Mechanics Of Machines PDF - Politeknik Port Dickson

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MECHANICAL ENGINEERING DEPARTMENT N DJJ5113 O KS IC D T R PO IK KN TE LI PO Razali bakri Azhar rostani Wan siti fatimah Unauthorized Besterfield: Qualitycopying, th ed.. Control, 8sharing or distribution of this copyrighted material © 2009 Pearson is strictlyUpper Education, prohibited. Saddle If you are interested to purchase this e-Book, please write to : [email protected] River, NJ 07458. All rights reserved  Politeknik Port Dickson http://www.polipd.edu.my N O KS IC MECHANICS OF MACHINES simple note and examples D T R PO IK KN TE LI PO Published by Politeknik Port Dickson, Km. 14 Jalan Pantai, 71050 Si Rusa, Port Dickson, Negeri Sembilan Darul Khusus, Malaysia. Copyright © 2018 by Politeknik Port Dickson. All right reserved. No part of this book may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, recording, or by any information storage and retrieval system, without written permission from the publisher.  Preface PREFACE This book is for students who are take Mechanics of Machines course in the Mechanical Engineering Program, Polytechnic Malaysia, Ministry of Higher Education. The book is published based on the curriculum for Course of Mechanics of Machines issued by the Curriculum Division, Department of Polytechnic Education (DPE), Ministry of Higher Education of Malaysia. Each chapter consists simple and conscise explanation, more easier for beginner engineering student to understand and get some information. Hopefully student can get some benefit about the theory of mechanic of machine and they can used as a basic knowledge when they involve in mechanical industries. The information ini this book suitable with real situation in mechanical engineering field. In study of engineering mechanics, student should learn how to construct and solve the problem by N using fundamental theory and mathematical method. The content of this book covers a whole range O of topics learned by the students. With the completion of the accompanying examples and exercises that include the answers. The examples contain from simple and introductory to intermediates KS problems to help the students gain the confidence and understanding for each chapter. Last semester examinations questions also included in this book. Hopefully students get the picture to solve the problem during final examination at the end of semester. IC With the publication of this book, students will be able to use it as one of their alternatives D references. Furthermore, this book is written based on the courses taken by the students. Therefore easier for students to adapt the information contained in this book based on the teaching and T learning process in the classroom. R PO IK KN TE LI PO i  Acknowledgement ACKNOWLEDGEMENT In the name of Allah, all praise and thanks to Allah due to time and good health during preparing this book. Peace and blessing to Prophet Muhammad SAW. Thanks to all those involved in the process of preparing this book. Hopefully the publication of this book will get benefit to all parties especially polytechnic student. N O KS IC D T R PO IK KN TE LI PO Team of Writers Lead Editor : Razali Bin Bakri (PPD) Chapter 1 & 2 : Azhar Bin Rostani (PPD) Chapter 3, 4 & 5 : Razali Bin Bakri (PPD) Chapter 6 : Wan Siti Fatimah Binti Wan Ab Rahman (PKS) ii  Table of Content TABLE OF CONTENT Preface i Acknowledgement ii List of Symbols v List of Formula vi Synopsis of Syllabus xiv General Introduction xv N O KS CHAPTER 1 : HOIST Introduction 1 IC Typical Cases for Hoisting System 2 Load Balancing System 19 Exercise D 40 T R CHAPTER 2 : SIMPLE HARMONIC MOTION PO Introduction 44 Displacement 45 IK Velocity 45 KN Acceleration 46 Amplitude 47 TE Periodic Time 