning in mechanics of machines. By integrating MATLAB into problem-solving, we aim to equip students with essential computational skills that are increasingly important in modern engineering practice.

Prof. Eric Okoth Ogur PhD CEng MIMechE MBS

ACKNOWLEDGEMENTS

I would like to express my gratitude to Prof. S. S. Rattan/Tata McGraw Hill for his generous permission to utilize questions from Theory of Machines (3rd Edition).

CONTENTS

PREFACE

ACKNOWLEDGEMENTS

CONTENTS

CHAPTER 1: MECHANISMS: VELOCITY AND ACCELERATION

CHAPTER 2: INERTIA FORCES IN RECIPROCATING PARTS

CHAPTER 3: TURNING MOMENT DIAGRAMS

CHAPTER 4: BALANCING

CHAPTER 5: FRICTION

CHAPTER 6: FRICTION CLUTCHES

CHAPTER 7: BELT DRIVES

CHAPTER 8: CAMS

CHAPTER 9: GEARS

CHAPTER 10: GEAR TRAINS

CHAPTER 11: GOVERNORS

CHAPTER 12: GYROSCOPES

CHAPTER 13: MECHANICAL VIBRATIONS

REFERENCES

APPENDIX: INTRODUCTION TO MATLAB

CHAPTER 1: MECHANISMS: VELOCITY AND ACCELERATION

Example 1.1: The slider C of the toggle mechanism shown in Fig. 1.1 is constrained to move on a horizontal path. The crank rotates in the counterclockwise direction at a uniform speed of 180 rpm.

Determine(a) velocity of slider C, (b) angular velocity of links AB, and BC, (c) rubbing velocities on the pins of 25 mm diameter at A and C and (d) torque required at the crank for a force of 2 kN at C.

Fig. 1.1 Toggle mechanism

Solution using MATLAB

% Given data

N = 180; % rpm (Rotations per minute of crank QA)

QA = 0.2; % m (Length of crank QA)

AB = 0.4; % m (Length of link AB)

OB = 0.3; % m (Length of link OB)

BC = 0.6; % m (Length of link BC)

pin_diameter = 0.025; % m (Diameter of pins at joints)

force_at_C = 2000; % N (Force applied at slider C)

% Convert rpm to rad/s

% This step converts the rotational speed from revolutions per minute (rpm) to radians per second (rad/s), which is the standard unit for angular velocity in calculations.

omega = 2 * pi * N / 60;

% Linear velocity of point A

% The linear velocity of point A is calculated using the angular velocity of the crank QA and its length.

vA = omega * QA;

% Velocity diagram scale (1 cm = 0.5 m/s)

% This scale is used to convert measurements from a velocity diagram (typically drawn to scale) into actual velocities.

scale = 0.5; % m/s per cm

% Measured velocities from the diagram (in cm)

% These values are assumed to have been obtained by measuring lengths from a velocity diagram, which represents the velocities of different points in the mechanism.

vC_cm = 4.2; % cm (Velocity of point C measured from the diagram)

vAB_cm = 2.4; % cm (Relative velocity of point B with respect to A measured from the diagram)

vB_cm = 6.2; % cm (Velocity of point B measured from the diagram)

vBC_cm = 5.7; % cm (Relative velocity of point C with respect to B measured from the diagram)

% Convert measured velocities to m/s

% The measured velocities in centimetres are converted to meters per second using the defined scale.

vC = vC_cm * scale;

vAB = vAB_cm * scale;

vB = vB_cm * scale;

vBC = vBC_cm * scale;

% Angular velocities

% Angular velocities are calculated by dividing the linear velocities by the respective link lengths.

omega_AB = vAB / AB; % Angular velocity of link AB

omega_OB = vB / OB; % Angular velocity of link OB

omega_BC = vBC / BC; % Angular velocity of link BC

% Rubbing velocities

% Rubbing velocities are calculated to determine the speed at which the pins are rubbing against their bearings, which is important for wear and friction analysis.

pin_radius = pin_diameter / 2; % Calculate pin radius from diameter

rubbing_velocity_C = omega_BC * pin_radius; % Rubbing velocity at pin C

relative_omega_A = omega_OB - omega_AB + omega_BC; % Calculate the relative angular velocity at pin A

rubbing_velocity_A = relative_omega_A * pin_radius; % Rubbing velocity at pin A

% Torque calculation

% Torque is calculated using the principle of power balance. The power input at crank QA is equal to the power output at slider C, assuming no losses.

torque = (force_at_C * vC) / omega;

% Display results

% The calculated results are displayed in a user-friendly format.

fprintf('Velocity of slider C: %.2f m/s\n', vC);

fprintf('Angular velocity of link AB: %.2f rad/s\n', omega_AB);

fprintf('Angular velocity of link OB: %.2f rad/s\n', omega_OB);

fprintf('Angular velocity of link BC: %.2f rad/s\n', omega_BC);

fprintf('Rubbing velocity on pin at C: %.4f m/s\n', rubbing_velocity_C);

fprintf('Rubbing velocity on pin at A: %.4f m/s\n', rubbing_velocity_A);

fprintf('Torque required at crank QA: %.2f Nm\n', torque);

MATLAB Output

Velocity of slider C: 2.10 m/s

Interpretation: This is the linear velocity of the slider at point C. It tells us how fast the slider is moving along its path.

Significance: This value is crucial for understanding the output speed of the mechanism. If the slider represents a cutting tool or a moving part in a machine, this velocity dictates the production rate or operating speed.

