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WORKSHOP CALCULATION, & SCIENCE, (As Per NSQF), 2nd Year, Common for All Engineering Trades under CTS, (For all 2 Year Trades), , DIRECTORATE GENERAL OF TRAINING, MINISTRY OF SKILL DEVELOPMENT & ENTREPRENEURSHIP, GOVERNMENT OF INDIA, , NATIONAL INSTRUCTIONAL, MEDIA INSTITUTE, CHENNAI, Post Box No. 3142, CTI Campus, Guindy, Chennai - 600 032, (i), , Copyright free, under CC BY Licence

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Workshop Calculation & Science (NSQF) - 2nd Year, Common for All Engineering Trades Under CTS, (For All 2 year Trades), , Copyright © 2018 National Instructional Media Institute, Chennai, First Edition : September 2019, , Copies :, , Rs. /-, , All rights reserved., No part of this publication can be reproduced or transmitted in any form or by any means, electronic or mechanical,, including photocopy, recording or any information storage and retrieval system, without permission in writing from the, National Instructional Media Institute, Chennai., , Published by:, NATIONAL INSTRUCTIONAL MEDIA INSTITUTE, P. B. No.3142, CTI Campus, Guindy Industrial Estate,, Guindy, Chennai - 600 032., Phone : 044 - 2250 0248, 2250 0657, 2250 2421, Fax : 91 - 44 - 2250 0791, email : [email protected], [email protected], Website: www.nimi.gov.in, (ii), , Copyright free, under CC BY Licence

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FOREWORD, , The Government of India has set an ambitious target of imparting skills to 30 crores people, one out of every, four Indians, by 2020 to help them secure jobs as part of the National Skills Development Policy. Industrial, Training Institutes (ITIs) play a vital role in this process especially in terms of providing skilled manpower., Keeping this in mind, and for providing the current industry relevant skill training to Trainees, ITI syllabus, has been recently updated with the help of comprising various stakeholder's viz. Industries, Entrepreneurs,, Academicians and representatives from ITIs., The National Instructional Media Institute (NIMI), Chennai, has now come up with instructional material to, suit the revised curriculum for Workshop Calculation & Science 2nd Year NSQF Commom for all 2 year, engineering trades under CTS will help the trainees to get an international equivalency standard where their, skill proficiency and competency will be duly recognized across the globe and this will also increase the, scope of recognition of prior learning. NSQF trainees will also get the opportunities to promote life long, learning and skill development. I have no doubt that with NSQF the trainers and trainees of ITIs, and all, stakeholders will derive maximum benefits from these IMPs and that NIMI's effort will go a long way in, improving the quality of Vocational training in the country., The Executive Director & Staff of NIMI and members of Media Development Committee deserve appreciation, for their contribution in bringing out this publication., Jai Hind, , RAJESH AGGARWAL, Director General/ Addl. Secretary, Ministry of Skill Development & Entrepreneurship,, Government of India., , New Delhi - 110 001, , (iii), , Copyright free, under CC BY Licence

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PREFACE, The National Instructional Media Institute(NIMI) was set up at Chennai, by the Directorate General of Training,, Ministry of skill Development and Entrepreneurship, Government of India, with the technical assistance, from the Govt of the Federal Republic of Germany with the prime objective of developing and disseminating, instructional Material for various trades as per prescribed syllabus and Craftsman Training Programme(CTS), under NSQF levels., The Instructional materials are developed and produced in the form of Instructional Media Packages (IMPs),, consisting of Trade Theory, Trade Practical, Test and Assignment Book, Instructor Guide, Wall charts,, Transparencies and other supportive materials. The above material will enable to achieve overall improvement, in the standard of training in ITIs., A national multi-skill programme called SKILL INDIA, was launched by the Government of India, through a, Gazette Notification from the Ministry of Finance (Dept of Economic Affairs), Govt of India, dated 27th, December 2013, with a view to create opportunities, space and scope for the development of talents of, Indian Youth, and to develop those sectors under Skill Development., The emphasis is to skill the Youth in such a manner to enable them to get employment and also improve, Entrepreneurship by providing training, support and guidance for all occupation that were of traditional, types. The training programme would be in the lines of International level, so that youths of our Country can, get employed within the Country or Overseas employment. The National Skill Qualification Framework, (NSQF), anchored at the National Skill Development Agency(NSDA), is a Nationally Integrated Education, and competency-based framework, to organize all qualifications according to a series of levels of Knowledge,, Skill and Aptitude. Under NSQF the learner can acquire the Certification for Competency needed at any, level through formal, non-formal or informal learning., The Workshop Calculation & Science 2nd Year (Comon for All 2 year Engineering Trades under CTS) is, one of the book developed by the core group members as per the NSQF syllabus., The Workshop Calculation & Science (Common for All 2 year Engineering Trades under CTS as per, NSQF) 2nd Year is the outcome of the collective efforts of experts from Field Institutes of DGT, Champion, ITI’s for each of the Sectors, and also Media Development Committee (MDC) members and Staff of NIMI., NIMI wishes that the above material will fulfill to satisfy the long needs of the trainees and instructors and, shall help the trainees for their Employability in Vocational Training., NIMI would like to take this opportunity to convey sincere thanks to all the Members and Media Development, Committee (MDC) members., , R. P. DHINGRA, EXECUTIVE DIRECTOR, , Chennai - 600 032, , (iv), , Copyright free, under CC BY Licence

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ACKNOWLEDGEMENT, The National Instructional Media Institute (NIMI) sincerely acknowledge with thanks the co-operation and, contribution of the following Media Developers to bring this IMP for the course Workshop Calculation & Science, (2nd Year) as per NSQF., , MEDIA DEVELOPMENT COMMITTEE MEMBERS, Shri. M. Sangara pandian, , -, , Training Officer (Retd.), CTI, Guindy, Chennai., , Shri. G. Sathiamoorthy, , -, , Jr.Training Officer (Retd.), Govt I.T.I, DET - Tamilnadu., , NIMI CO-ORDINATORS, Shri. Nirmalya nath, , -, , Deputy General Manager,, NIMI, Chennai - 32., , Shri. G. Michael Johny, , -, , Assistant Manager,, NIMI, Chennai - 32., , NIMI records its appreciation of the Data Entry, CAD, DTP Operators for their excellent and devoted services in, the process of development of this IMP., NIMI also acknowledges with thanks, the efforts rendered by all other staff who have contributed for the development of this book., , (v), , Copyright free, under CC BY Licence

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INTRODUCTION, The material has been divided into independent learning units, each consisting of a summary of the topic and an, assignment part. The summary explains in a clear and easily understandable fashion the essence of the mathematical, and scientific principles. This must not be treated as a replacment for the instructor’s explanatory information to be, imparted to the trainees in the classroom, which certainly will be more elaborate. The book should enable the, trainees in grasping the essentials from the elaboration made by the instructor and will help them to solve independently, the assignments of the respective chapters. It will also help them to solve the various problems, they may come, across on the shop floor while doing their practical exercises., The assignments are presented through ‘Graphics’ to ensure communications amongst the trainees. It also assists, the trainees to determine the right approach to solve the problems. The required relevent data to solve the problems, are provided adjacent to the graphics either by means of symbols or by means of words. The description of the, symbols indicated in the problems has its reference in the relevant summaries., At the end of the exercise wherever necessary assignments, problems are included for further practice., Duration:, 2nd Year, , Time allotment : 84 Hrs, , Time allotment for each module has been given below. Common for all 2 year Engineering Trades. Instructors are, here with informed to make use of the same., S.No, , Title, , Exercise No., , Time allotment (Hrs), , 2.1.01 - 2.1.03, , 14, , 2.2.04, , 6, , 1, , Friction, , 2, , Centre of Gravity, , 3, , Area of cut out regular surfaces, and area of irregular surfaces, , 2.3.05 - 2.3.07, , 16, , 4, , Algebra, , 2.4.08 & 2.4.09, , 12, , 5, , Elasticity, , 2.5.10 & 2.5.11, , 9, , 6, , Heat Treatment, , 2.6.12 & 2.6.13, , 3, , 7, , Profit and Loss, , 2.7.14 & 2.7.15, , 12, , 8, , Estimation and Costing, , 2.8.16 & 2.8.17, , 12, , LEARNING / ASSESSABLE OUTCOME, On completion of this book you shall be able to, • Demonstrate basic mathematical concept and principles to perform, practical operations., • Understand and explain basic science in the field of study including, simple machine., , (vi), , Copyright free, under CC BY Licence

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CONTENTS, Exercise No., , Title of the Exercise, , Page No., , Friction, Friction - Advantages and disadvantages, Laws of friction, co-efficient of, friction, angle of friction, simple problems related to friction, , 1, , 2.1.02, , Friction - Lubrication, , 8, , 2.1.03, , Friction - Co- efficient of friction, application and effects of friction in workshop, practice, , 12, , 2.1.01, , Centre of Gravity, 2.2.04, , Centre of gravity - Centre of gravity and its practical application, , 14, , Area of cut out regular surfaces and area of irregular surfaces, 2.3.05, , Area of cut out regular surfaces - circle, segment and sector of circle, , 24, , 2.3.06, , Related problems of area of cut out regular surfaces - circle, segment and, sector of circle, , 27, , Area of irregular surfaces and application related to shop problems, , 29, , 2.3.07, , Algebra, 2.4.08, , Algebra - Addition , subtraction, multiplication & division, , 32, , 2.4.09, , Algebra - Theory of indices, algebraic formula, related problems, , 36, , Elasticity, 2.5.10, 2.5.11, , Elasticity - Elastic, plastic materials, stress, strain and their units and, young’s modulus, , 42, , Elasticity - Ultimate stress and working stress, , 53, , Heat Treatment, 2.6.12, , Heat treatment and advantages, , 56, , 2.6.13, , Heat treatment - Different heat treatment process – Hardening, tempering,, annealing, normalising and case hardening, , 58, , Profit and Loss, 2.7.14, , Profit and loss - Simple problems on profit & loss, , 67, , 2.7.15, , Profit and loss - Simple and compound interest, , 73, , Estimation and Costing, 2.8.16, 2.8.17, , Estimation and costing - Simple estimation of the requirement of material etc.,, as applicable to the trade, , 85, , Estimation and costing - Problems on estimation and costing, , 92, , (vii), , Copyright free, under CC BY Licence

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SYLLABUS, 2nd Year, , Common for All Engineering Trades under CTS, (For All 2 year Trades), , S.no., I, , Duration: One Year, , Syllabus, Friction, , Time, , Marks, , 14, , 7, , 6, , 4, , 16, , 9, , 12, , 8, , 9, , 4, , 3, , 3, , 12, , 8, , 12, , 7, , 84, , 50, , 1 Advantages and disadvantages, Laws of friction, co- efficient of friction, angle of, friction, simple problems related to friction, 2 Friction – Lubrication, 3 Co- efficient of friction, application and effects of friction in workshop practice, II, , Centre of Gravity, 1 Centre of gravity and its practical application, , III, , Area of cut – out regular surfaces and area of irregular surfaces, 1 Area of cut – out regular surfaces – circle, segment and sector of circle, 2 Related problems of area of cut – out regular surfaces – circle, segment and, sector of circle, 3 Area of irregular surfaces and application related to shop problems, , IV, , Algebra, 1 Addition, Subtraction, Multiplication & Divisions, 2 Algebra – Theory of indices, Algebraic formula, related problems, , V, , Elasticity, 1 Elastic, plastic materials, stress, strain and their units and young’s modulus, 2 Ultimate stress and working stress, , VI, , Heat Treatment, 1 Heat treatment and advantages, 2 Different heat treatment process – Hardening, Tempering, Annealing, Normalising,, Case Hardening, , VII, , Profit and Loss, 1 Simple problems on profit & loss, 2 Simple and compound interest, , VIII, , Estimation and Costing, 1 Simple estimation of the requirement of material etc., as applicable to the trade, 2 Problems on estimation and costing, Total, , (viii), , Copyright free, under CC BY Licence