47 Frequency 47 LI Linear Elastic Oscillation : Mass-Spring 57 PO Simple Pendulum 60 Law of Simple Pendulum 62 Compound Pendulum 63 Exercise 80 iii  Table of Content CHAPTER 3 : VELOCITY AND ACCELERATION DIAGRAM Introduction 69 Velocity Diagram 71 Tangential Velocity 71 Radial Velocity 76 Crank, Connecting Rod and Piston 77 Acceleration Diagram 91 Acceleration of links – Radial/Centripetal Acceleration 91 N Acceleration of links - Tangential Acceleration 92 O Exercise 112 KS CHAPTER 4 : FRICTION IC Introduction 116 Limiting Friction D 117 T Laws of Static Friction 117 R Laws of Dynamic Friction or Kinetic Friction 118 PO Laws of Solid Friction 118 Laws of Fluid Friction 118 IK Coefficient of Friction 118 Angle of Friction 119 KN Minimum Force on Horizontal Plane 128 Movement on An Inclined Plane 130 TE Efficiency 137 LI Exercise 155 PO CHAPTER 5 : BALANCING Introduction 157 Balancing of Several Masses Rotating in One Plane/Same Plane 157 Balancing of Several Masses Rotating in Different Plane 168 Exercise 189 iv  Table of Content CHAPTER 6 : BELT DRIVES Introduction 194 Open Belt Drives 196 Crossed Belt Drive 197 Length of An Open Belt drive 198 Length of An Crossed Belt drive 200 Velocity Ratio of Belt Drive 204 Power Transmitted by a Belt 206 N Ratio of Driving Tensions for Flat Belt Drive 207 O Centrifugal Tension 211 KS Exercise 229 References IC D T R PO IK KN TE LI PO v  List of Symbols LIST OF SYMBOLS a - Acceleration, m/s2 IG - Moment of Inertia, kg/m2 r - Radius, m α (alpha) - Angular Acceleration, rad/s2 M (m) - mass, kg T - Torque, Nm T - Tension, N P - Force, N ω (omega) - Angular Velocity, rad/s g - Gravitational acceleration, m/s2 µ - Coefficient of Friction N β (beta) - Angle, degree (0) O φ (phi) - Angle, degree (0) π (pi) - - KS θ (theta) - Angle, degree (0) ς (sigma) - Stress, N/m2 ρ (rho) - density, kg/m3 IC d - Diameter, m L (l) - Length, m F Fr - - Force, N Frictional Force, N D T p (P) - Power, watt, Pressure, Pa, bar t - Time, s R v - Velocity, m/s PO tp - Periodic Time, s n - Frequency, Hz W - Weight, N k - Stiffness of Spring, N/m IK δ (delta) - Displacement,elongation, m η (eta) - Efficiency, % KN kG - Radius of Gyration about an Axis Through of Gravity, m RN - Normal Reaction C - Constant TE d - Distance, m G - Center of Gravity h - Height, m LI t - Thickness, m A - Area, m2 PO vi  List of Formula LIST OF FORMULA CHAPTER 1 : HOIST 1.1 Velocity , v  u  at 1.2 Displacement , s  1 u  v  t 2 1 1.3 Diplacemen t , s  ut  at 2 1.4 Velocity , v 2  u 2  2as N 1.5 Angular velocity, 2  1  t O 1 1.6 Angular displacement,   (1   2 )t 2 KS 1 1.7 Angular displacement,   1t  t 2 2 IC Angular velocity,  2  1  2 2 2 1.8 1.9 Velocity, v  r a D T 1.10 Angular acceleration,   r R Work 1.11 Power  PO Time 1.12 Power  F  v IK 1.13 Power  T   KN output 1.14 Efficeincy,    100% input TE 1.15 Inertia Force  ma LI 1.16 Inertia Couple  I PO 1.17 Moment of Inertia, I  mk 2 vii  List of Formula CHAPTER 2 : SIMPLE HARMONIC MOTION 2.1 Displacement, x  rcosθ 2.2 Displacement, x  rcost 2.3 Velocity , v   r 2  x 2 2.4 Max velocity , vmax  r 2.5 Accelerati on, a   2 x N Max accelerati on, amax   2 r O 2.6 KS 2 2.7 Time Period , t p   IC x 2.8 Time Period , t p  2 a 2.9 Time Period for angular motion, t p  2   D T 1 2.10 Frequency , n  R tp PO  2.11 Frequency , n  2 1 a 2.12 Frequency , n  2 IK x KN 2.13 Max Force  mamax 2.14 Spring force in static condition , k  mg TE 2.15 Total Spring Force  k  kx LI m 2.16 Time Period for mass - spring , t p  2 k PO  2.17 Time Period for mass - spring, t p  2 g 1 k 2.18 Frequency for mass - spring, n  2 m 1 g 2.19 Frequency for mass - spring, n  2  2.20 Amplitude , A  L sin  viii  List of Formula L 2.21 Time Period for Simple Pendulum , t p  2 g 1 g 2.22 Frequency for Simple Pendulum , n  2 L 2.23 Angular