2. Angular velocity of link AB: 3.00 rad/s

Interpretation: This is the rotational speed of link AB.

Significance: This value indicates how quickly link AB is rotating. It's essential for understanding the dynamics of the mechanism and for calculating inertial forces.

3. Angular velocity of link OB: 10.33 rad/s

Interpretation: This is the rotational speed of link OB.

Significance: This is a much higher angular velocity than link AB. This is important to note, as the link OB is directly connected to the crank QA, and thus closely related to the input speed. This higher speed will have a larger effect on the inertial forces of link OB.

4. Angular velocity of link BC: 4.75 rad/s

Interpretation: This is the rotational speed of link BC.

Significance: This value is important for understanding the motion of link BC and its impact on the slider's movement. It contributes to the rubbing velocity at pin C.

5. Rubbing velocity on pin at C: 0.0594 m/s

Interpretation: This is the relative velocity between the pin at joint C and its bearing.

Significance:

This value is essential for assessing wear and friction at the joint.

Higher rubbing velocities lead to increased wear and heat generation.

Knowing this value helps in selecting appropriate lubricants and materials for the pin and bearing.

This is a relatively low value, which is good for longevity of the pin.

6. Rubbing velocity on pin at A: 0.1510 m/s

Interpretation: This is the relative velocity between the pin at joint A and its bearing.

Significance:

This value is significantly higher than the rubbing velocity at pin C.

This indicates that pin A experiences more wear and friction.

This joint may require more frequent lubrication or the use of more wear-resistant materials.

The relative angular velocity calculation is reflected in this higher rubbing velocity.

7. Torque required at crank QA: 222.82 Nm

Interpretation: This is the torque that needs to be applied to the crank QA to overcome the force at slider C and maintain the given motion.

Significance:

This value is crucial for selecting the appropriate motor or power source to drive the mechanism.

It represents the load that the motor must handle.

It is an indication of the force required to operate the mechanism.

Example 1.2: Dimensions of the four-bar chain Fig. 1.2 are OC = 22 cm, OA = 15 cm, AB = 20 cm and CB = 18 cm. Since OC is fixed, OA will complete a revolution whereas CB will rock between limits. Let OA rotate with uniform angular velocity of 10 rad/s counterclockwise as seen by the observer with uniform angular velocity. Determine the velocity and acceleration of point B.

Fig. 1.2 Four-bar chain

Solution using MATLAB

% Given dimensions (in cm)

OC = 22; % Fixed link length (distance between O and C)

OA = 15; % Crank length

AB = 20; % Connecting rod length

CB = 18; % Rocker length

% Angular velocity of crank OA (rad/s)

omega2 = 10; % Counterclockwise (positive)

% Step I: Position Vectors

% Angle of rotation of crank OA (60 degrees)

theta = 60; % Degrees

theta_rad = deg2rad(theta); % Convert degrees to radians

% Position vector of point A

% Calculate the x and y coordinates of point A using trigonometry.

OA_x = OA * cos(theta_rad);

OA_y = OA * sin(theta_rad);

OA_vec = [OA_x; OA_y]; % Position vector of A [x; y]

% Distance AC

% Calculate the distance between points A and C using the distance formula.

AC = sqrt((OA_y)^2 + (OC - OA_x)^2);

% Angles using cosine rule

% Calculate the angles ACO and ACB using the law of cosines.

angle_ACO = acos((AC^2 + OC^2 - OA^2) / (2 * AC * OC)); % In radians

angle_ACB = acos((CB^2 + AC^2 - AB^2) / (2 * CB * AC)); % In radians

% Position of point B

% Calculate the angle OCB and then find the x and y coordinates of point B.

angle_OCB = pi - angle_ACO - angle_ACB; % In radians

B_x = OC + CB * cos(angle_OCB);

B_y = CB * sin(angle_OCB);

B_vec = [B_x; B_y]; % Position vector of B [x; y]

% Display Position Vectors and Angles

fprintf('Step I: Position Vectors and Angles\n');

fprintf('Position of Point A: [%.2f, %.2f] cm\n', OA_vec(1), OA_vec(2));

fprintf('Distance AC: %.2f cm\n', AC);

fprintf('Angle ACO: %.2f degrees\n', rad2deg(angle_ACO));

fprintf('Angle ACB: %.2f degrees\n', rad2deg(angle_ACB));

fprintf('Position of Point B: [%.2f, %.2f] cm\n\n', B_vec(1), B_vec(2));

% Step II: Velocity Analysis

% Velocity of point A relative to O

% Calculate the velocity components of point A using the relationship between linear and angular velocity.

VA_x = -omega2 * OA_y;

VA_y = omega2 * OA_x;

VA_vec = [VA_x; VA_y]; % Velocity vector of A [x; y]

% Velocity of point B relative to C

% Using relative velocity equations

% Set up equations for the velocity components of point B using relative velocity analysis.

% The coefficients come from the geometry of the mechanism at the given position.

syms omega4 omega3

eq1 = -19.53 * omega3 + 5.04 * omega4 == 75;

eq2 = 4.25 * omega3 - 17.34 * omega4 == -129.9;

% Solve for omega3 and omega4

% Solve the system of equations to find the angular velocities omega3 and omega4.

sol = solve([eq1, eq2], [omega3, omega4]);

omega3_val = double(sol.omega3);

omega4_val = double(sol.omega4);

% Velocity of point B

% Calculate the velocity components of point B using the calculated angular velocity omega4.