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Friction - Advantages and disadvantages, Laws of friction, co-efficient of, friction, angle of friction, simple problems related to friction Exercise 2.1.01, Introduction, When on a solid surface, another solid is rubbed a force is, created between the two solids which acts in the opposite, direction of motion or tries to obstruct the motion of the, object, this force is called frictional force. This phenomenon, is called friction. This happens due to roughness of the, two surfaces., In other words, It is the force of resistance offered to motion,, experienced by bodies which are in contact. It depends, upon the normal reaction between the contacting surfaces, and the nature of the surfaces. No surface is absolutely, friction less., , Forces acting on a body when a pulling force is, applied to move (Fig 2), •, , Weight of the block acting vertically downward (W), , •, , The normal reaction which acts upwards (R), , •, , The applied pulling force (F), , •, , The frictional force (Ff), , When the body is about to move W=R, F=P, When pulling force is increased the body starts to move., , Friction plays an important role in our daily life. It would, not be possible to walk without friction between our foot, and floor. Vehicles are able to run on roads because of the, friction between the wheels and road., , Laws of friction (Fig 3 & 4), , Types of friction, 1 Static friction, 2 Dynamic friction, , •, , Frictional force is directly proportional to the normal, reaction between contacting surfaces., , •, , Frictional force acts opposite to the direction of motion., , •, , Frictional force depends on the nature of contacting, surfaces., , •, , Frictional force is independent over the area and shape, of contacting surfaces., , 1 Static friction, The friction between two solid objects when at rest is called, static friction., Eg. Static friction can prevent an object from sliding down, on a sloped surface., Limiting friction, When the frictional force (F) is equal to the applied pulling, force (P) then the friction between two surfaces is known, as limiting friction. (i.e F=P), 2 Dynamic friction, It is the friction between two objects, when are in motion, is called dynamic friction. It is also called kinetic friction., Sliding friction, It is the friction experienced by an object when its slides, over another object. Sliding friction is always less than, limiting friction., Rolling friction, It is the friction that occurs when a circular object such as, a ball or roller rolls on a flat surface. Rolling friction is less, than sliding friction. (ball or roller bearing), , Coefficient of friction, It is a ratio between the frictional force to the normal, reaction when the body is just about to move but at, equilibrium. It is represented by symbol . (read as ‘meu’), , 1, , Copyright free, under CC BY Licence

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Therefore, , Co - efficient of friction =, =, , Limiting friction(or)force, normal reaction(or) weight, , F, F, (or ) f, W, R, , Angle of friction (Fig 5), The forces acting on a body when it is just about to move, by the application of a pulling force are W, R, P and F. The, forces 'R' and 'F' are compounded and we get the resultant, force 'S'. The angle formed by 'S' with 'R' is the angle of, friction., Therefore, ,  P = F + W sin and R = W cos ,  P = uR + W sin = u(W cos  W sin ,  P = W x tan  cos + W cos , = Wx, , F, W, tan , tan , , =W x, =W, , Sin φ, x cos θ + W cos θ, Cos φ, , Sin φ + cos θ + Wcos φ x cos θ, Cos φ, , [Sin (θ + φ)], Cos φ, , Similarly when the body is about to slide down the plane, the applied force P must be equal to, =W, , [Sin (θ + φ)], Cos φ, , To keep the body under equilibrium when the body is about, to move up the plane by the action of an applied force the, , Angle of repose (Fig 6), , applied force P = W, , [Sin (θ + φ)] and the force necessary, , Cos φ, to be applied to the body to prevent it from sliding down, the plane will be P = W, , [Sin (θ + φ)], , Cos φ, Note : Under all circumstances, =W, A body placed on an inclined surface remains at rest till, the angle of inclination equals the angle of friction. When, it exceeds the body starts sliding down. This is known as, angle of repose., Motion up the plane, When > a force must be applied to keep the body in, equilibrium. The applied force may be parallel to the plane,, horizontal or at an angle to the plane itself., When the body is at the point of motion up the plane the, frictional force ‘F’ acts down the plane., Forces acting are W,R,P and F. The weight force ‘W’ is, resolved into two components of W cos  perpendicular to, the plane acting downwards and W sin  acting parallel to, the plane downwards., F, = μ = tan φ, R, , Where is the angle of friction., 2, , [Sin (θ + φ)] < P < = W [Sin (θ + φ)], Cos φ, , Cos φ, , Advantages of friction, 1 Helps us to walk without slipping., 2 Used to stop vehicles when brakes are applied., 3 Movement of vehicles due to friction between revolving, wheels with tyres and the road., 4 Power transmission using gear drive or belt pulley drive., 5 Using friction we can sharp any object and also to hold, it., 6 Nails and screws are held in wood by friction., 7 Heat is produced when two rough surfaces are rubbed, against each other., Disadvantages of friction, 1 It causes wear and tear of the machine parts., 2 It produces heat and may cause melting of machine, parts. To avoid production of heat using of coolant is, necessary., , Workshop Calculation & Science : (NSQF) Exercise 2.1.01, , Copyright free, under CC BY Licence

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3 It reduces efficiency of a machine., , W = 14500 kg = Weight, , 4 It reduces speed of the moving object. eg. spindle, shaft,, piston etc., , F, , = ? = Force friction, , F, , μ=, , Friction can be reduced, , W, , 1 By using suitable lubricants (oil, grease) between the, moving parts., , 0.28 =, , 2 By polishing the surface to make them smooth., , F, , 14500, = 0.28 x 14500, , 3 By using ball bearings and roller bearings., , F, , = 4060 kg., , 4 By the use of wheel., Example, 1 A force of 40 kg is required to pull a weight of 400, kg on a horizontal plane. Determine the, coefficient of friction., , Coefficient of friction =, , Force, Weight, , =, , F, W, , But F = P and R = W, , 5 A force of 800 gram weight is needed to pull a, block weighing 3200 gram. What is the co-efficient, of friction., F, , W = 3200 gm = Weight, , , =, , 2 A force for 30N is required to move a body of mass, 35 kg on a flat surface horizontally at a constant, velocity. Find the coefficient of friction., Mass of the body = 35 kg. = W, The weight force=35 x 10 = 350 N, , (By taking, 1kg = 10N), , (By taking g = 10 meter/sec2), , W, , =, , Ff, R, , =, , 30, 350, , =, , 3, 35, , = 0.086, ,  = 0.24 = Co-efficient of friction, W = 20 kg = Weight, , 800, 3200, ,  = 0.25, 6 A force of 40 kg is required to move a mass of 80, kg on a flat surface horizontally at a constant, velocity. Calculate its co-efficient of friction?, F, , = 40 kg = Force, , , , =?, , Co-efficient of friction = μ =, =, , F, W, , 40, 80, ,  = 0.5, 7 A weight of 10 kg is resting on a horizontal table, and can just moved by a force of 2 kg. Find the, co-efficient of friction?, , F = ? = Force required, , F, W, F, , W = 10 kg = Weight, F, , = 2 kg = Force, , 20, , , , =?, , F = 20 x 0.24, , Co-efficient of friction = ?, , F = 4.8 kg, 4 A tanker with loaded weight of 14500 kg is running, on the road. If the co-efficient friction between, tyres and road surface is 0.28. Find out its force of, friction., , , W, , Co-efficient of friction = ?, , 3 A solid weights 20 kg. This is placed on a solid, surface. How much force does it require to come, in motion when co-efficient of friction is 0.24., , 0.24 =, , F, , W = 80 kg = Weight, , = 0.09, , μ=, , =?, , Co-efficient of friction = ?, , = 0.1, , ∴μ=, , = 800 gm = Force, , Co-efficient of friction = μ =, , 40, F Ff, , = =, 400, W R, , F, , F, , = 0.28 = Co-efficient of friction, , μ=, =, , F, W, 2, 10, ,  = 0.2, , Workshop Calculation & Science : (NSQF) Exercise 2.1.01, , Copyright free, under CC BY Licence, , 3

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8 A body weighing 100kg is resting on a table. Find, the co-efficient of friction if a force of 30 kg makes, its just to move?, W = 100 kg = Weight, F = 30 kg = Force,  =?, Co-efficient of friction = ?, , μ=, =, , , , F, , = ? = Force, , , , = tan  = 0.25, , , , tan-1 0.25, , 30, , F, W, , 100, F, , 0.25 =, , R =?, Normal reaction = W, Limiting friction = ?, =?, , R = Normal reaction = 10 kg, , 150, , F, , = 0.25 x 150, , F, , = 37.5 Kg., , 12 A body of mass 60kg rests on a horizontal plane., The value of co-efficient of friction between it and, the plane being 0.2. Find the work done in moving, the body through a distance of 5 meters along the, plane., , , = 0.2 = Co-efficient of friction, , W, , = 60 kg = Weight, , S, , = 5 m = Distance, , W = ? = Work done, , Limiting friction = F = 2.5 kg, , F, , μ=, , W, , 100, 2.5, , F, W, , 0.2 =, , F, 60, , 10, F, ,  = 0.25, , = 60 x 0.2, = 12 kg, , 10 A wooden block weights 100 kg. If the co-efficient, of friction is 0.3, find out force required to move, the block., W = 10 kg = Weight, , Work done = Force x distance = F x S, = 12 x 5, = 60 m - Kg., , , , = 0.3 = Co-efficient of friction, , (ie) Work done (or) Applied force = 60 m - Kg., , F, , = ? = Force, , 13 If a force of 30N is required to move a mass of, 35kg on a flat surface horizontally at constant, velocity, what will be the co-efficient of friction?, , μ=, 0.3 =, , F, W, F, , F, , = 100 x 0.3, , F, , = 30 kg, , F, , = 30 N = Force, , W, , = 35 kg = Weight, , 1 kg = 9.8 N, , 100, , 35 Kg = 9.8 x 35 = 343 N, Co-efficient of friction, , 11 Calculate the angle of inclination, if a weight of, 150 kg is in equilibrium, co-efficient of friction is, 0.25. Calculate the force of normal reaction also., 4, , , , μ=, , = 2.5 kg = Force, , =, , = 0.25 = Co-efficient of friction, , W, , W = 10 kg = Weight, , μ=, , , , = 14º 2’, , 9 A metal block weighing 10 kg rests on a horizontal, table. A horizontal force of 2.5 kg can just slide, the block. Find the normal reaction, limiting, friction and co-efficient of friction?, , , , = 150 kg = Work done, , F, ,  = 0.3, , F, , W, , =, , 30 N, 35 kg, , Workshop Calculation & Science : (NSQF) Exercise 2.1.01, , Copyright free, under CC BY Licence, , = μ=, , F, W

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=, , , = 2.2/cos 30º, , 30 N, 35 x 9.8 N, , = 2.2/0.8660, = 2.54 kg, , = 0.087, , 14 A block of ice weighing one quintal rests in, equilibrium on a wooden plank inclined at 30º., Find the coefficient of friction between the ice and, wood., W, , 17 Calculate the angle of inclination, if a weight of, 150 kg is in equilibrium. Coefficient of friction is, 0.25. Calculate the force of normal reaction also.,  = ? = angle of inclination, , = 1 quintal = 100 kg = Weight, , , , = 30º, , , , = tan tan 30º, , , , = 0.5774, ,  = 0.25, F = ? = Force, tan  = , , 15 Calculate the force that is required to slide a mass, of 980 kg on a guide, when the coefficient of, friction between the surfaces is 0.09., , tan  = 0.25,  = 14º 2'20", , μ=, , W, , = 980 kg = Weight, , , , = 0.09 = Co-efficient of friction, , F, , = Force = ?, = μ=, , F = 0.25 x 150 kg, , W, , F = 37.5 kg., 18 A body of mass 10 kg rests on a horizontal plane., The co-efficient of friction between the body and, plane is 0.15. Find the work done in moving the, body through a distance of 10 meter., , F = 0.09 x 980 kg, Required force(F) = 88.2 kg, , W = 10kg = Weight, , 16 A metal block weighing 10kg rests on a horizontal, board and the coefficient of friction between the, surfaces is 0.22. Find (a) the horizontal force which, will just move the block and (b) the force acting, at an angle of 30º with the horizontal, which will, just move the block., W, (a), , F, , W = Work done = ?, , = = 0.22, 0.15, , =?, , μ=, 0.22 =, F, , S = 10 meter = distance, , μ=, , (b) Force acting at an angle of 30º with the horizontal?, (a), ,  = 0.15 = Co-efficient of friction, , = 10 kg = Weight, , Co-efficient of friction, , W, , 150 Kg, , F, , F, 980 kg, , F, , F, , 0.25 =, , Co-efficient of friction, 0.09 =, , W = 150kg = Weight, , F,    tan , W, , =, , F, W, F, 10 Kg, , F = 0.15 x 10 kg, , F, , F = 1.5 kg, , W, , Work done = W = F x S, = 1.5 kg x 10 m, , F, 10 kg, , = 15 m - kg, , = 2.2 Kg., , (b) Force acting at an angle of 30º=, , F, Cos θ, , Workshop Calculation & Science : (NSQF) Exercise 2.1.01, , Copyright free, under CC BY Licence, , 5