Velocity for Simple Pendulum,     2   2 2.24 Max Angular Velocity for Simple Pendulum,  max   2.25 Angular Acceleration for Simple Pendulum,     2 N 2.26 Max Angular Acceleration for Simple Pendulum,  max   2 O (k G  h 2 2 2.27 Time Period for Compound Pendulum, t p  2 KS gh 1 gh 2.28 Oscillation for Compound Pendulum, n  IC 2 (k G  h 2 ) 2 D T R PO IK KN TE LI PO ix  List of Formula CHAPTER 3 : VELOCITY AND ACCELERATION DIAGRAM 3.1 Linear Velocity of Linkage AB, vba   (AB ) vba 3.2 Angular Velocity of Linkage AB,  ba  AB bp BP 3.3 Corresponding Point,p on Linkage BC,  bc BC Radial Acceleration for Linkage AB, (a R ) ab   ab ( AB ) 2 3.4 2 N v ab 3.5 Radial Acceleration for Linkage AB, (a R ) ab  AB O 3.6 Tangential Acceleration, (aT ) ab   ab ( AB ) KS IC D T R PO IK KN TE LI PO x  List of Formula CHAPTER 4 : FRICTION 4.1 Friction Force, Fr  RN 4.2 Angle of Friction, Tan   4.3 Minimum Force, Pmin  mg sin(   ) 4.4 Newton’s Second Law, F  ma required force without friction 4.5 Efficiency,    100% reqiured force with friction N O KS IC D T R PO IK KN TE LI PO xi  List of Formula CHAPTER 5 : BALANCING 5.1 Static Balance for One Plane,  mr  0 Dynamic Balance for Different Plane,  mr  0 2 5.2 5.3 Static Balance for Different Plane,  mrl  0 Dynamic Balance for Different Plane,  mrl  0 2 5.4 N O KS IC D T R PO IK KN TE LI PO xii  List of Formula CHAPTER 6 : BELT DRIVES r1  r2 6.1 Pulley Angle – Open Belt Drives, kos 1  L 6.2 Pulley Angle – Open Belt Drives, 1   2 6.3 Angle of Lap for Driver Pulley – Open Belt Drives, 1  2(  1 ) 6.4 Angle of Lap for Driven Pulley – Open Belt Drives,  2  2 2 N 6.5 Length Of Belt – Open Belt Drives, Length of Belt  r11  r2 2  2L sin 1 O r1  r2 6.6 Pulley Angle – Cross Belt Drives, kos 1  KS L 6.7 Angle of Lap for Pulley – Cross Belt Drives,  1   2  2  2 1 IC 6.8 Length Of Belt – Cross Belt Drives, Length of Belt  r1θ1  r2 θ 2  2Lsinα1 N 2 r1  D T 6.9 Velocity Ratio, N 1 r2 R T1  T2 PO 6.10 Initial Tension, T0  2 6.11 Velocity of Pulley, v  r IK 2Nr 6.12 Velocity of Pulley, v  KN 60 6.13 Torque  (T1  T2 )r TE 6.14 Power  (T1  T2 )v LI 2NT 6.15 Power  PO 60 T1 6.16 Ratio of Tension for Flat Belt,  e T2  1  6.17 Power Transmitted by Flat Belt, Power  T1 1  v  e   T1 6.18 Ratio of Tension for “V” Belt,  e sin  T2 xiii  List of Formula   Power Transmitted by “V” Belt, Power  T1 1  v 1 6.19   sin    e  6.20 Centrifugal Tension, Tc  mv 2 T1  Tc 6.21 Ratio of Tension for Flat Belt with Centrifugal Tension,  e T2  Tc T1  Tc  6.22 Ratio of Tension for “V” Belt with Centrifugal Tension,  e sin  T2  Tc N  1  6.23 Power Transmitted by Flat Belt with Centrifugal Tension, Power  (T1  Tc )1  v O  e  KS   Power Transmitted by “V” Belt with Centrifugal Tension, Power  (T1  Tc )1  v 1 6.24   sin     IC e 1 6.25 Centrifugal Tension when Power is Maximum, Tc  T1 6.26 Velocity when Power is Maximum, v  T1 3 D T 3M R PO IK KN TE LI PO xiv  Synopsis of Syllabus SYNOPSIS OF SYLLABUS MECHANICS OF MACHINES exposes the students with knowledge on basic techniques and concepts of mechanics of machines. This course also gives knowledge on how to create and use simple methods to solve problem in relation to hoists, friction, simple harmonic motion, velocity and acceleration diagram, friction and belt drives. PROGRAM LEARNING OUTCOMES (PLO) Upon completion of the programme, graduates should be able to: 1. Apply knowledge of mathematics, science and engineering fundamentals to well-defined N mechanical engineering procedures and practice. O 2. Analyse well-defined mechanical engineering with respect to operation & maintenance, including troubleshooting. KS 3. Conduct investigations & assist in the design of solutions for mechanical engineering system. 