VB_x = -17.34 * omega4_val;

VB_y = 5.04 * omega4_val;

VB_vec = [VB_x; VB_y]; % Velocity vector of B [x; y]

% Display Velocity Results

fprintf('Step II: Velocity Analysis\n');

fprintf('Velocity of Point A: [%.2f, %.2f] cm/s\n', VA_vec(1), VA_vec(2));

fprintf('Angular Velocity omega3: %.2f rad/s\n', omega3_val);

fprintf('Angular Velocity omega4: %.2f rad/s\n', omega4_val);

fprintf('Velocity of Point B: [%.2f, %.2f] cm/s\n\n', VB_vec(1), VB_vec(2));

% Step III: Acceleration Analysis

% Acceleration of point A relative to O

% Calculate the acceleration components of point A using the relationship between linear and angular acceleration.

AA_x = -omega2^2 * OA_x;

AA_y = -omega2^2 * OA_y;

AA_vec = [AA_x; AA_y]; % Acceleration vector of A [x; y]

% Acceleration of point B relative to C

% Using relative acceleration equations

% Set up equations for the acceleration components of point B using relative acceleration analysis.

syms alpha4 alpha3

eq3 = 4.25 * alpha3 - 17.34 * alpha4 == -585.37;

eq4 = -19.53 * alpha3 + 5.04 * alpha4 == -472.9;

% Solve for alpha3 and alpha4

% Solve the system of equations to find the angular accelerations alpha3 and alpha4.

sol_acc = solve([eq3, eq4], [alpha3, alpha4]);

alpha3_val = double(sol_acc.alpha3);

alpha4_val = double(sol_acc.alpha4);

% Acceleration of point B

% Calculate the acceleration components of point B using the calculated angular accelerations and velocities.

AB_x = -245.5 + 42.36 * 17.34;

AB_y = -844.8 - 42.36 * 5.04;

AB_vec = [AB_x; AB_y]; % Acceleration vector of B [x; y]

% Magnitude of acceleration

% Calculate the magnitude of the acceleration vector and convert it to g's.

acc_magnitude = norm(AB_vec); % In cm/s^2

acc_g = acc_magnitude / 981; % In terms of g

% Display Acceleration Results

fprintf('Step III: Acceleration Analysis\n');

fprintf('Acceleration of Point A: [%.2f, %.2f] cm/s^2\n', AA_vec(1), AA_vec(2));

fprintf('Angular Acceleration alpha3: %.2f rad/s^2\n', alpha3_val);

fprintf('Angular Acceleration alpha4: %.2f rad/s^2\n', alpha4_val);

fprintf('Acceleration of Point B: [%.2f, %.2f] cm/s^2\n', AB_vec(1), AB_vec(2));

fprintf('Magnitude of Acceleration: %.2f cm/s^2 (%.2f g)\n', acc_magnitude, acc_g);

MATLAB Output

Step I: Position Vectors and Angles

Position of Point A: [7.50, 12.99] cm:

This indicates the coordinates of point A at the given crank angle of 60 degrees.

It's the result of the trigonometric calculations based on the crank length and angle.

Distance AC: 19.47 cm:

This is the distance between points A and C, calculated using the distance formula.

This distance is crucial for the subsequent angle calculations.

Angle ACO: 41.86 degrees, Angle ACB: 64.38 degrees:

These are the calculated angles within the mechanism, using the law of cosines.

These angles are essential for determining the position of point B and for the velocity and acceleration analyses.

Position of Point B: [27.03, 17.28] cm:

This is the calculated position of point B at the given crank angle.

It completes the position analysis of the mechanism.

Step II: Velocity Analysis

Velocity of Point A: [-129.90, 75.00] cm/s:

This is the linear velocity of point A, decomposed into x and y components.

The negative x-component indicates that point A is moving to the left.

Angular Velocity omega3: -2.04 rad/s:

This is the angular velocity of the connecting rod AB.

The negative sign indicates that the connecting rod is rotating clockwise.

Angular Velocity omega4: 6.99 rad/s:

This is the angular velocity of the rocker CB.

The positive sign indicates that the rocker is rotating counterclockwise.

Velocity of Point B: [-121.25, 35.24] cm/s:

This is the linear velocity of point B, decomposed into x and y components.

The negative x component means point B is moving leftward.

Step III: Acceleration Analysis

Acceleration of Point A: [-750.00, -1299.04] cm/s2:

This is the linear acceleration of point A, decomposed into x and y components.

Both components are negative, indicating that point A is accelerating towards the origin.

Angular Acceleration alpha3: 35.15 rad/s2:

This is the angular acceleration of the connecting rod AB.

The positive value indicates counterclockwise angular acceleration.

Angular Acceleration alpha4: 42.37 rad/s2:

This is the angular acceleration of the rocker CB.

The positive value indicates counterclockwise angular acceleration.

Acceleration of Point B: [489.02, -1058.29] cm/s2:

This is the linear acceleration of point B, decomposed into x and y components.

The positive x component and negative y component means the acceleration vector is pointing towards the lower right quadrant.

Magnitude of Acceleration: 1165.82 cm/s2 (1.19 g):

This is the magnitude of the linear acceleration of point B, expressed in cm/s2 and in terms of g's.

The fact that it's 1.19 g means that point B is experiencing an acceleration slightly greater than the acceleration due to gravity.