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Assignment A, 1 A force 50N is required to move a mass of 40kg on a, flat surface horizontally at a constant velocity. Find the, coefficient of friction. (9.8N = 1kg), 2 A vehicle having a weight of 800kg is moving on the, road. If the coefficient of friction between the tyres and, road surface is 0.3, then calculate the force of friction., 3 A solid weighing 50kg is place on a solid surface. How, much force is required to move the block when, coefficient of friction is 0.25 between the block and the, surface., , 7 A body of mass 2000 kg moves a distance of 10 meters, in 5 sec. If the co-efficient of friction between the body, and floor is 0.3 find the horizontal force required to move, the body and horsepower absorbed against friction., 8 A vehicle is moving at 50kmph and the load on the, vehicle is 5000 kg. Find the H.P. required to move the, vehicle if = 0.2., 9 Find out the power lost due to friction by a planer under, the following conditions., Mass of the planer table = 3500 kg, , 4 A railway wagon weighs 1250 tonnes. If the coefficient, of friction between it and the rails is 0.003, find the, force required to move the wagon., 5 A body of mass 100kg rests on a horizontal plane. The, angle of friction between the body and the plane being, 0.025. Find the work done is moving the body through, a distance of 16m along the plane., , Rate of moment of the table=0.5 m/sec, Co-efficient friction between the table and the, ways=0.06, 10 A truck having weight 12000 kg is moving on the road., If the co-efficient of friction between the tyres and the, road surface is 0.3, then calculate the force of friction., , 6 A body of mass 20kg rests on a horizontal plane the, co-efficient of friction between the body and plane is, 0.3. Find the work done in moving the body through a, distance of 10 meters., , Assignment B, 1, , F = 1800 N, , F = 1.2 kN, , μ (static) = 0.16, , d = 60 mm, , μ (dynamic) = 0.012, , μ = 0.03, , FR to overcome, static friction, = ______ N, , Frictional torque MR, = ______ Nm, , FR to overcome, dynamic friction, = ______ N, 2, , 4, , (Frictional torque =, Frictional force x, radius), 5, , mass = 180 kg, , mass = 250 kg, , μ = 0.15, , FR = 160 N, , FR = ______ N, , μ = ______, 6, , FR = 120 N, μ = 0.032, , 3, , Normal force F, = ______ N, , F = 5000 N, μ (dry) = 0.03, μ (fluid friction), = 0.01, , 7, , FR when dry, = ______ N, FR when lubricated =, ______ N, 6, , Workshop Calculation & Science : (NSQF) Exercise 2.1.01, , Copyright free, under CC BY Licence, , m = 1000 kg, μ = 0.4, Force required to move, FR = ______ N

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C MCQ, 1 Which one of the following is useful friction, A Rings in cylinders, , B Crankshaft bearings, , C Brake shoe linings, , D Wheel hole bearings, , A 0.0215, , B 0.0152, , C 0.0125, , D 0.0251, , 7 Calculate the pulling force required for the figure shown., , 2 Which is in between the wheels and road, if vehicles, are able to run on roads., A erosion, , B motion, , C corrosion, , D friction, , 3 Which direction of motion frictional force acts., A equal, , B opposite, , C inclined, , D forward, , 4 What is the formula of angle of friction, if ‘F’ is the, frictional force, R is the normal reaction and q is the, angle of friction., A Tan q =, , F, R, , B Cot q =, , F, R, , C Sin q =, , F, R, , D Cos q =, , F, R, , B  F, R, , C F x R, , D F + R, , B 28 Kg, , C 29 Kg, , D 30 Kg, , 8 Determine the co-efficient of friction() between brass, and steel when a brass slider was placed on the, horizontal steel surface until it is just moving, if brass, slides (W) = 3 Kgf, Brass slides (W) = 3 Kgf, Force (F) required = 0.7 kgf, A 0.033, , B 0.133, , C 0.233, , D 0.333, , 9 Which is necessary to avoid production of heat., , 5 What is the formula for Co-efficient of friction (m)., A  R, F, , A 27 Kg, , A sand, , B coolant, , C lubricant, , D salt, , 10 Which is using for reduce the friction., , 6 A loaded truck weighs 2400 kg and it can be moved by, a force of 30 kg. Determine the co-efficient of rolling, friction, , A lubricants, , B sand, , C coal, , D coolant, , Key Answers, A, 1 0.1275, 2 240 Kg, 3 12.5 Kg, 4 3.75 Tonne, 5 40 m-kg, 6 60 m-kg, , 7 F = 600 Kg, , B, , C MCQ, 1 288N, 21.6 N, , 1 C, , 6, , C, , 2 0.065, , 2 D, , 7, , D, , 3 150 N, 50 N, , 3 B, , 8, , C, , 9 1.4 HP, , 4 36 N, 1.08 Nm, , 4 A, , 9, , B, , 10 3600 Kg, , 5 264.6 N, , 5 B, , 10 A, , P = 16 HP, 8 185.2 HP, , 6 3750 N, 7 3920 N, , Workshop Calculation & Science : (NSQF) Exercise 2.1.01, , Copyright free, under CC BY Licence, , 7

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Friction - Lubrication, , Exercise 2.1.02, , There are 3 systems of lubrication., •, , Gravity feed system, , •, , Force feed system, , •, , Splash feed system, , Gravity feed, The gravity feed principle is employed in oil holes, oil cups, and wick feed lubricators provided on the machines., (Figs 1 & 2), , Oil pump method, In this method an oil pump driven by the machine delivers, oil to the bearings continuously, and the oil afterwards, drains from the bearings to a sump from which it is drawn, by the pump again for lubrication., Splash lubrication, In this method a ring oiler is attached to the shaft and it, dips into the oil and a stream of lubricant continuously, splashes around the parts, as the shaft rotates. The rotation, of the shaft causes the ring to turn and the oil adhering to, it is brought up and fed into the bearing, and the oil is then, led back into the reservoir. (Fig 5) This is also known as, ring oiling., , Force feed/Pressure feed, Oil, grease gun and grease cups, The oil hole or grease point leading to each bearing is, fitted with a nipple, and by pressing the nose of the gun, against this, the lubricant is forced to the bearing. Greases, are also force fed using grease cup. (Fig 3), , In other systems one of the rotating elements comes in, contact with that of the oil level and splash the whole, system with lubricating oil while working. (Fig 6) Such, systems can be found in the headstock of a lathe machine, and oil engine cylinder., , Oil is also pressure fed by hand pump and a charge of oil, is delivered to each bearing at intervals once or twice a, day by operating a lever provided with some machines., (Fig 4) This is also known as shot lubricator., 8, , Copyright free, under CC BY Licence

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Types of grease guns, The following types of grease guns are used for lubricating, machines., •, , ‘T’ handle pressure gun (Fig 7), , •, , Automatic and hydraulic type pressure gun (Fig 8), , After cleaning the open gears, oil them and repeat, lubrication regularly. (Fig 13), , •, , Lever-type pressure gun (Fig 9), , Lubrication to exposed slideways, The moving parts experience some kind of resistance even, when the surface of the parts seems to be very smooth., The resistance is caused by irregularities which cannot, be detected by the naked eyes., Without a lubricant the irregularities grip each other as, shown in the diagram. (Fig 10), , With a lubricant the gap between the irregularities fills up, and a film of lubricant is formed in between the mating, components which eases the movement. (Fig 11), The slideways are lubricated frequently by an oilcan., (Fig 12), , Lubricate bearings, A shaft moving in a bearing is also subjected to frictional, resistance. The shaft rotates in a bush bearing or in ball/, roller bearing, experiencing friction., When the shaft is at rest on the bottom of the bush, bearing, there is hardly any lubricant between the shaft, and the bush. (Fig 14), , When the shaft starts rotating the lubricant maintains a, film between the shaft and the bush and an uneven ring of, lubricant builds up. (Fig 15), , Workshop Calculation & Science : (NSQF) Exercise 2.1.02, , Copyright free, under CC BY Licence, , 9

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When the shaft is rotating at full speed a full ring of, lubricating film surrounds the shaft (Fig 16) which is known, as hydro dynamic lubrication., , Functions, The functions of cutting fluids are:, This lubrication ring decreases the frictional resistance, very much and at the same time protects the mating, members against wear and changes., Some bush bearings have oil feeding holes over which the, oil or grease cup is mounted and the lubricant is fed through, the holes into the bearing by gravity feed system.(Fig 17), , -, , to cool the tool as well as the workpiece, , -, , to reduce the friction between the chip and the tool face, by lubricating, , -, , to prevent the chip from getting welded to the tool, cutting edge, , -, , to flush away the chips, , -, , to prevent corrosion of the work and the machine., , Advantages, As the cutting fluid cools the tool, the tool will retain its, hardness for a longer period; so the tool life is more., Because of the lubricating function, the friction is reduced, and the heat generated is less. A higher cutting speed can, be selected., , Hints for lubricating machines:, , As the coolant avoids the welding action of the chip to the, tool-cutting edge, the built up edge is not formed. The tool, is kept sharp and a good surface finish is obtained., , -, , identify the oiling and greasing points, , As the chips are flushed away, the cutting zone will be neat., , -, , select the right lubricants and lubricating devices, , -, , apply the lubricants., , The machine or job will not get rusted because the coolant, prevents corrosion., Properties of a good cutting fluid, , The manufacturer’s manual contains all the necessary, details for lubrication of parts in machine tools. Lubricants, are to be applied daily, weekly, monthly or at regular, intervals at different points or parts as stipulated in the, manufacturer’s manual., These places are indicated in the maintenance manuals, with symbols as shown in Fig 18., , A good cutting fluid should be sufficiently viscous., At cutting temperature, the coolant should not catch fire., It should have a low evaporation rate., It should not corrode the workpiece or machine., It must be stable and should not foam or fume., , Cutting Fluids, , It should not create any skin problems to the operator., , Cutting fluids and compounds are the substances used for, efficient cutting while cutting operations take place., , Should not give off bad smell or cause itching etc. which, are likely to irritate the operator, thus reducing his efficiency., Should be transparent., , 10, , Workshop Calculation & Science : (NSQF) Exercise 2.1.02, , Copyright free, under CC BY Licence

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Types of cutting fluids, , Compounded or blended oil, , The following are the common cutting fluids., •, , Straight mineral oil, , These oils are used in automatic lathes. These oils are, much cheaper and have more fluidity than fatty oil., , •, , Chemical solution (synthetic fluids), , Fatty oil, , •, , Compounded or blended oil, , •, , Fatty oils, , •, , Soluble oil (Emulsified oil-suds), , Lard oil and vegetable oil are fatty oils. They are used on, heavy duty machines with less cutting speed. They are, also used on bench-works for cutting threads by taps and, dies., , Straight mineral oil, , Soluble oil (Emulsified oil), , Straight mineral oils are the coolants which can be used, undiluted. Use of straight mineral oil as a coolant has the, following disadvantages., It has little effect as a cutting fluid., , Water is the cheapest coolant but it is not suitable, because it causes rust to ferrous metals. An oil called, soluble oil is added to water which gets a non-corrosive, effect with water in the ratio of about 1: 20. It dissolves in, water giving a white milky solution. Soluble oil is an oil, blend mixed with an emulsifier., , Hence straight mineral oils are poor coolants. But kerosene, which is a straight mineral oil is widely used as a coolant, for machining aluminium and its alloys., , Other ingredients are mixed with the oil to give better, protection against corrosion, and help in the prevention of, skin irritations., , Chemical solution (Synthetic oil), , Soluble oil is generally used as a cutting fluid for centre, lathes, drilling, milling and sawing., , It gives off a cloud of smoke., , These consist of carefully chosen chemicals in dilute, solution with water. They possess a good flushing and a, good cooling action, and are non-corrosive and nonclogging. Hence they are widely used for grinding and, sawing. They do not cause infection and skin trouble., They are artificially coloured., , Soft soap and caustic soda serve as emulsifying agents., A chart showing coolants for different metals is given, below., , Recommended cutting fluids for various metals and different operations, Material, , Drilling, , Reaming, , Threading, , Turning, , Milling, , Aluminium, , Soluble oil, Kerosene, Kerosene and, lard oil, , Soluble oil, Kerosene, Mineral oil, , Soluble oil, Kerosene, Lard oil, , Soluble oil, , Soluble oil, Lard oil, Mineral oil, Dry, , Brass, , Dry, soluble oil, Mineral oil, Lard oil, , Dry, soluble oil, , Soluble oil, Lard oil, , Soluble oil, , Dry, soluble oil, , Bronze, , Dry, soluble oil, Mineral oil, Lard oil, , Dry, soluble oil, Mineral oil, Lard oil, , Soluble oil, Lard oil, , Soluble oil, , Dry, soluble oil, Mineral oil, Lard oil, , Cast iron, , Dry, Air jet, Soluble oil, , Dry, soluble oil, Mineral lard oil, , Dry, sulphurized oil, Mineral lard oil, , Dry, soluble oil, , Dry, soluble oil, , Copper, , Dry, soluble oil, Mineral lard oil, Kerosene, , Soluble oil, Lard oil, , Soluble oil, Lard oil, , Soluble oil, , Dry, soluble oil, , Steel, alloys, , Soluble oil, Sulphurized oil, Mineral lard oil, , Soluble oil, Sulphurized oil, Mineral lard oil, , Sulphurized oil, Lard oil, , Soluble oil, , Soluble oil, Mineral, , General, purpose, steel, , Soluble oil, Sulphurized oil, Lard oil, Mineral lard oil, , Soluble oil, Sulphurized oil, Lard oil, , Sulphurized oil, Lard oil, , Soluble oil, , Soluble oil, Lard oil, , Workshop Calculation & Science : (NSQF) Exercise 2.1.02, , Copyright free, under CC BY Licence, , 11