4. Apply appropriate techniques, resources, & engineering tools to well-defined mechanical engineering activities, with an awareness of the limitations. IC 5. Demonstrate awareness and consideration for societal, health, safety, legal and cultural issues and their consequent responsibilities. D 6. Communicate effectively with the engineering community & society at large. 7. Function effectively as an individual & as a member in diverse technical teams. T 8. Demonstrate an understanding of professional ethics, responsibilities and norms of engineering practices. R 9. Demonstrate an awareness of management, business practices & entrepreneurship. PO 10. Demonstrate an understanding of the impact of engineering practices, taking into account the needs for sustainable development. 11. Recognize the needs for professional development & to engage in independent & lifelong learning. IK KN COURSE LEARNING OUTCOMES (CLO) Upon completion of this course, students should be able to:- TE 1. Analyze problems related to the mechanics of machines and data from the experiments in relation to the theoretical aspects. LI 2. Organize appropriately experiments in groups according to the Standard Operating Procedures. 3. Demonstrate ability to work in team to complete assigned tasks during practical work sessions. PO xv  General Introduction GENERAL INTRODUCTION Mechanics can be defined as an area of science concerned with the behaviour of physical bodies when subjected to forces or displacements, and the subsequent effects of the bodies on their environment. During the early modern period, scientists such as Galileo, Kepler, and Newton laid the foundation for what is now known as classical mechanics. It is a branch of classical physics that deals with particles that are either at rest or are moving with velocities significantly less than the speed of light. It can also be defined as a branch of science which deals with the motion of and forces on objects. Mechanics always divided into three parts which are mechanics of rigid bodies, mechanics of deformable bodies and mechanics of fluids. The mechanics of rigid bodies is divided into statics and dynamics. It is dealing with bodies at rest and then with bodies in motion. The basic concepts used in mechanics are space, time, mass and force. N O Machines may be defined as that branch of Engineering-science, which deals with the study of relative motion between the various parts of a machine, and forces which act on them. The KS knowledge of this subject is very essential for an engineer in designing the various parts of a machine. It is a device which receives energy in some available form and utilises it to do some particular type of work. IC Machines employ power to achieve desired forces and movement (motion). A machine has a power D source and actuators that generate forces and movement, and a system of mechanisms that shape the actuator input to achieve a specific application of output forces and movement. Modern T machines often include computers and sensors that monitor performance and plan movement, and are called mechanical systems. R PO Theory of Machines may be sub-divided into the following four branches : i. Kinematics Deals with the relative motion between the various parts of the machines. IK ii. Dynamics Deals with the forces and their effects, while acting upon the machine parts in motion. KN iii. Kinetics Deals with the inertia forces which arise from the combined effect of the mass and motion of the machine parts. TE iv. Statics Deals with the forces and their effects while the machine parts are at rest. The mass of the parts is assumed to be negligible. LI Mechanics of Machines always used Newton’s Three Fundametal Laws to solve the problem. These PO law can be stated as follows : First Law. If the resultant force acting on a particle is zero, the particle will remain at