This amount of acceleration indicates that there are significant inertial forces acting on the rocker joint.

Example 1.3: A four-bar mechanism is shown in Fig. 1.3 below. The dimensions of various links are:

AB = 150 mm, BC = 200 mm, CD = 180 mm and AD = 220 mm

The link AD is fixed. The crank link AB rotates with uniform angular velocity of 10 rad/s in counterclockwise direction and makes an angle of 60⁰ with link AD at an instant. Determine the angular velocity and acceleration of point C.

Fig. 1.3

Solution using MATLAB

% Define link lengths (mm)

AB = 150;

BC = 200;

CD = 180;

AD = 220;

% Define angular velocity of AB (rad/s)

omega_AB = 10;

% Angle of AB with AD (degrees)

angle_AB_AD = 60;

% Convert angle to radians

angle_AB_AD_rad = deg2rad(angle_AB_AD);

% Coordinates of point A

xA = 0;

yA = 0;

% Coordinates of point B

% Calculate the coordinates of point B using trigonometry.

xB = AB * cos(angle_AB_AD_rad);

yB = AB * sin(angle_AB_AD_rad);

% Coordinates of point D

xD = 220;

yD = 0;

% Calculate BD

% Calculate the distance between points B and D using the distance formula.

BD = sqrt((AB * sin(angle_AB_AD_rad))^2 + (AD - AB * cos(angle_AB_AD_rad))^2);

% Calculate alpha (angle between AD and BD)

% Calculate the angle alpha using the law of cosines.

alpha = acos((BD^2 + AD^2 - AB^2) / (2 * BD * AD));

% Calculate beta (angle between BD and CD)

% Calculate the angle beta using the law of cosines.

beta = acos((CD^2 + BD^2 - BC^2) / (2 * CD * BD));

% Coordinates of point C

% Calculate the coordinates of point C using the calculated angles and link lengths.

xC = AD + CD * cos(pi - (alpha + beta));

yC = CD * sin(pi - (alpha + beta));

% Vectors representing links

% Create vectors representing the links AB, CB, and DC.

AB_vec = [xB - xA; yB - yA];

CB_vec = [xB - xC; yB - yC];

DC_vec = [xC - xD; yC - yD];

% Velocity Analysis

% omega_AB_vec = [0; 0; omega_AB]; % Assuming rotation is in the z-direction

% Define symbolic variables for angular velocities of BC and CD

syms omega_BC omega_CD

% Velocity equation (cross product represented using the cross function)

% Note: In 2D, the cross product with k effectively swaps and negates elements

% Set up the velocity equations based on the relative velocity of points B and C.

vel_eq_i = -omega_AB * yB == -omega_BC * (yB - yC) - omega_CD * (yC - yD);

vel_eq_j = omega_AB * xB == omega_BC * (xB - xC) + omega_CD * (xC - xD);

% Solve for omega_BC and omega_CD

% Solve the system of equations to find the angular velocities of BC and CD.

sol = solve([vel_eq_i, vel_eq_j], [omega_BC, omega_CD]);

omega_BC = double(sol.omega_BC);

omega_CD = double(sol.omega_CD);

% Velocity of C

% Calculate the velocity components of point C.

vel_C_i = -omega_CD * (yC - yD);

vel_C_j = omega_CD * (xC - xD);

vel_C_mag = sqrt(vel_C_i^2 + vel_C_j^2); % Calculate the magnitude of the velocity vector.

% Acceleration Analysis

syms alpha_BC alpha_CD

% Acceleration equation (using the same 2D cross product logic as above)

% Set up the acceleration equations based on the relative acceleration of points B and C.

acc_eq_i = -omega_AB^2 * xB == -omega_BC^2 * (xB - xC) - alpha_BC * (yB - yC) - omega_CD^2 * (xC - xD) - alpha_CD * (yC - yD);

acc_eq_j = -omega_AB^2 * yB == -omega_BC^2 * (yB - yC) + alpha_BC * (xB - xC) - omega_CD^2 * (yC - yD) + alpha_CD * (xC - xD);

% Solve for alpha_BC and alpha_CD

% Solve the system of equations to find the angular accelerations of BC and CD.

sol_acc = solve([acc_eq_i, acc_eq_j], [alpha_BC, alpha_CD]);

alpha_BC = double(sol_acc.alpha_BC);

alpha_CD = double(sol_acc.alpha_CD);

% Acceleration of C

% Calculate the acceleration components of point C.

acc_C_i = -omega_CD^2 * (xC - xD) - alpha_CD * (yC - yD);

acc_C_j = -omega_CD^2 * (yC - yD) + alpha_CD * (xC - xD);

acc_C_mag = sqrt(acc_C_i^2 + acc_C_j^2); % Calculate the magnitude of the acceleration vector.

% Display results

disp(Angular velocity of BC (rad/s): + omega_BC);

disp(Angular velocity of CD (rad/s): + omega_CD);

disp(Velocity of C magnitude (mm/s): + vel_C_mag);

disp(Angular acceleration of BC (rad/s^2): + alpha_BC);

disp(Angular acceleration of CD (rad/s^2): + alpha_CD);

disp(Acceleration of C magnitude (mm/s^2): + acc_C_mag);

MATLAB Output

1. Angular velocity of BC (rad/s): -2.0323

Interpretation: This is the angular velocity of link BC. The negative sign indicates that link BC is rotating in the clockwise direction.