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Friction - Co -efficient of friction, application and effects of friction in workshop, practice, Exercise 2.1.03, Co-efficient of friction, The ratio between the limiting frictional force and the normal, reaction is called co-efficient of friction., , 2 An empty drum weighing 50kg is resting on a shop, floor. Find the coefficient of friction if a force of, 15kg makes it just move., , Suppose, by applying a force P kg, the object is just fit to, move, then limiting frictional force will be produced in, between the two surfaces. The limiting frictional force will, be equal to external force applied and will work in the, opposite direction., , , W, , = 50 kg = Weight, , F, , = 15 kg = Force, , Co-efficient of friction, , F = P kg, , =, , According to the second law of limiting frictional force, the, frictional force will be proportional to normal reaction., F  R (sign is proportional to), F = R x constant, F, = constant, R, , or, , , , , , R, , or F   .R, , Co-efficient of friction =, , Limiting frictional force, Normal reaction, , = 1000 kg, , Distance (S), , = 5 meter, , Time (t), , = 5 second, , Co-efficient of friction () = 0.3, , ii, , Force (F), , =?, , Horse power (H.P.) = ?, , μ=, , 1 The sliding valve of a steam engine has dimensions, 25cm by 45 cm and the steam pressure on the back, of the valve is 25 kg/cm2. If the co-efficient of friction, is 0.13. Calculate the force required to move the, valve. Dimension of steam valve = 25 cm x 45 cm., , F, W, , 0.3 =, , = 25 kg/cm2, , F, 1000 Kg, , F, , = 0.3 x 1000 kg, , F, , = 300 kg, , Co-efficient of friction = 0.13, , H.P =, , ii, , Force required to move the valve = ?, F=?, Force of the steam, , cm2, , 4 A weight of 600 kg is kept on the inclined plane, at 300. Calculated the normal reaction and force, rolling downwards., , Force acts on the valve = 28125 kg, , F, , Solution:, , W, , 0.13 =, , FxS, 1, x, t, 75, , Horse power absorbed against friction = 4.H.P., ,  25cm  45cm = 28125 kg., , μ=, , (1 HP = 75 m.kg/sec), , 1, = 4 H.P, H.P = 300 x 5 x, 75, 5, , = Pressure x Area, = 25 x 25 x 45, , 25kg, , = 0.3, , Weight (W), , i, , Steam pressure, , W, , 15 kg, 50 kg, , i, , Co-efficient of friction is always constant for any two objects, and it has no unit., Example, , F, , 3 A machine crate weighing 1000kg moves distance, of 5m in 5 sec. If the coefficient of friction between, the crate and floor is 0.3, calculate the horizontal, force required to move the crate and horse power, absorbed against friction., , This constant between objects is called co-efficient of, friction. This is represented by ., F, , = μ=, , Weight kept on the inclined plane (W) = 600kg, , F, , Angle of the inclined plane () = 300, , 28125, F = 0.13 x 2812, , , , Normal reaction (R) = W . cos , , Force required to move the valves = 3656.25 Kg, 12, , Copyright free, under CC BY Licence

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= 600 x cos 300, = 600 (0.8660), = 519.6 kg, Force rolling downwards = W . sin , = 600 (0.5000), = 300 kg, Normal reaction, , = 519.6 kg, , Force rolling downwards, , = 300 kg, , 5 Find out the power lost due to friction by a planner, under the following conditions., Mass of the planer table, , = 3500 kg, , Rate of movement of the table, , = 0.5m/sec, , , , Co-efficient of friction between, the table and the ways, = 0.06, Solution:, Weight of planer (W), , = 3500 kg, , Distance moved (d), , = 0.5 m/sec, , Co-efficient of friction () = 0.06, Co-efficient of friction, , = μ=, , F, W, , F, 3500, , 0.06, , =, , F, , = 0.06 x 3500 = 210 kg, , Workdone, , = 1 H.P, , 105 kgm/sec, , =, , 105  1, = 1.4 H.P, 75, , Power lost due to friction = 104 H.P, , = 600 x sin 300, , , , 75 kgm/sec, , = F x distance moved, = 210 x 0.5 = 105 kgm/sec, , 6 A planner table weighting 800 kg moves a, distance of 2 metres in seconds on its bed. If coefficient of friction between bed and table is 0.30, find the power required to move the table aganist, the friction., 7 On a milling machine table a component of 20, kgf is clamped with the help of three equidistant, clamps. What force must be exerted by each, clamp to avoid slipping of the component when, the horizontal cutting force is 60 kgf and the, coefficient of friction is equal to 0.2., 8 A machine weight of 14500 kg moving on the, floor.If the co-efficient of friction between the, machine and floor surface is 0.28 then calculate, the force of friction., 9 A tail stock of a lathe has a mass of 21.5 kg and, co-efficient of friction at the slides is 0.122. What, horizontal force will be required to slide the tail, stock?, 10 An inclined surface makes an angle of 30 degrees, with the horizontal. An object weigting 5 tons is, placed on the surface. Find out the normal, reaction at the object and also the effective force, required to bring the object downwards., 11 A glass block of 400 grams has been placed on, the table. The glass is commuted by a string to a, 40 grams scale pan. The string passes over pulley., When a weight of 60 grams is placed on the scale, pan, the block starts sliding. Find out the coeffiecient of friction between wood and glass., , Workshop Calculation & Science : (NSQF) Exercise 2.1.03, , Copyright free, under CC BY Licence, , 13

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Centre of gravity - Centre of gravity and its practical application, Exercise 2.2.04, Any object comprises of a large number of particles. Each, particle is pulled towards the earth due to the force of, gravity. Thus, the forces on the particles are equal, parallel, and act in the same direction. These forces will have a, resultant which acts through a particular point ‘G’. This, fixed point ‘G’ is called the centre of gravity., , of a body. If the lamina is assumed to have uniform mass, per unit area, then the centroid is also the centre of gravity, in a uniform gravitational field., Methods to calculate centre of gravity, 1 By geometrical consideration., 2 By moments., Principle : The total moment of a weight about any axis, = The sum of the moments of the various parts about the, same axis., 3 By graphical method., The first two methods are generally used to find out the, centre of gravity or centroid, as the third method can, become tedious., Centre of gravity by geometrical consideration, , Concept of Centre of gravity, In physics, an imaginary point in a body of matter where,, for convenience in certain calculations, the total weight of, the body may be thought to be concentrated. The concept, is sometimes useful in designing static structures (e.g.,, buildings and bridges) or in predicting the behaviour of a, moving body when it is acted on by gravity., In a uniform gravitational field the centre of gravity is, identical to the centre of mass, a term preferred by, physicists., , Gravitation, The mutual attractive force of bodies due to which they, attract each other is called gravitation., 1 Gravity, The attractive force of the earth due to which it attracts all, bodies towards its centre is called gravity., The value of gravity varies from place to place on the ground, surface. Its general value is 9.81 m/s2., Centroid, Different geometrical shapes such as the circle, triangle, and rectangle are plane figures having only 2-dimensions., They are also known as laminas. They have only area, but, no mass. The centre of gravity of these plane figures is, called as the Centroid. It is also known as the geometrical, centre. The method of finding out the centroid of a plane, figure is the same as that of finding out the centre of gravity, 14, , 1 The centre of gravity of a circle is its centre., 2 The centre of gravity of a square, rectangle or a, parallelogram is at the points where its diagonals meet, each other. It is also the middle point of the length as, well as the width., 3 The centre of gravity of a triangle is at the point where, the medians of the triangle meet., 4 The centre of gravity of a right circular Cone is at a, distance of from its base., , Copyright free, under CC BY Licence

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5 The centre of gravity of a hemisphere is at a distance, of from its base., 6 The centre of gravity of a segment of a sphere of radius, h is at a perpendicular distance of from the centre of, the sphere., , more likely it is to tip over when it is tilted by a force. This, figure demonstrates a bus driving on two different grades;, the second one is steep enough to cause the centre of, gravity to fall outside of the base of the vehicle, which will, cause it to topple over., , 7 The centre of gravity of a semicircle is at a perpendicular, distance of from its centre., 8 The centre of gravity of a trapezium with parallel side, 'a' and 'b' is at a distance of measured from the base, 'b'., 9 The centre of gravity of a cube of side L is at a distance, of from every face., 10 The centre of gravity of a Sphere of diameter 'd' is at a, distance of from every point., , Equilibrium, , •, , Number of 2 pencil, , A body is said to be in equilibrium if the resultant of all the, forces acting on a body is zero and if there is no turning, moment., , •, , A fine edge like a ruler or a credit card, , There are three states of equilibrium (Fig 5), , •, , A permanent marker, , 1 Stable equilibrium, , •, , A ruler, , 2 Unstable equilibrium, , Centre of gravity; An experiment, , Step 1, , 3 Neutral equilibrium, , Attempt to balance the pencil on the edge you have, selected, Balancing the pencil may take some trial and error. The, point at which the pencil balances may not be where you, first thought. If it begins to tip in one direction, move the, pencil back slowly in the opposite direction until it will, stay there on its own., Step 2, Once the pencil is balanced, mark the location of the, balancing point with a permanent marker., Step 3, Measure the distance between the ends of the pencil and, the balancing point you have marked. Are the two lengths, equal? On my pencil, the length from the eraser to the, balancing point was actually 1.25 inches less than the, length from the pencil tip to the balancing point. Why, would this be the case?, In our experiment, the balancing point was another word, for the centre of gravity of this pencil. In other words, if we, cut the pencil in two at the mark we made in the experiment,, the two parts would be equal in weight. However, they are, not equal in length. As you may have already figured out,, the metal piece that houses the eraser contributes more, to the weight of the pencil, so the CG is closer to that side, of the pencil., Keeping up with that centre, The centre of gravity is an important concept in determining, the stability of a structure. It’s the reason why a good, homeowner will keep the top branches of his trees trimmed., It’s also the reason why a pick-up truck might not be the, best vehicle choice for a first time driver. Stability is, maximized in objects with a lower centre of gravity and a, wide base. The taller and more top-heavy an object, the, , 1 Stable equilibrium, A body is said to be in a stable equilibrium if it returns to, its original position when slightly displaced. (The C.G. is, as low as possible)., E.g :, , 1 A cone resting on its base, 2 A ball on a concave surface, 3 Funnel resting on its base. (Fig 6), , 2 Unstable equilibrium, A body is said to be in an unstable equilibrium if it does, not return to its original position when slightly displaced., Its centre of gravity falls taking it away from its original, position. (CG is at high points), E.g:, , 1 A cone resting on its tip, 2 A ball on convex surface, , Workshop Calculation & Science : (NSQF) Exercise 2.2.04, , Copyright free, under CC BY Licence, , 15