rest (if originally at rest) or will move with constant speed in a straight line (if originally in motion) Second Law. If the resultant force acting on a particle is not zero, the particle will have an acceleration proportional to the magnitude of the resultant and in the direction of this resultant force. This law can be stated as F=ma Third law. The forces of action and reaction between bodies in contact have the same magnitude, same line of action and opposite sense. xvi  General Introduction Fundamental Quantity and Units The measurement of physical quantities is one of the most important operations in engineering. Every quantity is measured in terms of some arbitrary, but internationally accepted units, called fundamental units. International System of Units (SI Units) is the units used in this book. The base unit of length, mass and time and are called respectively meter (m), kilogram (kg) and the second (s). Length is the distance between two points or place, time is interval between the two events and mass is the quantity of matter contained by a body. All physical quantities, met within this subject, are expressed in terms of the following three fundamental quantities is shown in Table 1. Fundamental Quantities Fundamental Units N Length (L) Meter (m) O Time (t) Second (s) KS Mass (m) Kilogram (kg) IC Table 1 : Fundamental Quantity and Units D Another fundamental quantity and unit such as Temperature (C/K), Current (A), light intensity (cd) and amount of material (mol) is not discussed in this book. T Derived Quantity and Units R PO The derivative unit for derived quantity is derived from the relationship between the fundamental quantities. Table 2 shows the derived quantity and units. Derived Quantity Derived Units IK Area (A) m2 KN Volume (V) m3 Weight (W) Newton TE Density (ρ) kg/m3 Velocity (v) m/s LI Acceleration (a) m/s2 PO Force (F) Newton Work (w) Joule Energy (E) Joule Power (P) Watt Pressure (P) Pascal Angle rad Angular Acceleration rad/s2 Angular Velocity rad/s xvii  General Introduction Frequency Hz Impulse Ns Moment of Inertia Kgm2 Spring constant N/m Table 2 : Derived Quantity and Units Dimensions Dimensions is the definition or terms that used to describe the fundamental units. Symbol for quantity is : [quantity]. Table 3 shows the quantity and dimension used in engineering. N O Quantity Dimension Length L KS Time T Mass M Temperature K IC Table 3 : Dimensions D All physical relations must be dimensionally homogenous, which mean the dimensions of all T terms in an equation must be the same. Example of dimension application in equation : R i. Area  Length  Width PO  L L  L2 Therefore : IK Dimension for Area, [Area] = L2 Unit for area = m2 KN Displacement ii. Velocity  TE Time L  T LI  LT 1 PO Therefore : Dimension for velocity, [velocity] = LT-1 Unit for velocity = ms-1 xviii  General Introduction mass iii. Density  volume M  L3  ML3 Therefore : Dimension for density, [density] = ML3 Unit for density = kgm-3 N Multification Factor in SI Units O Multiplication Prefix Symbol KS Factor terra T 1012 109 IC giga G mega M 106 kilo k D 103 102 T hecto h R deka da 10 PO deci d 10-1 centi c 10-2 milli m 10-3 IK micro  10-6 KN nano n 10-9 pico p 10-12 TE femto f 10-15 ato a 10-18 LI Table 4 : Multification Factor PO xix  General Introduction Imperial and SI Unit In engineering field both unit is used during the calculation. So that engineering personnel, technicians and engineer need to convert from Imperial unit to SI units or vice versa. Table 5 Show the conversion factor for both units. Imperial Conversion Factor Quantity SI Unit Unit Imperial - SI SI - Imperial Length inches milimeter 1 in = 25.4 mm 1 m = 0.039 in foots centimeter 1 in = 2.54 cm 1 cm = 0.394 in miles meter 1 foots = 0.305 m 1 m = 3.28 foot N