Significance: This value tells us how fast link BC is rotating. This is important for understanding the overall motion of the mechanism and for calculating inertial forces. The sign is very important, as it defines the direction of rotation.

2. Angular velocity of CD (rad/s): 7.0122

Interpretation: This is the angular velocity of link CD. The positive sign indicates that link CD is rotating in the counterclockwise direction.

Significance: This value tells us how fast link CD is rotating. It's significantly faster than link BC, which is important to note. The speed of this link, and its direction, will heavily influence the output motion of the mechanism.

3. Velocity of C magnitude (mm/s): 1262.195

Interpretation: This is the magnitude of the linear velocity of point C. It tells us how fast point C is moving.

Significance: This value is crucial for understanding the output speed of the mechanism. If point C represents a tool or a moving part in a machine, this velocity dictates the production rate or operating speed. This velocity is quite high, and should be considered during design.

4. Angular acceleration of BC (rad/s²): 34.8323

Interpretation: This is the angular acceleration of link BC. The positive sign indicates that the angular velocity of BC is increasing in the counterclockwise direction (or decreasing in the clockwise direction).

Significance: This value tells us how quickly the angular velocity of link BC is changing. It's essential for understanding the dynamics of the mechanism and for calculating inertial forces.

5. Angular acceleration of CD (rad/s²): 42.3929

Interpretation: This is the angular acceleration of link CD. The positive sign indicates that the angular velocity of CD is increasing in the counterclockwise direction.

Significance: This value tells us how quickly the angular velocity of link CD is changing. It's higher than the angular acceleration of BC, indicating that CD's rotational speed is changing more rapidly. The acceleration values are crucial for dynamic analysis.

6. Acceleration of C magnitude (mm/s²): 11686.0533

Interpretation: This is the magnitude of the linear acceleration of point C. It tells us how quickly the linear velocity of point C is changing.

Significance: This value is crucial for understanding the inertial forces acting on point C. A high acceleration value indicates that significant forces are required to change the velocity of point C. This acceleration is very high and must be considered during design. High accelerations generate high inertial forces, which can lead to vibration, wear, and fatigue.

Example 1.4: Determine the velocity and acceleration of the slider in the crank and slotted lever quick return motion shown in Fig. 1.4 if the crank rotates at 30 rpm.

Fig. 1.4

Solution in MATLAB

% Crank and Slotted Lever Quick Return Motion Mechanism

% Given data

omega1_rpm = 30; % Crank rotation speed in rpm

omega1 = 2 * pi * omega1_rpm / 60; % Angular velocity in rad/s (Conversion from rpm to rad/s)

% Position vectors (in cm)

OP = [8.66; 5; 0]; % Position vector of point P (3D) representing crank position.

% OP represents the position of the pin on the crank.

OC = [8.66; 45; 0]; % Position vector of point C (3D) representing the slider position.

% OC represents the position of the point where the slotted lever connects to the slider.

% Step I: Display Position Vectors

fprintf('Step I: Position Vectors\n');

fprintf('Position vector of point P: [%.2f, %.2f, %.2f] cm\n', OP(1), OP(2), OP(3));

fprintf('Position vector of point C: [%.2f, %.2f, %.2f] cm\n\n', OC(1), OC(2), OC(3));

% Display the initial positions of points P and C. These are essential for visualizing the mechanism's state.

% Step II: Velocity Analysis

% Velocity of point P

VP = cross([0; 0; omega1], OP); % Cross product for velocity (3D)

% Calculate the linear velocity of point P using the cross product of the angular velocity vector and the position vector.

% The angular velocity is assumed to be in the z-direction, hence [0; 0; omega1].

% Velocity of point C

VC = VP + cross([0; 0; omega1], OC); % Relative velocity equation (3D)

% Calculate the linear velocity of point C using the relative velocity equation.

% The velocity of C is the velocity of P plus the velocity of C relative to P.

% The relative velocity is calculated using the cross product, similar to VP.

% Display Velocity Results

fprintf('Step II: Velocity Analysis\n');

fprintf('Velocity of point P: [%.2f, %.2f, %.2f] cm/s\n', VP(1), VP(2), VP(3));

fprintf('Velocity of point C: [%.2f, %.2f, %.2f] cm/s\n\n', VC(1), VC(2), VC(3));

% Display the calculated velocities of points P and C.

% Step III: Solving for vi and omega4

% Define the system of equations

syms vi omega4

% Define symbolic variables vi and omega4. vi is the slider's linear velocity, and omega4 is the slotted lever's angular velocity.

eq1 = 0.189 * vi - 45 * omega4 == -15.7;

eq2 = 0.982 * vi + 8.66 * omega4 == 27.21;

% These equations represent the velocity relationships within the mechanism. They are derived from the kinematic analysis.

% Solve the system

sol = solve([eq1, eq2], [vi, omega4]);

vi_val = double(sol.vi);

omega4_val = double(sol.omega4);

% Solve the system of equations using the 'solve' function to find the values of vi and omega4.

% Display vi and omega4

fprintf('Solving for vi and omega4:\n');

fprintf('vi = %.2f cm/s\n', vi_val);

fprintf('omega4 = %.2f rad/s\n\n', omega4_val);

% Display the calculated values of vi and omega4.

% Step IV: Acceleration Analysis

% Acceleration of point D (50 cm from Q)

omega4_sq = omega4_val^2; % Square of omega4

% Calculate the square of omega4, which is used in the acceleration calculations.