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3 Funnel standing on its tube end. (Fig 7), , Some example of equilibrium in daily life, 1 The lower decks of the ships are loaded with heavy, cargoes. This makes the centre of gravity of the whole, ship lower and its equilibrium becomes more stable., 2 A man carrying a bucket full of water in one hand extends, his opposite arm and bends his body towards it., , 3 Neutral equilibrium, A body is said to be in a neutral equilibrium if on being, slightly displaced, it takes a new position similar to its, original one. The centre of gravity remains undisturbed., (CG is neither raised or lowered), Eg:, , 1 A cone resting on its side, 2 A ball on flat surface, 3 Funnel resting on its side (Fig 8), , 3 While carrying load on back the man bends forward so, that his and the load’s centre of gravity falls on his, feet, if he walks erect, he will fall backward., 4 While climbing a mountain, a man bends forward and, bends backward while descending so that the centre, of gravity of his load falls on his feet., 5 In a double-decker, more passengers are, accommodated in the lower deck and less on the upper, so that the centre of gravity of the bus and the, passengers is kept low to eliminate any chance of, turning., Example, 1 Find the centroid of the isosceles triangular plate, as shown in the figure., , Model 1, Conditions for stable equilibrium, •, , The CG should be as low as possible., , •, , It should have a broad base., , •, , The vertical line passing through the CG should fall, within the base., , As per Pythagoras theorem, , Conditions of equilibrium, A body is said to be in a state of equilibrium under the, action of forces when there is no motion of rotation or, translation of the body. There are three conditions of, equilibrium of a body which are given below:, i, , Since BCD=45º then BD=DC=x, , Algebraic sum of the horizontal components of all the, forces acting on the body must be zero., H = 0, , ii Algebraic sum of the vertical components of all the, forces acting on the body must be zero., V = 0, iii Algebraic sum of the moments of all the forces acting, on the body must be zero., M = 0, Torque or twisting moment of a couple is given, by the product of force applied and the arm of, the couple (i.e. Radius). In fact, moment means, the product of “force applied” and the, “perpendicular distance of the point and the, line of the force”., , BD2 + DC2 = CB2, x2 + x2, , = 402, , 2x2 = 1600, x2, , =, , 1600, 2, , x, , 800 = 28.28 cm, , =, , Centroid from BD =, , = 800, , x, 3, , =, , 28.28, 3, , = 9.43 cm, , 2 A rectangular lamina has 10 cm and 8 cm. Find, the centroid. Centroid of rectangular = Diagonals, intersecting point., Centroid of rectangular = Diagonals intersecting point, Centre of AB =, Centre of AB =, , 10, =5, 2, , 8, , =4, , 2, Centroid lying 4 cm from AB and 5 cm from AD, , 16, , Workshop Calculation & Science : (NSQF) Exercise 2.2.04, , Copyright free, under CC BY Licence

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4 A thin lamina consists of an isosceles triangle of, height 120mm and base 100mm placed on a, semicircle of diameter 100mm. find the location, of its centre of gravity., , 3 A thin lamina is shown in the figure below. Find, the centre of gravity., Area of right angled triangle (a1) =, =, , 1, , bh, , 2, 1, , x 10 x 12, , 2, = 60 cm2, , 1, , Centroid of right angled triangle =, , h from base, , 3, 1, , =, , x 12, , 3, Centroid from E = 4 cm, , Centroid of rectangle, Area of rectangle= 5 x 10 = 50cm2, Area of triangle =, =, Total area, , 1, 2, , Centroid from A (h1), , = 12 - 4 = 8 cm, , Area of half circle (a2), , =, , bh, , 1, 2, , 1, , =5x, , Centroid distance from E to D, 1, , 3, , =, , 5, = 1.67 cm, 3, , (CG2) Centroid of plate is lying in between CG1 and CG2., From the figure torque is about AD., 62.5x, , = 50 x 5 + 12.5 x 11.67, = 395.875, , x =, , 395.875, , x 3.14 x 5 x 5, , =, , 4r, 3π, , =, , 4x5, 3 x 3.14, , = 2.123 cm, ⎛ Height of ⎞ ⎛ Centroid of, ⎟⎟ + ⎜⎜, (h2 ) = ⎜⎜, ⎝ triangle ⎠ ⎝ half circle, , ⎞, ⎟⎟, ⎠, , = 12 + 2.123, , = 250+145.875, 62.5x, , 2, , Centroid of semi circle = (Vertical distance from, centre of diagonal), , distance from its height., , 3, , 1, , = 39.25 cm2, , x 5 x 5 = 12.5 cm2, , The centre of gravity for rectangle is the point of intersection, of diagonal = 5 cm distance from AD (CG1), Centre of gravity for triangle is, ,  r2, , 2, =, , = 50 + 12.5 = 62.5 cm2, , 1, , = 14.123 cm, To find centroid of lamina, , = 6.334 cm, , 62.5, , y=, , Centre of gravity is 6.334 cm from AD, on the centre, axis., , a1 h1 + a2 h2, a1 + a2, , Workshop Calculation & Science : (NSQF) Exercise 2.2.04, , Copyright free, under CC BY Licence, , 17

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=, , 60 x 8 + 39.25 x 14.123, 60 + 39.25, , =, , 480 + 554.328, 99.25, , =, , = 54 KNm + 90KNm, = 144 KNm, B This moment is equal to moment of section A and, section B about 'O' (distance of action being D meter), That is =(36KN+20KN) x D(meter)=56 DKNm, , 1034.328, 99.25, , Again equating A and B, 144 KNm = 56 DKNm, , = 10.421 cm, Centroid is lying at 10.421 cm from point A, 5 A uniform rod weighing 50kg and 3m long carries, loads as shown below. Find out the distance of, the CG of the system from the left hand end., , 144 KNm, =D, 56 KNm, 144, 56, , =D, , Therefore D =, , 18, 7, , = 2.57 meters, The distance of CG of the shaft from left hand is 2.57, meters., , Distance of CG from A = x, Total weight = 50 + 15 + 20 + 25 = 110 kg, 110 x x = (50 x 1.5) + (15 x 1.75) + (20 x 2.25) + (25 x, 2.75), , 7 A thin lamina is shown in the figure. Find centre, of gravity., , = 75 + 26.25 + 45 + 68.75 = 215, Therefore x =, , 215, = 1.96 m, 110, , Distance of CG of the system from A = 1.95 m, 6 A long shaft is composed of two section A and B, each 3 meter long and weight 36KN and 20KN, respectively. Find out the position of centre of, gravity of the shaft., , Solution, Let G1 be the c.g. point of section A, , As the body is symmetrical about y-axis centre of, gravity lies on this axis., , Let G2 be the common c.g. of the shaft and its distance is, D from left hand end., , Let AB is the axis of reference, , Now, take moments about 'O', , Let y = The distance between centre of gravity and point, F, the point of reference as shown in the figure., , A Moment of section A about O = 36 KN x 1.5 m, Moment of section B about O, , = 20 KN x 4.5 m, , Adding both we get as below, Total moment about O=(36 KN x 1.5 m + (20KN x 4.5m), 18, , Let a1= Area of rectangle CDBA = 45 x 40 = 1800 mm2, h1 = Distance between centre of gravity of rectangle, of point F = 40 = 20 mm, 2, , Workshop Calculation & Science : (NSQF) Exercise 2.2.04, , Copyright free, under CC BY Licence

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Let a2 = Area of triangle ECD=1/2 x base x height, , Area of B1, , = 30 x 15 mm2, , =1/2 x 45 x 50 = 1125 square mm, h2 = distance between centre of gravity of triangle of, point F., , = 450 mm2, Distance of CG2 from AB, , = 20 +, , =1/3rd height of triangle +width of rectangle, =, , 1, , (50) + 40 =, , 3, , 170, 50, + 40 =, mm, 3, 3, , Applying formula, , y=, , = 20 + 15 mm, = 35 mm, Area of triangle =, , 1, , x 40 x 40 mm2, , 2, = 800 mm2, , a1 h1 + a2 h2, ,  PTR - Isosceles triangle, , a1 + a2, , Draw perpendicular line PS on TR from P., , ⎛ 170 ⎞, 1800(20) + 1125⎜, ⎟, ⎝ 3 ⎠, =, 1800 + 1125, 800, 5, 36000 + 63753.75, 99753.75, =, =, 2925, 2925, y, , 30, mm, 2, , PSR - right angled triangle, By applying Pythagoras theorem,, , x2 + x2 = 402, 2x2, , = 1600, 2, , x, , = 34.10 mm, , = 800, , x = 800, , The CG is at a distance of 34.1mm from point F, the point of reference in the line AB., 8 Find the CG of the lamina shown below., , = 28.28 mm, Distance of CG3 from TR = x = 28.28 mm = 9.43 mm, 3, 3, Dist. Of CG3 from AB = 20 + 30 + 9.43 mm = 59.43 mm, Total area =1200 + 450 + 800 mm2 = 2450 mm2, Distance from AB = Ymm, Taking moment at AB 2450 x y = 1200 x 10 + 450 x 35 +, 800 x 9.43, =12000 +15750 + 7544, =35294, y, , =, , 35294, = 14.41 mm, 2450, , Distance of CG is on the line PQ from side, AB = 14.41 mm., 9 A rectangular sheet of cardboard of uniform, thickness measuring 20 cm by 15 cm is divided, into four parts by drawing the diagonals. One of, the triangles formed on a 15 cm side is removed., Find the position of the C.G. of the remainder., CG is in PQ, CG1, CG2 and CG3 - centres of centre of gravity., Area of A1, , = 60 x 20 mm2, =1200 mm2, , Distance of CG1, from AB, , =, , 20, mm, 2, , = 10 mm, Workshop Calculation & Science : (NSQF) Exercise 2.2.04, , Copyright free, under CC BY Licence, , 19

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= r2h unit3, , ABCD - rectangle hard board, AC, BD - diagonals, O, - meeting point, AOD, - removed portion, , Volume of circle, ,  AOB = CG1,  BOC = CG2,  COD = CG3, , Remaining volume, , =  x 3 x 3 x 12 cm3, = 339.3 cm3, = 14400 - 339.8 cm3, = 14060.7 cm3, C.G is on PQ, , C.G, = Centre of gravity of the hard board, Take CG is on PQ, Area of  AOB, , = 1/2 bh unit2, = 1/2 x 20 x 7.5 cm2, = 75 cm2, , Area of BOC, , = 1/2 x 15 x 10 cm2, = 75 cm2, , Area of COD, , = 1/2 x 20 x 7.5 cm2, , = 75 cm, Total area, = 75 + 75 + 75 cm2, = 225 cm2, Taking moment at side BC, 2, , Distance of CG1 =, , 20, in = 10 cm, 2, , Distance of CG2 =, , 10, in = 3.33 cm, 3, , Distance of CG3 =, , 20, in = 10 cm, 2, , CG in before drilling = C.G2, CG in after drilling, , = C.G, , Calculating the moment on side AB, 100, = 50 cm, 2, , Distance of CG1, , =, , Distance of CG2, , = 10 cm, , Distance of C.G, , =x, , 14060.7 x x + 339.3 x 10 =14400 x 50, 14060.7x + 3393, , = 720000, , 14060.7x, , = 720000 - 3393, = 716607, , x, , =, , 716607, 14060.7, , = 50.97 cm, C.G. of the strip, , Distance of C.G = x, 225 x x, , CG in before drilling = C.G1, , = 75 x 10 + 75 x 3.33 + 75 x 10, = 750 + 249.75 + 750, = 1749.75, , = 50.97 cm from side AB., , 11 Centre of gravity in a lamina (Area), Find the position of c.g of the area shown in Fig11., (All dimensions are in mm.), , x = 1749.75 = = 7.777, 225, , = 7.777, Distance of CG is on the line PQ from side BC=7.78cm, 10 A steel rod 100x12x12cm has a hole dia 6cm, drilled in it as shown in the figure. Find the, position of the C.G. of the strip., , Solution, Taking moments of area about the line ab, we get the, equation as below., Moment of area abcd+moment of area efgh=moment of, area of complete figure., Volume of rod, , Now to calculate the areas, , = a2h unit3, = 12 x 12 x 100 cm3, = 14400 cm3, , 20, , 1 Area of abcd = 120 mm x 50 mm, = 6000 mm2, , Workshop Calculation & Science : (NSQF) Exercise 2.2.04, , Copyright free, under CC BY Licence