kilometer 1 miles = 1.61 km 1 km = 0.621 miles O Mass ounce gram 1 aun = 28.3 g 1 g = 0.0353 aun KS tael gram 1 tael = 37.8 g 1 g = 0.0265 tael pound kilogram 1 lb = 0.605 kg 1 kg = 1.65 lb IC ton ton 1 ton = 1.02 t 1 t = 0.984 ton Area inches2 cm2 D 1 in2 = 6.45 cm2 1 cm2 = 0.155 in2 T foots2 m2 1 ft2 = 0.0929 m2 1 m2 = 10.76 ft2 R miles2 km2 1 mil2 = 2.59 km2 1 km2 = 0.386 mil2 PO Table 5 : Imperial and SI Unit Some example is shown below how to convert from imperial unit to SI Unit. IK Example 1 KN Convert 80 km/h to m/s TE Solution : 1 km = 1000 m LI 1 hour = 3600 second PO km 1000m 1h 80km/h  80   h 1km 3600s = 22.22 m/s xx  General Introduction Example 2 Convert 0.282 Ib/in3 to kg/m3 Solution : 1 ib = 0.4536 kg 1 in = 25.4 mm = 0.0254 m 1 in3 = (0.0254)3 m3 ib 1 in 3 0.4536 kg 0.282 ib/in 3  0.282 3  3 3  in (0.0254) m 1 ib N = 7809.86 kg/m3 O KS IC D T R PO IK KN TE LI PO xxi  HOIST CHAPTER 1 : HOISTING Introduction  Simple hoisting machine  Linear and angular motion  Action force in hoisting motion : a) Inertia force (ma) b) Inertia couple (Iα) c) Torque drive (T) d) Torque brake (Tb) e) Friction couple (Tf) f) Torque couple (Pr) N O Inertia force (ma) KS  Action force from mass and acceleration  Act at linear motion  Opposite direction of linear acceleration IC Inertia couple (Iα)  D Action force from moment of inertia and angular acceleration T  Act at angular motion  R Opposite direction of angular acceleration PO Torque drive (T)  To drive pulley  Power transmitted from motor IK  Same direction with pulley motion KN Torque brake (Tb)  Opposite direction of Torque TE  Act when brake is apply Friction couple (Tf) LI  PO Friction between cable and pulley  Opposite direction of pulley motion Torque couple (Pr)  Torque that produced by the action of force from cable that rotate the pulley  Same direction with cable tension  Torque Couple = Force (P) x radius (r) 1  HOIST TYPICAL CASES FOR HOISTING SYSTEM 1. Load upward with acceleration I T r  Tf Angular Motion N P O Linear a Motion KS ma IC mg W here: a  Linear acceleration m / s 2 D T ma  Inertia Force N R   Angular Acceleration rad / s 2 I  Inertia Couple Nm PO T  Torque ( produceby pulley) Nm T f  FrictionCouple Nm P  Tension in cable N IK mg  W eightof mass N Pr  Torque Couple Nm KN P Linear motion: TE F y  0 P  ma  mg  0 LI P  ma  mg ma P  m( a  g ) PO mg Iα Angular motion: T r Fcw  Fccw Tf Iα  Pr  T f  T Pr 2  HOIST 2. Load upward with constant velocity (a=0) T r Angular Tf motion P N O KS Linear IC Motion mg P D T Linear motion: R F y  0 PO P  mg  0 P  mg IK mg KN T Angular motion: r TE Fcw  Fccw Tf Pr  T f  T LI PO Pr 3  HOIST 3. Load upward and stopped by torque brake and friction couple I(-α) Tb r -α Angular Motion Tf P N Linear O -a Motion KS m(-a) IC mg P Linear motion: D T F y  0 R P  [m( a )]  mg  0 PO P  ma  mg  0 m(-a) P  mg  ma P  m( g  a ) IK mg KN I(-α) Angular motion: TE Tb Fcw  Fccw r Tb  Pr  T f  [ I ( )]  0 Tf Tb  Pr  T f  I  0 LI Tb  Iα  Pr  T f PO Pr 4  HOIST 4. Load downward with acceleration Iα T α r Angular Tf Motion N O Linear a Motion KS ma IC mg P D T Linear motion: R F y  0 P  ma  mg  0 PO ma P  mg  ma P  m( g  a ) IK mg KN I Angular motion: TE T ΣFcw  ΣFccw r Pr  T  Iα  T f Tf LI T  Iα  T f  Pr PO Pr 5  HOIST 5. Load downward stopped by torque brake and friction couple I(-) Tb r - Angular Tf Motion N O m(-a) KS -a Linear IC Motion mg P D T Linear motion: R F y  0 PO P  [m(a )]  mg  0 m(-a) P  ma  mg  0 P  mg  ma IK P  m( g  a ) mg KN I(-α) Angular motion: TE Tb Fcw  Fccw r Tf Pr  Tb  T f  [ I ( )] LI Pr  Tb  T f  I Tb  Pr  Iα  T f PO Pr 6  HOIST 6. Load freely fall (T = 0) I r α Tf Angular Motion P r N a O KS Linear ma Motion IC mg D T P Linear motion: R F y  0 PO P  ma  mg  0 P  mg  ma ma P  m( g  a ) IK mg KN TE Iα Angular motion: r Fcw  Fccw