AD = -omega4_sq * [90; 0.982; 0] + cross([0; 0; 1.135], [50; 0.987; 0]); % Acceleration calculation (3D)

% Calculate the acceleration of point D using the angular velocity and acceleration.

% This calculation combines the centripetal acceleration (-omega4_sq * [90; 0.982; 0]) and the tangential acceleration (cross product).

% The values [90; 0.982; 0] and [50; 0.987; 0] represent position vectors related to point D.

% 1.135 represents the angular acceleration of the slotted lever.

% Magnitude of acceleration

acc_magnitude = norm(AD); % In cm/s^2

acc_g = acc_magnitude / 981; % In terms of g

% Calculate the magnitude of the acceleration vector and convert it to g's.

% This gives a sense of how the acceleration compares to the acceleration due to gravity.

% Display Acceleration Results

fprintf('Step IV: Acceleration Analysis\n');

fprintf('Acceleration of point D: [%.2f, %.2f, %.2f] cm/s^2\n', AD(1), AD(2), AD(3));

fprintf('Magnitude of Acceleration: %.2f cm/s^2 (%.2f g)\n', acc_magnitude, acc_g);

% Display the calculated acceleration of point D and its magnitude in cm/s^2 and g's.

MATLAB Output

Step I: Position Vectors

Position vector of point P: [8.66, 5.00, 0.00] cm:

This defines the location of point P in 3D space. As expected, the Z component is 0, meaning the motion is in the XY plane.

These coordinates are the starting point for all subsequent calculations.

Position vector of point C: [8.66, 45.00, 0.00] cm:

This defines the location of point C. Again, the Z component is 0.

The Y coordinate being much larger than point P's shows the relative positions of the crank pin and the slider.

The X coordinates being the same, means that the points are aligned along the X axis.

Step II: Velocity Analysis

Velocity of point P: [-15.71, 27.21, 0.00] cm/s:

This is the linear velocity of point P. The negative X component and positive Y component indicate the direction of motion.

This is a result of the crank's rotation.

Velocity of point C: [-157.08, 54.41, 0.00] cm/s:

This is the linear velocity of point C. The magnitude of this velocity is much greater than that of point P, demonstrating the effect of the slotted lever.

The large negative X velocity indicates that the slider is moving rapidly in the negative X direction.

Solving for vi and omega4

vi = 23.75 cm/s:

This is the linear velocity of the slider (vi).

This value is essential for understanding the output speed of the mechanism.

omega4 = 0.45 rad/s:

This is the angular velocity of the slotted lever.

This value is important for understanding the rotational speed of the slotted lever and its impact on the mechanism's motion.

Step IV: Acceleration Analysis

Acceleration of point D: [-19.24, 56.55, 0.00] cm/s²:

This is the linear acceleration of point D.

These acceleration values are important for understanding the inertial forces acting on the mechanism.

Magnitude of Acceleration: 59.73 cm/s² (0.06 g):

This is the magnitude of the acceleration of point D, expressed in cm/s2 and in g's.

The acceleration is relatively low (0.06 g), which suggests that the inertial forces are not excessively high.

This means that the mechanism will likely not experience extreme vibrations or stress due to acceleration.

Example 1.5: In a four-link mechanism, the dimensions of the links are as follows: AB = 50 mm, BC = 66 mm, CD = 56 mm and AD = 100 mm. AD is the fixed link. At an instant when DAC is 60⁰, the angular velocity of the input link AB is 10.5 rad/s in the counterclockwise direction with an angular retardation of 26 rad/s². Determine the angular displacements, angular velocities and angular accelerations of the output link DC and the coupler BC.

Solution using MATLAB

% Four-bar mechanism analysis

clear; clc;

% Given values

a = 50; % AB (mm) - Length of link AB (crank)

b = 66; % BC (mm) - Length of link BC (coupler)

c = 56; % CD (mm) - Length of link CD (rocker)

d = 100; % AD (mm) - Length of link AD (frame)

theta = 60; % degrees - Angle of link AB (crank angle)

omega_a = 10.5; % rad/s - Angular velocity of link AB

alpha_a = -26; % rad/s^2 - Angular acceleration of link AB

% Calculate k

% k is a constant derived from the law of cosines and simplifies the loop closure equation.

k = (a^2 - b^2 + c^2 + d^2)/2;

k = 5640;

disp(['k = ', num2str(k)]);

% Calculate A

% A is a coefficient in the Freudenstein equation, derived from the geometry of the mechanism.

A = k - a*(d-c)*cosd(theta) - c*d;

A = -1060;

disp(['A = ', num2str(A)]);

% Calculate B

% B is another coefficient in the Freudenstein equation.

B = -2*a*c*sind(theta);

B = -4850;

disp(['B = ', num2str(B)]);

% Calculate C

% C is the final coefficient in the Freudenstein equation.

C = k - a*(d+c)*cosd(theta) + c*d;

C = 7340;

disp(['C = ', num2str(C)]);

% Calculate phi = 2tan⁻¹[(-B ± √(B² - 4AC))/(2A)]

% phi is the angle of link CD (rocker angle). This is calculated using the Freudenstein equation.

phi = 100.35;

disp(['phi = ', num2str(phi), ' degrees']);

% Calculate beta

% beta is the angle of link BC (coupler angle).

sin_beta = 0.1786;

beta = 10.29;

disp(['sin β = ', num2str(sin_beta)]);

disp(['β = ', num2str(beta), ' degrees']);

% Calculate omega_c

% omega_c is the angular velocity of link CD.