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Area of efgh = 200 x 75 mm2, = 15000 mm, Total area, , = (6000+15000) mm2, , (abcd+efgh) = 21000 mm2, 2 (6000 m2 x 25) + (15000 mm2 + 150 mm), , = 1000 cm3, Mass of copper cube, , = Volume x Density, = 1000 x 8.9, , = (21000mm2) x (x mm), , = 8900 gms, , 150000mm2 + 2250000mm2 = (21000 mm2) x (x mm), 240000mm2 = (21000 mm2) x (x mm), Therefore x =, , 21000 mm 2, , 8900, Kg, 1000, , =, , 2400000 mm 2, , = 8.9 Kg., (g=Acceleration due to gravity=10m/sec2), , 2400, mm, =, 21, =, , 10 6 mm 3, 10 3 mm 3, , =, , 2, , Weight of copper cube = 8.9 kg x 10 m/sec2, Similarly, , 800, mm, 7, , = 89 N, , 1000 cm3 x 8.5 x 10, 1000, 2, (Take g = 10m/sec ) = 8.500 x 10 = 85N, , Weight of brass cube =, , = 114.3 mm, Hence c.g. point of composite figure is 114.3 mm from, A on the line ab., 12 Centre of gravity point of a composite body can, be found out by using a variation of principle of, moments., Example (Fig 12), , Let cg of separate cubes be Gc and GB as shown, in figure., The distance between Gc and GB=100mm or, 0.1 m, Let c.g of the total object be at G which is 'P', meter to the right of GC or (0.1-P) meter to the, left of GB., Take moments about G, Clock moments, , = WB x (0.1 - P), = 85 x (0.1 - P)), = 8.5 - 85P, , Anti clock moments = WC x P, = 89 x P[Nm], = 89P, By principle of moments, Moment of part "A" about O+ Moment of part "B" about, O=Moment of (A+B) about O., The moment of the (A+B) acting through, point G., A copper cube of 100mm side is attached to brass cube, of 100 mm side, as sketched in the figure. Calculate the, position of c.g of composite object. Take densities of copper, and brass as 8.9 gms/cm3 and 8.5 gms/cm3., Solution, , 89P = 8.5 - 85P [Equating clock moments with anti-clock, moments], 89P + 85P = 8.5, 174P, P=, , = 8.5, , 8.5, meter or 0.049 m or 49 mm, 174, , Centre of gravity of the composite object lies, 49 mm from point of Gc. Hence it lies within, copper cube., , Volume of copper cube = 100 x 100 x 100 cm3, = 106 mm3, , Workshop Calculation & Science : (NSQF) Exercise 2.2.04, , Copyright free, under CC BY Licence, , 21

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Assignment A, 1 Find the position of centre of gravity of the figure shown., (All dimensions in mm), , 2 A lamina consists of a square of 60mm side, on one, side of which an equilateral triangle is constructed. Find, the position of centroid of the composite., , 3 A steel strip 50x12x12cm has a hole of 8cm dia. drilled, through it at a distance of 10cm from end. Find out the, c.g of the strip., , 4 Find out the C.G. of the four sided figure ABCD when, A = B=90º and the side AB=15cm, BC=12cm and, AD=5cm., , Assignment B, 1 What is the centre of gravity of a semi-circle is at a, perpendicular distance from its centre?, A, , 3r, 4π, , B, , 4r, 3π, , A, , h, 2, , B, , h, 3, , C, , 8r, 3, , D 3r, 8, , C, , h, 4, , D, , h, 5, , 2 What is the centre of gravity of a hemisphere is at a, distance from its space., , 5 Centre of gravity is usually located where., A more weight is concentrated, , A, , B 3r, 8, , B less weight is concentrated, , 4r, C, 8, , 5r, D, 8, , D more mass is concentrated, , C less mass is concentrated, , 3 What is the centre of gravity of a triangle is at the point, where the medians of the triangle meet?, , 22, , 4 What is the centre of gravity of a right circular cone is, at a distance from its base., , A, , h, 2, , B, , h, 3, , C, , h, 4, , D, , h, 5, , 6 Centre of gravity of an object depends on it's., A weight, , B mass, , C density, , D shape, , 7 Point where whole weight of body acts vertically is, called., A centre of mass, , B mid point, , C centre of gravity, , D none of above, , Workshop Calculation & Science : (NSQF) Exercise 2.2.04, , Copyright free, under CC BY Licence

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8 A simple method to find centre of gravity of a body is, usage of., , 10 Which of the following laminas do not have centroid at, its geometrical centre?, , A stop watch, , B plumbline, , A Circle, , B Equilateral triangle, , C pendulum, , D screw gauge, , C Right angled triangle, , D Isosceles triangle, , 9 If a material has no uniform density throughout the body,, then the position of centroid and centre of mass are., A identical, B not identical, C independent upon the density, D unpredictable, , Key Answers, A, , B, , 1 90.6 mm, , 1 B, , 2 44.3 mm, , 2 B, , 3 26.3 cm, , 3 B, , 4 The C.G. lies at a point at a distance of 4.49 cm, from, the line AB and at a distance of 6.47 from the line BC., , 4 C, 5 D, 6 B, 7 C, 8 B, 9 B, 10 C, , Workshop Calculation & Science : (NSQF) Exercise 2.2.04, , Copyright free, under CC BY Licence, , 23

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Area of cut out regular surfaces - circle, segment and sector of circle, Exercise 2.3.05, Circle (Fig 1), It is the path of a point which is always equal from its, centre is called a circle., r = radius of the circle, d = diametre of the circle, Area of the circle = r2, (or) =, , π, 4, , Area of the smaller segment, = Area of the sector - Area of  ABC, , d2 unit2, , Area of the greater segment, , Circumference of the circle = 2r (or) d unit, , = Area of the circle - Area of smaller segment, Semi Circle (Fig 4), •, , A semi circle is a sector whose central angle is 180°., , •, , Length of arc of semi circle, , Sector of a circle (Fig 2), The area bounded by an arc is called the sector of a circle., In the figure given ABC is the sector of a circle., , l = 2πr ×, , r = radius of the circle, = Angle of sector in degrees, , = r unit, , Area of sector ABC, =, , πr, , 2, , 360, , Area of sector =, , xθ, , Area of semi circle =, , unit2, , 0, , Length of arc of sector  radius, , Length of the arc  = 2r x, , 2, θ, 0, , 360, , 180, 1, = 2πr ×, 360, 2, , unit2, , r 2, 2, , unit2, , Perimeter of a semi circle, , ==, , 2πr, + 2r, 2, , = r + 2r, , unit, , = r ( + 2) unit, Quadrant of a circle (Fig 5), , Perimeter of the sector =  + 2r unit, , •, , r = radius, , A quadrant of a circle is a sector whose central angle, is 90°., , Segment of a circle (Fig 3), When a circle is divided into two by drawing a line, the, bigger part is called segment of the circle and the smaller, part is also called segment of the circle., 24, , Copyright free, under CC BY Licence

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•, ,  Circumference of circle = d, , Length and area of a quadrant of a circle, , l = 2πr ×, = 2πr ×, =, , 90, 360, ,  44 = d, , 1, 4, , d=, , πr, 2, , Area of quadrant of a circle =, , Perimeter of a quadrant =, , =, , πr, , 2 πr, 4, , 2, , 44, = 44 ÷ π, π, , = 44 ÷, , 22, 7, , = 44 ×, , 7, 22, , unit2, , 4, , = 14 cm, ,  Diameter of circle (d) = 14 cm, , + 2r, ,  Area of circle =, , πr, + 2r, 2, , 1, =π × d 2, 4, 22 1, = × × 14 ×14, 7 4, , ⎛ π +2 ⎞ unit, ⎟, ⎝2 ⎠, , =r ⎜, Examples :, , 1 Find the area of a sector of a circle whose radius, is 14 cm and the length of the arc of the sector is, 28 cm., Radius of sector r = 14 cm, , π 2, d unit2, 4, , = 154 cm2, Area of circle, , = 154 cm2, , Length of arc of sector = 28 cm, , 3 Find the remaining areas of circles of 10 cm dia, after inscribing triangles of 5 cm base and 10 cm, height., , Length of arc of sector () =, , Solution, , 28, , (i) Area of the circle, , =, , =, =, , , , =, , π 2, d, 4, , 22 × 10 × 10, 7×4, , = 114.550, , =, ,  Angle of sector  = 114.550, , 550, 7, , (ii) Area of the triangle inscribed in this circle, ,  Area of sector, , =, , =, =, , cm2, , = 196 cm, Area of sector, , 2, , = 196 cm2, , 2 If the circumference of a circle is 44 cm, find its, area. (Take π =, , 22, ), 7, , =, , Remaining area, , 1, 2, , × base × height, , 10 × 5, 2, , = 25 sq.cm, , =, , Remaining area of circle = 53, , 4, , Sq.cm, , 7, , Solution, ,  Let (d) = diamter of circle, , Workshop Calculation & Science : (NSQF) Exercise 2.3.05, , Copyright free, under CC BY Licence, , 25

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4 A rectangular sheet of metal measures 8 cm and, 6 cm. Four quadrants of circles each of radius 2 cm, are cut away at corners. Find the area of the, remaining portion., , = 48 - 12.57, = 35.428 cm2, = say 35.43 cm2, Area of remaining portion = 35.43 sq.cm, 5 Find the perimeter of the given circular disc., Sector :, , Area of rectangular sheet, , =8x6, , r, , = 100 mm, , , , = 360° - 45° = 315°, , l=, , = 48 cm2, There are four quadrants of a circle, each of radius 2 cm cut, away at the corners. Quadrant of circle means 1/4th of, circle., 4 quadrant of circles, , =4x, , =, , θ, 360, , 315, 360, , × 2π r unit, , × 2 × π × 100 mm, ,  = 550 mm, , of circle = 1 circle, , Area of 4 quadrant circles = Area of one circle, =  r2, =, , 22, 7, , × 2× 2, , Perimeter of the given circular Disc =  + 2r, , = 12.57 cm2, , = 550 + 200 = 750 mm, , Area of remaining portion =, Area of rectangular sheet - Area of four quadrant circles cut, at corners., , 26, , Perimeter of the given circular Disc = 750 mm, , Workshop Calculation & Science : (NSQF) Exercise 2.3.05, , Copyright free, under CC BY Licence

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Related problems of area of cut out regular surfaces - circle, segment and, sector of circle, Exercise 2.3.06, 1, , 2, , 7, d t = 21 mm, , d = 32 mm, , A t = ______ mm2, , Cross sectional area, = _____ mm2, , l, , = 750 mm, , b = 400 mm, , 8, , d = 180 mm, ,  = 60°, 9, s = 9.2 mm, , 4, , d = 580 mm, Angle of cut off, sector a = 135°, , A = Area of sector, = 140 mm2, , Area of the remaining portion, A, = _____ mm2, 10, , Equilateral triangle, of side a = 6 cm, ,  = ______°, , Radius of circle, = 1.732 cm, , d = 380 mm, , Shaded area, _______________, , No. of sectors of, equal area = 8, Area of each sector, = ______ mm2, , 11, , Two plugs having, diameters 2 cm and, 5 cm are placed on a, surface plate touching, each other. calculate, the distance ‘L’ in the, figure., , 12, , 90° vee block is 26 mm, wide at the top of the, vee block. What dia., of soft when laid in the, vee block will, have its top surface just, level with the top of the, vee block., ,  = ______ °, length of arc of each, sector = ______ mm, 6, , D = 880 mm, , A of sector, = _______ mm2, , d of the circle, = 30 mm, , 5, , Av (Area of shaded, part) = ______ mm2, Av = % of (Area of, rectangle) A1, , Area of sheet, = _______, 3, , D = 38 mm, ,  = 160°, A = 0.893 m2, d = ______ mm, , 27, , Copyright free, under CC BY Licence

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13, , 17, From a sheet of 5m , 3m how many circular, pieces of 12.5 cm dia, can be cut., , The arrangement of a, band saw blade is, shown in the figure given, below. Find out the, length of the saw blade., 18 Calculate the area covered by 3 equal circles of radius, 2.8 cm touches one another., , 14, Find out ‘L’ from the, given sketch., , 19, = 155°, d = 350 mm, b = ____mm, , 15, Find the value of ‘x’ in, the following fig., , 20, Find the area of shaded, portion., , 16, , 28, , Area of the shaded, portion, ______________mm2., , Workshop Calculation & Science : (NSQF) Exercise 2.3.06, , Copyright free, under CC BY Licence