LI Tb Pr  Iα  T f PO Pr 7  HOIST Example 1.1 : Final Examination Session June 2013 A hoisting system with a drum diameter of 0.65 m has a moment of inertia of 75 kgm2. A hoist was used to raise 1.2 tonnes lift with acceleration 1.2m/s2. Calculate the : i) Drum driven torque ii) Power output after the lift accelerated for 5 seconds from rest iii) Power required, if the lift moving upward with a uniform velocity in (ii), after it accelerates for 5 seconds from rest Solution Example 1.1 i) Driven Torque N O Free Body Diagram : Iα KS T r    0.65m IC Angular I  75kgm2 Motion D T P R a  1.2m/s 2 PO m  1200kg ma Linear IK mg Motion KN a α r 1.2 α  3.69 rad/s 2 TE 0.325 LI P ΣFy  0 PO P  ma  mg  0 P  ma  mg ma P  m(a  g) P  1200(1.2 9.81) mg P  13212N 8  HOIST I ΣFcw  ΣFccw T Iα  Pr  T r T  (75  3.69) (13212 0.325) T  4570.65Nm Pr ii) Power Given, a  1.2 m/s 2 , u  0 m/s, t  5 s N  v  u  at v  0  (1.2)(5) O v  6 m/s KS v 6    18.46 rad/s IC r 0.325 Power  Tω Power  4570. 65  18.46 D T Power  84374.2 Watt R PO iii) Power required for moving upward with uniform velocity IK T r Angular Motion KN TE P LI m  1200kg PO Linear Motion mg P ΣFy  0 P  mg  0 P  mg P  (1200)(9.81) P  11772 N mg 9  HOIST T ΣFcw  ΣFccw r Pr  T T  11772 0.325 T  3825.9 Nm Pr Given, v  6 m/s, ω  18.46 rad/s 2 Power  Tω Power  3825.9  18.46 N Power  70626.11 Watt O KS IC D T R PO IK KN TE LI PO 10  HOIST Example 1.2 Simple hoisting machine is used to lift up 5 ton of mass with 1.2 m/s2 acceleration. The pulley mass is 1.5 ton, diameter 1.8 m and radius of gyration is 630 mm. Find the torque to lift up the mass if the friction couple between cable and pulley is 1.9 kNm. What is the power to move the mass from rest until 5 second. Solution Example 1.2 Free Body Diagram : I N T O r α Angular   1.8m Tf KS Motion m p  1500kg k  0.63m IC P r D T a R PO ma Linear Motion mg IK I  mk 2 I  (1500)(0.63 2 ) KN I  595.35kgm 2 TE a α r 1.2 LI α 0.9 PO α  1.333 rad/s 2 P ΣFy  0 P  ma  mg  0 P  ma  mg ma P  m(a  g) P  5000(1.2 9.81) mg P  55050 N 11  HOIST Iα ΣFcw  ΣFccw T Iα  Pr  T f  T r Tf T  (595.35 1.333) (55050 0.9)  1900 T  52239 Nm Pr Given; a  1.2 m/s 2 , u  0 m/s, t  5 s  v  u  at v  0  (1.2)(5) v  6 m/s N O v 6 ω   6.667 rad/s KS r 0.9 Power  Tω IC  52239  6.667  348260 Watt D T R PO IK KN TE LI PO 12  HOIST Example 1.3 The mass pulley of hoisting machine is 950 kg and the diameter is 0.8 m. One mass with 3.7 ton is release from rest and it take 2 second to move along 4 m. The friction couple between cable and pulley is 3.2 kNm. Find the radius of gyration for this pulley. Solution Example 1.3 Free Body Diagram : I N O r α Tf Angular Motion KS IC P r a D T R Linear ma PO Motion Given : mg mpulley = 950 kg φpulley = 0.8 m IK m = 3.7 ton t =2s KN s =4m Tf = 3.2 kNm U = 0 m/s TE Find a : LI 1 2 s  ut  at PO 2 1 4  (0  2)  a(2)2 2 a  2 m/s2 a α r 2 α 0.4 α  5 rad/s2 13  HOIST P ΣFy  0 P  ma  mg  0 P  mg  ma P  m (g  a) ma P  3700(9.81 2) P  28897 N mg Iα ΣFcw  ΣFccw Pr  Iα  T f N r Tb Pr  T f I O α (28897 0.4)  3200 KS I 5 I  1672 kgm2 Pr IC I  mk 2 k I D T m R 1672 k  1.33 m 950 PO IK KN TE LI PO 14  HOIST Example 1.4 A pulley of mass 550 kg of hoisting machine is used to lowered a mass of 200 kg with acceleration of 0.8 m/s2. The diameter of pulley is 950 mm and a radius of gyration of 450 mm. The friction couple during the operation is 1.8 kNm. Calculate, a) The cable tension b) The driven torque c) The power required when the load has velocity of 1.5 m/s. Solution Example 1.4 Free Body Diagram N Iα O T α KS r Angular Tf Motion IC D T Linear R a Motion PO ma IK mg Given : KN mpulley = 550 kg φpulley = 0.95 m, r = 0.475 m k = 0.45 m TE m = 200 kg a = 0.8 m/s2 Tf = 1800 Nm LI PO a α r 0.8 α  1.684 rad/s 2 0.475 I  mk 2 I  550  0.45 2 I  111.375 kgm2 15  HOIST a) Cable Tension