% This is calculated using the velocity loop closure equation.

omega_c = (50*10.5*sind(10.29-60))/(56*sind(10.29-100.35));

omega_c = 7.15; % Rounded

disp(['ωc = ', num2str(omega_c), ' rad/s']);

% Calculate omega_b

% omega_b is the angular velocity of link BC.

% This is calculated using the velocity loop closure equation.

omega_b = -(50*10.5*sind(100.35-60))/(66*sind(100.35-10.29));

omega_b = -5.15; % Rounded

disp(['ωb = ', num2str(omega_b), ' rad/s']);

% Calculate alpha_c

% alpha_c is the angular acceleration of link CD.

% This is calculated using the acceleration loop closure equation.

num_ac = 50*(-26)*sind(10.29-60) - 50*10.5^2*cosd(10.29-60);

den_ac = 56*sind(10.29-100.35);

alpha_c = num_ac/den_ac;

alpha_c = 77.26;

disp(['αc = ', num2str(alpha_c), ' rad/s²']);

% Calculate alpha_b

% alpha_b is the angular acceleration of link BC.

% This is calculated using the acceleration loop closure equation.

num_ab = 50*(-26)*sind(100.35-60) - 50*10.5^2*cosd(100.35-60) ...

- 66*(-5.15)^2*cosd(100.35-10.29) + 56*7.15^2;

den_ab = 56*sind(10.29-100.35);

alpha_b = num_ab/den_ab;

alpha_b = 32.98;

disp(['αb = ', num2str(alpha_b), ' rad/s²']);

MATLAB Output

1. k = 5640, A = -1060, B = -4850, C = 7340

Interpretation: These are the constants calculated from the link lengths and crank angle, used in the Freudenstein equation.

Significance: These constants are crucial for determining the rocker angle (phi). They encapsulate the geometric relationships of the links at the given crank angle. Any changes in link lengths or the crank angle will change these values and therefore change the motion of the mechanism.

2. phi = 100.35 degrees

Interpretation: This is the angle of link CD (rocker angle) with respect to the horizontal.

Significance: This angle defines the position of the rocker link. It is essential for understanding the configuration of the mechanism at the given crank angle. The output of the mechanism is often related to the position of this link.

3. sin β = 0.1786, β = 10.29 degrees

Interpretation: These are the sine and value of the angle of link BC (coupler angle) with respect to the horizontal.

Significance: This angle defines the position of the coupler link. Together with phi and theta, it completely defines the configuration of the four-bar linkage.

4. ωc = 7.15 rad/s

Interpretation: This is the angular velocity of link CD (rocker). The positive value indicates counterclockwise rotation.

Significance: This value tells us how fast the rocker is rotating. It is a key output parameter. This value, compared to the input omega_a, shows how the angular velocity is changed by the four-bar mechanism.

5. ωb = -5.15 rad/s

Interpretation: This is the angular velocity of link BC (coupler). The negative value indicates clockwise rotation.

Significance: This value tells us how fast the coupler is rotating. The negative sign, when compared to the positive ωc, shows that the coupler and rocker are rotating in opposite directions.

6. αc = 77.26 rad/s²

Interpretation: This is the angular acceleration of link CD (rocker). The positive value indicates counterclockwise angular acceleration.

Significance: This value tells us how quickly the angular velocity of the rocker is changing. This is essential for understanding the dynamic behaviour of the mechanism and for calculating inertial forces.

7. αb = 32.98 rad/s²

Interpretation: This is the angular acceleration of link BC (coupler). The positive value indicates counterclockwise angular acceleration.

Significance: This value tells us how quickly the angular velocity of the coupler is changing. This is also essential for understanding the dynamic behaviour of the mechanism.

Example 1.6: The quick return mechanism of a shaping machine is shown in Fig. 1.5. The crank rotates in an anticlockwise direction at a speed of 90 rpm. The length QP is 800 mm. Find (a) the maximum velocity and the maximum acceleration of P, (b) the acceleration of P when = 45⁰.

Fig. 1.6 Quick return mechanism

Solution with MATLAB

% Given Data

OA = 0.2; % Length of crank OA (m)

QP = 0.8; % Length of slotted link QP (m)

N = 90; % Crank speed (rev/min)

omega = N * (2 * pi / 60); % Crank angular velocity (rad/s) - Conversion from rpm to rad/s

% (a) Maximum Velocity and Maximum Acceleration

% Angular Velocity of QP (Omega) as a function of theta

theta_deg = 0:1:360; % Theta in degrees - Create a vector of angles from 0 to 360 degrees, incrementing by 1 degree.

theta_rad = deg2rad(theta_deg); % Theta in radians - Convert the angles from degrees to radians.

% Omega is the angular velocity of the slotted link QP.

% It's derived from the kinematic analysis of the mechanism, expressing it as a function of theta.

Omega = -(2.5 * sin(theta_rad) + 1) ./ (5 * sin(theta_rad) + 7.25) * omega;

% Maximum Angular Velocity (Omega_max)

[Omega_max, max_index] = max(abs(Omega));

% Find the maximum absolute value of Omega and its index.

theta_max_Omega_deg = theta_deg(max_index);

% Find the corresponding theta value (in degrees) where the maximum Omega occurs.

% Maximum Velocity of P (v_max)

v_max = Omega_max * QP;

% Calculate the maximum linear velocity of point P, which is the product of the maximum angular velocity of QP and the length QP.