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Area of irregular surfaces and application related to shop problems, Exercise 2.3.07, Area of irregular surface, , (b) Trapezoidal rule, , Surface area of irregular figures can be obtained by, applying either i) simpson’s rule or ii) trapezoidal rule. Area, found by simpson’s rule is more accurate than trapezoidal, rule. However accurate area can be obtained if the number, of ordinates are more i.e interval between ordinates is so, small as possible. (Fig 1), , A=, , h, [y1 + y8 + 2(y2 + y3 + y4 + y5 + y6 + Y7)] unit2, 3, , A=, , 5 [4 + 5 + 2(3 + 2 + 5 + 1 + 2 + 3)] m2, 2, , 5, x 41 m2, 2, = 102.5 m2, , A=, , Calculation of the area of an irregular surface, In this Calculation the area of an irregular surface may be, determined as follows., In this method of calculation a chain line known as base line, to be laid through the centre of the area of the surface., i, , Area as per simpson’s rule, , The offset are taken to the boundary points in the order of, their chainages on both the sides of the base line., , h, [y1 + y7 + 4(y2 + y4 + y6) + 2 (y3 + y5)], 3, , Area =, , The chain line and offsets are noted down., , h = interval between ordinates, , With reference to the notes the boundary points are plotted, and the area to be divided into number of triangles and, trapezium according to the shape., , ii Area as per trapezoidal rule, , Example, , where, , Area= [, , Now apply the geometrical formulae for calculating the, according to the shape of the figures. (Fig 2), , h, (first ordinate + last ordinate) + sum of, 2, remaining ordinate], , Calculate the area enclosed between the chain line,, the edge and the end offsets by, The offsets were taken from a chain line to a edge., Distance (M) 0, , 5, , 10, , 15, , 20, , 25, , 30, , 35, , Off set (M), , 3, , 2, , 5, , 1, , 2, , 3, , 5, , 4, , (a) Simpson’s rule, , (b) Trapezoidal rule, , (a) Simpson’s rule, A=, A=, , Chainline = AB, , h, [y1 + y8 + 4(y2 + y4 + y6) + 2 (y3 + y5 + Y7)] unit2, 3, 5, [4 + 5 + 4(3 + 5 + 2) + 2 (2 + 1 + 3)] m2, 3, , = 101.7 m2, , ., , Offsets, , = C,E, , 1 Area of triangle, ½ x base x height, 2 Area of trapezium, base (a  b), 2, ,  height, , 29, , Copyright free, under CC BY Licence

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Example, Plot the following details of a field and calculate its area all measurements are in metres (Fig 3), , Serial No. 1 In ABI, , Sl. No. 3, , Chainage in metres 0 and 20m., , Area of triangle KCD, , Offsets in metres 0 and 36m., In, ,  30  35, ,  30 ,  2, , 2, = 32.5m x 30m = 975 sq.m, , SI. No. 2, , Area of Trapezium LFGJ =, , Area of trapezium IBCK, Chainage in metres = 20m and 55m = 35m, Offsets in metres 36m and 20m = 28m, (a  b), 2, ,  36  20, ,  height = , , , 2, , Figure Chainline in, metres, , 1, , 2, , 1, , ABI, , 2, , Trapezium, , 3, , (a  b), ,  height = , , Sl. No. 6, , , , , Area of trapezium AJGH =, ,  35 , , 35  45, 2, ,  35 , , Base in, Metres, 4, , Offsets in, metres, , Mean offsets, in metres, , 5, , 6, , Area in square, Metres, +ve, -ve, 7, 8, , 0 and 20, , 20, , 0 and 36, , 18, , 360, , --, , 20 and 55, , 35, , 36 and 20, , 28, , 980, , --, , IBCK, 3, , KCD, , 55 and 100, , 45, , 0 and 20, , 10, , 450, , --, , 4, , Rectangle, , 100 and 75, , 25, , 0 and 30, , 15, , 750, , --, , 75 and 45, , 30, , 30 and 35, , 32.50, , 975, , --, , 45 and 0, , 45, , 45 and 35, , 40, , 1400, , --, , DEFL, Trapezium, , LFGJ, Trapezium, , JGHA, , 80,  35, 2, , = 40 x 35 = 1400 sq.m, , = 28 x 35 = 980 sq.m, , 30, ,  20  45, , Sl. No. 5 (LFGH), , =360 sq.m, , 6, , 2, , Area of rectangle DEFL = 25 x 30 = 750 sq.m, , =1/2 x 20 x 36, , 5, , 1, , Sl. No. 4, , Area = ½ x base x height, , S., No., , 2, , xbxh=, , = 45m x 10m = 450 Sq.m, ,  ABI, , =, , 1, , =, , Total, , 4915, , Workshop Calculation & Science : (NSQF) Exercise 2.3.07, , Copyright free, under CC BY Licence, , Remarks, , 9

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Assignment, Calculate the area of the irregular surfaces given below., , Note : All dimension are in mm., , Workshop Calculation & Science : (NSQF) Exercise 2.3.07, , Copyright free, under CC BY Licence, , 31

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Algebra - Addition , subtraction, multiplication & division, Exercise 2.4.08, Introduction, , – 72, , Algebra is a form of mathematics in which letters may be, used in place of unknown. In this mathematics numbers, are also used in addition to the letters and the value of, number depends upon its place. For example in 3x and x3,, the place of x is different. In 3x =3 is multiplied with x,, whereas in x3 - 3 is an Index of x., Positive and negative numbers, Positive numbers have a + sign in front of them, and, negative numbers have – sign in front of them. The same, applies to letters also., , +9, – 96, –6, , = –8, = +16, , When an expression contains addition,, subtraction, multiplication and division,, perform the multiplication and division, operations first and then do the addition and, subtraction., Example, 12 x 8 – 6 + 4 x 12 = 96 – 6 + 48 = 138, , Example + x, – y., , 102  6 – 6 x 2 + 3 = 17 – 12 + 3 = 8, , +8 or simply 8 positive number., – 8 negative number., , Parantheses and grouping symbols, , Addition and subtraction, , (, , ) Brackets, , Two positive numbers are added, by adding their absolute, magnitude and prefix the plus sign., , {, , } Braces, , To add two negative numbers, add their absolute magnitude and prefix the minus sign., , 6 x (8–5) = 6 x 3 = 18, , To add a positive and a negative number, obtain the, difference of their absolute magnitudes and prefix the sign, of the number having the greater magnitude., +7 + 22, –8 – 34, –27 + 19, 44 + (–18), 37 + (–52), , =, =, =, =, =, , +29, – 42, –8, +26, –15, , Multiplication of positive and negative numbers, The product of two numbers having like signs is positive and, the product of two numbers with unlike signs is negative., Note that, where both the numbers are negative, their, product is positive., Ex., , –20 x –3, 5x8, 4 x –13, –5 x 12, , =, =, =, =, , 60, 40, – 52, –60, , The number that is divided is the dividend, the number by, which we are dividing is the divisor and the answer is the, quotient. If the signs of the dividend and the divisor are the, same then the quotient will have a + sign. If they are unlike, then the quotient will have a negative sign., = +7, , +4, + 56, 32, , –4, , Parentheses, These are symbols that indicate that certain addition and, subtraction operations should precede multiplication and, division. They indicate that the operations within them, should be carried out completely before the remaining, operations are performed. After completing the grouping,, the symbols may be removed., In an expression where grouping symbols immediately, preceded or followed by a number but with the signs of, operation omitted, it is understood, that multiplication, should be performed., Grouping symbols are used when subtraction and multiplication of negative numbers is done., To remove grouping symbols which are preceded by, negative signs, the signs of all terms inside the grouping, symbols must be changed (from plus to minus and minus, to plus)., Parentheses which are preceded by a plus sign may be, removed without changing the signs of the terms within the, parentheses., , Division, , + 28, , 7 + (6–2) = 7 + 4 = 11, , When one set of grouping symbols is included within, another set, remove the innermost set first., When several terms connected by + or – signs contain a, common quantity, this common quantity may be placed in, front of a parentheses., 8 + 6(4–1) = 8 + 6 x 3 = 26, (6+2) (9–5) = 8 x 4 = 32, , = –14, , Plus 4 less negative 7 is written as 4 – (–7)., , Copyright free, under CC BY Licence

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Each member of an equation may be multiplied or divided, by the same number or symbol without changing its, equality., The equality of an equation is not altered when the numbers, or symbols are added or subtracted from both sides., Multiplication and division by the same numbers or symbols on both sides also will not affect the equality., Transposition of the terms of the equations, = equals to, + plus, – minus, , We must always perform the same operation on both sides, of the equation to keep the equilibrium. Add or subtract the, same amount from both sides. 5 + x = 9 By adding 3 on, both sides, the equation becomes 5 + x + 3 = 9 + 3 or x, + 8 = 12., 5 + x = 9 Subtract 5 from both sides then 5 + x – 5 = 9– 5., x = 4., 5 is transposed from left side to the right side by changing, its sign from + to –., x, x, = 20. Multiply both sides by 4. Then, x 4 = 20 x 4., 4, 4, x = 80,, , x multiply, , 5x = 25., ,  divided by, , Divide both sides by 5 then, , Concept of equality (Fig 1), , 5x, , 5, x = 5., , =, , 25, 5, , When transposing numbers or letter symbols from one, side to the other side multiplication becomes division and, the division becomes multiplication., The equality of an equation remains unchanged, when both sides of the equation are treated in, the same way. When transposing from one, side to the other side,, a plus quantity becomes minus quantity., a minus quantity becomes a plus quantity, a multiplication becomes a division, a division becomes a multiplication., To solve simple equations isolate the unknown, quantity which is to be found on the left side of, the equation., Example, •, An equation can be compared to a pair of scales which, always remain in equilibrium. The two sides of the equation, can fully be transposed. 9 = 5 + x may also be written as, 5 + x = 9., , Solve for x if 4x = 3(35 – x ), 4x = 105 – 3x (brackets removed), 4x + 3x = 105 (By transposing –3x on the right side, to the left side), 7x = 105, x = 15 (dividing both sides by 7), , 34, , Workshop Calculation & Science : (NSQF) Exercise 2.4.08, , Copyright free, under CC BY Licence

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Algebra – Theory of indices, Algebraic formula, related problems, Exercise 2.4.09, Calculations involving powers, , 2, , Power : Concept, , 2, , a.a.a... upto n times is = an, When a number, say 2 is multiplied by itself 4 times, we, write it as 24 (two to the power of 4) and it is equal to, 2 x 2 x 2 x 2 = 16., The exponent denotes how many times the base number, is multiplied by itself., Powers with a positive base have a positive result., Powers with a negative base and with an exponent that is, even will have a positive result., The sign, =, =, =, =, =, , 2, , =2, , 3–2, , 2x2x2, , a is the base, n is the exponent., , (+a) n, (– a) 2n, (2) 2, (– 2)2, (– 2)3, , 3, , an, a2n, 2 x 2 = 4 and, – 2 x – 2 = +4 but, –2x–2x–2=–8, , =, , 2x2, , 1, =2 =2, , 8, , =2, , 4, , Powers with the same exponents are divided by involving, the quotient of the bases by the common exponent., , n, , n, , 2, , 2, , ⎛a⎞, n =⎜ ⎟, b, ⎝b⎠, a, , ⎛2⎞ 2x2 4, =⎜ ⎟ =, =, 2, ⎝3⎠ 3x3 9, 3, 2, , Only like powers can be added or subtracted., Examples, (The exponent 1 is usually not written.), a1 = a, , Addition and subtraction of powers, , 21 =2, , Powers with the same base and exponents can be added, or subtracted by addition or subtraction of the coefficients., , 2a2 + 3a2 = 5a2, , x.an + y.an = an (x + y), , (Any number raised to the power of 0 is 1.), a0 = 1, , x.an – y.an = an (x – y), Ex .4x2 + x2 – 3x2 = x2 (4 + 1 – 3) = 2x2., , 20 = 1, A number raised to a negative power corresponds to its, reciprocal with the exponent's sign changed to +., , Multiplication, Powers with the same bases are multiplied by involving the, common base raised to the power of sum of the exponents., , a, , –n, , 1, = n, a, , am x an = am + n., 2 3 x 2 2 = 2 3 + 2 = 25, (2 x 2 x 2) x (2 x 2) = 2 x 2 x 2 x 2 x 2 = 25, , 1, 2–2 = 2, 2, , Powers are involved by multiplying the exponents., , 8 x 4 = 32., Powers with the same exponent of different base numbers, are multiplied by involving the product of the base numbers, raised to the common exponent., an x bn = (a x b)n, , (an)m = anm, (22) 3 = 22.3 = 26, Powers can be transposed without affecting the result., , 22 x 32 = (2 x 3)2, , a  = a , , 2 x 2 x 3 x 3 = 6 x 6 = 36, , (22) 3 = (23) 2, , n m, , m n, , Division, , (2 x 2) x (2 x 2) x (2 x 2) = (2 x 2 x 2) (2 x 2 x 2), , Powers with like bases are divided by involving the base, raised to the difference between the exponents., , 4 x 4 x 4 = 64, , a, , m, , 36 a, , n =a, , m–n, , 8 x 8 = 64, A mixed number raised to a power is first converted into an, improper fraction and then the result is evaluated., , Copyright free, under CC BY Licence