P ΣF y  0 P  ma  mg  0 P  mg  ma P  m(g  a) ma P  200(9.81  0.475) P  1867 N mg b) Driven Torque N I O T ΣFcw  ΣFccw r KS Tf T  Pr  T f  Iα T  T f  Iα  Pr IC T  1800  (111.375  1.684)  (1867  0.475) T  1100.731 Nm Pr D T c) Power R PO Given, v  1.5 m/s v  IK r 1.5 KN   0.713 rad/s 0.475 TE Power  Tω  1100.731  0.713 LI  784.821 Watt PO 16  HOIST Example 1.5 A hoist drum has a moment of inertia of 115 kgm2 and is used to stop a mass of 100 kg that move downward with deceleration of 0.2 m/s2 by mean of light cable. If the drum diameter is 950 mm and neglect the friction torque, calculate the torque brake required to stop the mass. Solution Example 1.5 Free Body Diagram I(-) Tb - N r Angular O Motion KS IC m(-a) D T -a R Linear Motion PO Given : mg φpulley = 0.95 m; r = 0.475 m I =115 kgm2 m = 100 kg IK a = -0.2 m/s KN a α r TE  0.2 α  0.421 rad/s 2 0.475 LI PO P Linear motion : ΣFy  0 P  [m( a)]  mg  0 P  (100  0.2)  (100  9.81)  0 m(-a) P  20  981  0 P  1001 N mg 17  HOIST I(-α) Angular motion : Tb ΣFcw  ΣFccw r Pr  Tb  [I( α)] Tb  Pr  (I( α)] Tb  (1001  0.475)  (115  0.421) Tb  475.475  ( 48.415) Pr Tb  523.89 Nm N O KS IC D T R PO IK KN TE LI PO 18  HOIST LOAD BALANCING SYSTEM  In this hoisting system there is a balance mass used to balance the hoisting machine during the operation.  Balance mass is connected to a load with a cable and the acceleration both of it is the same TYPICAL CASES FOR HOISTING SYSTEM WITH BALANCING LOAD 1. Load upward with acceleration I T  Angular N r Tf Motion O KS P2 P1 IC a m2 m1 a D T m2a m1a R Linear Linear m2g m1g PO Motion Motion P1 IK Linear motion: F y  0 KN P1  m1 a  m1 g  0 m1a P1  m1 a  m1 g P1  m1 (a  g ) TE m1g P2 Linear motion: LI F y  0 PO P2  m 2 a  m 2 g  0 m2a P2  m 2 g  m 2 a P2  m 2 ( g  a ) m2g Iα Angular motion: T r ΣFcw  ΣFccw Tf Iα  P1 r  T f  T  P2 r T  Iα  P1 r  T f  P2 r P2r P1r 19  HOIST 2. Load upward with constant velocity (a=0) T r Angular Tf Motion P2 P1 Linear N Linear Motion Motion O m2 m1 KS m2g m1g IC P1 Linear motion: D T F y  0 R P1  m1 g  0 PO P1  m1 g m1g IK P2 KN Linear motion: F y  0 TE P2  m 2 g  0 P2  m 2 g LI PO m2g T r Angular motion: Tf ΣFcw  ΣFccw P1 r  T f  T  P2 r P2r P1r T  P1 r  T f  P2 r 20  HOIST 3. Load upward and stopped by torque brake and friction couple I(-) Tb - Angular r Tf Motion P2 P1 N -a -a O m2 m1 KS m2(-a) m1(-a) Linear Linear Motion m2g m1g Motion IC P1 Linear motion: D T F y  0 R P1  [m1 (a )]  m1 g  0 PO m1(-a) P1  m1 a  m1 g  0 P1  m1 g  m1 a m1g P1  m1 ( g  a ) IK KN P2 Linear motion: F y  0 P2  [m 2 (a )]  m 2 g  0 TE m2(-a) P2  m 2 a  m 2 g  0 P2  m 2 g  m 2 a LI m2g P2  m 2 ( g  a ) PO I(-α) Angular motion: Tb Fcw  Fccw r Tb  [ I ( )]  P1 r  T f  P2 r Tf Tb  I  P1 r  T f  P2 r Tb  P2 r  Iα  P1 r  T f P2r P1r 21  HOIST 4. Load downward with acceleration I T  Angular r Tf Motion P2 P1 N O a a m2 m1 KS m2a m1a Linear Linear IC Motion m2g m1g Motion P1 Linear motion: D T F y  0 R P1  m1 a  m1 g  0 PO m1a P1  m1 g  m1 a P1  m1 ( g  a ) m1g IK P2 Linear m otion: KN  Fy  0 P2  m2 a  m2 g  0 TE m2a P2  m2 g  m2 a P2  m2 ( g  a ) LI m2g PO Iα Angular motion: T ΣFcw  ΣFccw r Tf P1r  T  Iα  T f  P2 r T  Iα  T f  P2 r  P1r P2r P1r 22  HOIST 5. Load downward stopped by torque brake and friction couple I(-) Tb r - Angular Tf Motion P2 P1 N -a O -a m2 m1 KS m2(-a) m1(-a) Linear Linear IC Motion m2g m1g Motion P1 Linear motion: D T F y  0 R P1  [m1 (a )]  m1 g  0 P1  m1 a  m1 g  0 PO m1(-a) P1  m1 g  m1 a m1g P1  m1 ( g  a ) IK P2 Linear motion: KN F y  0 P2  [m 2 (a )]  m 2 g  0 TE m2(-a) P2  m 2 a  m 2 g  0 P2  m 2 g  m 2 a m2g P2  m 2 ( g  a ) LI PO I(-α) Angular motion: Tb Fcw  Fccw r Tf P1 r  Tb  T f  P2 r  [ I(- )] P1 r  Tb  T f  P2 r  I Tb  P1 r  Iα  T f  P2 r P2r P1r 23  HOIST 6. Load freely fall (T = 0) I r  Angular Tf Motion P2 P1 N a