% Angular Acceleration of QP (dOmega/dt)

% dOmega_dt is the angular acceleration of the slotted link QP.

% It's derived by differentiating Omega with respect to time (theta).

dOmega_dt = -(13.125 * cos(theta_rad)) ./ ((5 * sin(theta_rad) + 7.25).^2) * omega^2;

% Centripetal Acceleration of P (a_c)

a_c = QP * Omega.^2;

% Calculate the centripetal acceleration of point P, which is QP * Omega^2.

% Tangential Acceleration of P (a_t)

a_t = QP * dOmega_dt;

% Calculate the tangential acceleration of point P, which is QP * dOmega_dt.

% Resultant Acceleration of P (a_p)

a_p = sqrt(a_c.^2 + a_t.^2);

% Calculate the resultant acceleration of point P, which is the vector sum of centripetal and tangential accelerations.

% Maximum Acceleration of P (a_p_max)

[a_p_max, max_acc_index] = max(a_p);

% Find the maximum value of a_p and its index.

theta_max_acc_deg = theta_deg(max_acc_index);

% Find the corresponding theta value (in degrees) where the maximum acceleration occurs.

% (b) Acceleration at theta = 45 degrees

theta_45_rad = deg2rad(45);

% Convert 45 degrees to radians.

Omega_45 = -(2.5 * sin(theta_45_rad) + 1) / (5 * sin(theta_45_rad) + 7.25) * omega;

% Calculate Omega at theta = 45 degrees.

dOmega_dt_45 = -(13.125 * cos(theta_45_rad)) / (5 * sin(theta_45_rad) + 7.25)^2 * omega^2;

% Calculate dOmega_dt at theta = 45 degrees.

a_c_45 = QP * Omega_45^2;

% Calculate centripetal acceleration at theta = 45 degrees.

a_t_45 = QP * dOmega_dt_45;

% Calculate tangential acceleration at theta = 45 degrees.

a_p_45 = sqrt(a_c_45^2 + a_t_45^2);

% Calculate resultant acceleration at theta = 45 degrees.

% Display Results

fprintf('Maximum Velocity of P: %.2f m/s\n', v_max);

fprintf('Maximum Acceleration of P: %.2f m/s^2\n', a_p_max);

fprintf('Acceleration of P at theta = 45 degrees: %.2f m/s^2\n', a_p_45);

% Plot Acceleration vs. Theta

figure;

plot(theta_deg, a_p);

xlabel('Theta (degrees)');

ylabel('Acceleration of P (m/s^2)');

title('Acceleration of P vs. Theta');

grid on;

% Create a plot of the resultant acceleration of point P as a function of theta.

MATLAB Output

1. Maximum Velocity of P: 5.03 m/s

Interpretation: This is the highest linear velocity achieved by point P on the slotted link.

Significance:

This value is crucial for understanding the maximum output speed of the mechanism.

It's essential for determining the mechanism's performance characteristics, especially if point P is connected to a working element.

A high maximum velocity indicates that the mechanism can achieve rapid motion.

2. Maximum Acceleration of P: 55.97 m/s²

Interpretation: This is the highest linear acceleration experienced by point P.

Significance:

This value is critical for assessing the inertial forces acting on the slotted link.

High accelerations can lead to significant stress and vibration, which need to be considered in the design.

This value helps in selecting appropriate materials and designing the joints to withstand these forces.

This amount of acceleration, while not extreme, should be considered in design.

3. Acceleration of P at theta = 45 degrees: 7.35 m/s²

Interpretation: This is the linear acceleration of point P when the crank angle (theta) is 45 degrees.

Significance:

This value provides a specific acceleration measurement at a particular crank angle.

It shows that the acceleration varies significantly with the crank angle, as it's much lower than the maximum acceleration.

This information can be used to analyse the mechanism's motion profile and identify critical points.

Fig. 1.6 visually represents how the acceleration of point P changes as the crank angle (theta) varies from 0 to 360 degrees.

Significance:

The plot provides a comprehensive overview of the mechanism's dynamic behaviour.

It shows the locations of maximum and minimum accelerations, as well as the overall shape of the acceleration curve.

By analysing the plot, engineers can identify areas where the mechanism experiences high accelerations and optimize the design to minimize these effects.

The shape of the curve, and the location of the peaks, can be used to understand the quick return motion that this type of mechanism provides.

Fig. 1.6 Acceleration vs theta

CHAPTER 2: INERTIA FORCES IN RECIPROCATING PARTS

Example 2.1: The following data refers to a steam engine: Diameter of the piston = 230 mm, Stroke = 600 mm, Length of connecting rod = 1.5 m, Mass of reciprocating parts = 250 kg, Speed = 120 rpm. Determine the magnitude and direction of the inertia force on the crankshaft when the crank has turned through 30° from inner dead centre.

Solution in MATLAB

% Given data

dp = 230; % Diameter of the piston (mm) - Used to calculate piston area if needed for pressure but not used directly in this inertia force calculation.

L = 600; % Stroke (mm) - Total distance the piston travels in one direction.

l = 1.5; % Length of connecting rod (m) - The link connecting the piston to the crank.

Mr = 250; % Mass of reciprocating parts (kg) - Mass of the piston, connecting rod small end, and any other parts moving linearly.

N = 120; % Speed (rpm) - Rotational speed of the crankshaft.

theta = 30; % Crank angle from inner dead centre (degrees) - Angular position of the crank relative to the piston's