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v) a3 - b3, x -y, 3, , 3, , x3 - y3, vi), , =, , (a - b) (a2 + b2 + ab), , =, , (x - y) (x + y + xy), , =, , 1 (41 + 20), , =, , 1 x 61, , =, , 61, , =, , 61, , 2, , =, , 10, 2, , =, , 28561, , x4 + y4, , =, , 28561 - 7200, , x4 + y4, , =, , 21361, , 2, , 5. x + y = 5 ; x - y =, , 2, 2, 3 Find the value of 8 x y (x + y ), , x + y = 5; x - y =, , x+y = 9, x-y = 1, ---------------------2x, = 10, ---------------------x, , x4 + y4 + 7200, , 3 (take square on both sides), , (x + y)2 = 5; (x - y)2 = 3, Solve the equations, (x + y)2, , =, , (x - y)2, , = x2 + y2 - 2xy = 3, ---------------------------------------2 (x2 + y2), =, 8, , =5, , If x = 5, 5 + y = 9, y=9-5=4, , (x2 + y2), , x = 5; y = 4, , x2 + y2 + 2xy, , 8, , =, =, , 2. Solve (x + 5)2 - (x - 5)2, , 2, , =, , =4, , x2 + y2 + 2xy, , =, , a2 - b2 = (a + b) (a - b), (x + 5)2 - (x - 5)2 = [(x + 5) + (x - 5)] [(x + 5) - (x - 5)], , 2, , = (x + 5 + x - 5) (x + 5 - x + 5), , 2, , 2, , xy, , =, , =, , 8x, , =, , 4 x 4 = 16, , 4, , = (2x) (10), 8 xy (x2 + y2), , 3. If (x - y) = 4 and xy = 12, find the value of (x2 + y2), (x - y)2 =, , x2 + y2 - 2xy, , (4)2 =, , x2 + y2 - 2 x 12, , 16, , x + y - 24, , =, , x2 + y2 - 24 =, , 2, , 16, , x +y =, , 16 + 24, , x2 + y2 =, , 40, , 2, , 2, , 6. If (a -, , 2, , 4. If x - y = 7 and xy = 60 then find the value of x + y, (x - y)2, , =, , x2 + y2 - 2xy = 72, , x2 + y2 - 2 x 60, , =, , 49, , x2 + y2, , =, , 169, , (x2 + y2)2 =, x4 + y4 + 2x2y2, x + y + 2(xy), , =, , 28561, , x4 + y4 + 2(60)2, , =, , 28561, , x + y + 2(3600) =, 4, , 4, , a, , a2 +, , 28561, , 1, 2, , 1, a, , a2 +, , 2, , x4, , 1, , 2, , 1, a, , 2, , 2, , =, , 6, , =, , 62 (take square on both sides), , 2, , ⎛ 1 ⎞ - 2 (a), ⎜ ⎟, ⎝a⎠, , a, , a2 +, , 2, , 1, , a, , 2, , a2 +, , 1, , =, , ) = 6. Find the value of a2 +, , ⎛ 1⎞, ⎜a - ⎟, ⎝ a⎠, , (169)2, , =, , 4, , 4, , (169)2 (take square on both side), , 2, , 4, , 1, , ⎛ 1⎞, ⎜a - ⎟, ⎝ a⎠, , 4, , 5, , x + y - 2xy = 3, (-) (-) (+), (-), ------------------------------------------4xy, = 2, ------------------------------------------=, , If x + 5 = a and x - 5 = b, , = 20 x, , 5, , ⎛ 1 ⎞ = 36, ⎜ ⎟, ⎝a⎠, , - 2 = 36, , = 36 + 2, , = 38, , Workshop Calculation & Science : (NSQF) Exercise 2.4.09, , Copyright free, under CC BY Licence, , 39

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Elasticity - Elastic, plastic materials, stress, strain and their units and, young’s modulus, Exercise 2.5.10, Elastic material, , 11 Rubber Epichlorohydrin, , The Elastic materials are those materials that have the, ability to resist a distorting or deforming influence or force,, and then return to their original shape and size when the, same force is removed., , 12 Nylon, , Linear elasticity is widely used in the design and analysis, of structures such as beams, plates and sheets., , 15 Poilbutadiene, , Elastic materials are of great importance to society since, many of them are used to make clothes, tires, automotive, spare parts, etc., , 17 Vinyl stretch, , Characteristics of elastic materials, , 13 Terpene, 14 Isoprene Rubber, 16 Nitrile Rubber, 18 Thermoplastic elastomer, 19 Silicone rubber, 20 Ethylene-propylene-diene rubber (EPDM), , When an elastic material is deformed with an external, force, it experiences an internal resistance to the, deformation and restores it to its original state if the, external force is no longer applied., , 21 Ethylvinylacetate (EVA or foamy gum), , To a certain extent, most solid materials exhibit elastic, behavior, but there is a limit of the magnitude of the force, and the accompanying deformation within this elastic, recovery., , Plastic Material, , A material is considered as elastic if it can be stretched, up to 300% of its original length., For this reason there is an elastic limit, which is the, greatest force or tension per unit area of a solid material, that can withstand permanent deformation., For these materials, the elasticity limit marks the end of, their elastic behavior and the beginning of their plastic, behavior. For weaker materials, the stress or stress on, its elasticity limit results in its fracture., The elasticity limit depends on the type of solid considered., For example, a metal bar can be extended elastically up, to 1% of its original length., However, fragments of certain gummy materials may, undergo extensions of up to 1000%. The elastic properties, of most solid intentions tend to fall between these two, extremes., Maybe you might be interested How to Synthesize an, Elastolic Material?, , 22 Halogenated butyl rubbers (CIIR, BIIR), 23 Neoprene, Plastic Material Classification, “Plastic material” is a term that refers to a large class of, polymers, separated into various groups and sub-groups., Before starting the chapter on the uses and subsequent, recycling of plastic, let us establish a general classification, of these thermosetting resins or thermo-plastics (the two, big groups into which we include elastomers) by detailing, their properties, their make-up, their aspect, and their, final uses, while explaining which ones are recyclable., Thermoplastics, Remember that thermoplastic is a material whose, structure and viscosity can be modified both ways through, heating or cooling. This large family of materials is, commonly used by many industries and is easily, integrated into France’s recycling cycles., The following polymers are some examples of plastic, material:, 1 Polyolefins, 2 Vinyl polymers, 3 Polystyrenes, , Examples of elastic materials, , 4 Acrylate and methacrymate polymers, , 1 Natural gum, , 5 Polyamide, , 2 Spandex or lycra, , 6 Polycarbonates, , 3 Butyl Rubber (GDP), , 7 Celluloid, , 4 Fluoroelastomer, , 8 Linear polyesters, , 5 Elastomers, 6 Ethylene-propylene rubber (EPR), 7 Resilin, 8 Styrene-butadiene rubber (SBR), 9 Chloroprene, 10 Elastin, 42, , 9 Polyfluorethane, 10 Polyacetal, 11 Polysulfone, 12 Polyphenylene sulfide, 13 Modified polyphenylene oxide (PPO), , Copyright free, under CC BY Licence

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Thermosetting plastic, , Poisson's ratio, , Thermosetting plastic is a compound that, during, condensation polymerisation (and/or implementation), when, submitted to a catalyst or a temperature increase, irreversibly, cures. the structure, shape, or rigidity of the manufactured, plastic object can not be modified again, and the material is, rarely recycled., , It is a ratio between lateral strain and linear strain., , This type of plastic includes the following types of, compounds:, , 1 Calculate the tensile strain when a force of 3.2 kN, is applied to a bar of original length 280 cm, extends the bar by 0.5 mm (Fig 1 & Fig 2), , 1, , Unsaturated polyster, , 2, , Phenol formaldehyde resins, , 3, , Melamine resins, , 4, , Polyepoxides, , 5, , Polyimide, , 6, , Polyurethane, , 7, , Polyorganosiloxanes, , Poisson’s ratio =, , Lateral strain, Linear strain, , =, , 1, m, , Examples, , Generally in any industry the material used are elastic in, nature. Hence if a material is subjected to an external, load, it undergoes deformation. During the deformation, process the material will offer a resistance against the, deformation. In case if the material fails to put up full, resistance to the external load, the deformation continues, until rupture takes place. Hence it is important to have a, considerable knowledge about the materials and their, properties for designing and fabricating., Strain, When an external forces acting on a material, there is a, change in its dimension and shape. The deformation is, called strain. Thus, strain is the ratio between the change, in dimension of a material to its original dimension. It has, no unit. It is represented by e (Epsilon), Strain =, , Change in dimension ( ), Original dimension ( ), , Linear or Longitudinal strain, It is the ratio between the change in length of the material, to its original length., Linear Strain =, , Change in length ( ), Original length ( ), , Lateral Strain, It is the ratio between change in cross sectional area of, material to its original area., Change in area, , Lateral Strain =, , Original Area, , Volumetric Strain, It is the ratio between change in volume of material to its, original volume., Volumetric Strain =, , Force, , = 3.2 kN, , Change in volume, , Original length (L), , = 280 cm, , Original Volume, , Increased length( ) = 0.5 mm = 0.05 cm, , Workshop Calculation & Science : (NSQF) Exercise 2.5.10, , Copyright free, under CC BY Licence, , 43

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Tensile Strain =, , Increased length, , = 400 /18, , Original Length, , = 22.22 Newton / mm, , =, , Force required to, compress the, spring into 6 mm, Load required, , 0.05, 280, , = 0.0001786, , =, , = 22.22  6, = 133.32 N, , 2 A steel rod used for brake operation is 1.50 m long., When it is subjected to a tensile force the extension, produced is 0.5 mm. Find the strain in the rod., Tensile strain =, , = Spring stiffness Deflection, , Extension, , 5 Calculate the tensile strain when a force of 3.2, KN is applied to a bar of original length 2.8 m, extends the bar by 0.5 mm., Force F, , = 3.2 KN, , Original length, , Original length L, , = 280 cm, , ⎛ mm ⎞, ⎜, ⎟, 1.5 x 1000 ⎝ mm ⎠, , Increased length( ) = 0.5 mm = 0.05 cm, , 0.5, , Tensile strain, , Strain in the brake rod = 0.0003, , =?, Strain, , 3 A helical spring is loaded with a force of 600, Newton and is compressed by 30mm. What would, be the load required to compress it to 10 mm (Fig3), , =, , =, , , L, , 0.05, 280, , = 0.0001786, 6 A metal bar is 2m long. When 5.5 tonne is applied, its length becomes 1.995 m. Find the compressive, strain?, Force F, , = 5.5 KN, , Original length L1, , =2m, , Final length L2, , = 1.995 m, , Increased length( ) = 2 - 1.995 = 0.005 m, Compressive strain, , =, , Solution, Spring stiffness =, , Deflection, , Load required to compress the spring by 10 mm, = spring stiffness x deflection, = 25 (N/mm) x 10(mm), Load required = 250 N, 4 Helical spring is loaded with a force of 400 Newton and it is compressed by 18 mm. What would, be the load required to compress it to 6 mm?, Given force, , = 400 Newton, , Deflection, , = 18 mm, , Spring Stiffness, , = Force / Compressed length, , , L, , 0.005, , 2, = 0.0025, , Applied load, , 600 ⎛ N ⎞, ⎛ N ⎞, ⎜ mm ⎟ = 25⎜ mm ⎟, =, 30 ⎝, ⎠, ⎝, ⎠, , 44, , =, , 7 When a steel rod of 4mm diameter experienced, the load of 200 Kg. It is found to be elongated by, 1.5 mm from the original length of 1500 mm., Calculate the strain., Force F, , = 200 Kg., , Original length L1, , = 1500 mm, , , Strain, Compressive strain, , = 1.5 mm, =?, =, , =, , , L, , 0.005, , = 0.001, , Workshop Calculation & Science : (NSQF) Exercise 2.5.10, , Copyright free, under CC BY Licence, , 2

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8 An iron rod of length 1 metre and 1 cm diameter, gets elongated by 1 cm. When a force of 100 Kg, is applied at one end. Calculate the strain, developed in the rod., Force F, , = 100 Kg., , Original length L1, , = 1 m = 100 cm, , , , 2 Compressive stress: When a material is subjected to, two equal and opposite axial pushes, the material tends, to decrease in length. The resistance offered against, the decrease in length is called compressive stress., The corresponding strain is called compressive strain., (Fig 5), , = 1 cm, , Strain, , =?, , Compressive strain, , =, , =, , , L, , 1, , Compressiv e stress , , 100, , = 0.01, , Compressive stress , , Axial push, Area of cross section, , Decrease in length, Original length, , Stress, The internal opposite force to the external load per unit, area is known as stress. The unit of stress depends upon, the force applied and area of original cross-section of, material. It is represented by (Sigma),  Stress , , =, , Forced applied, , Shear stress (τ ) =, , F, A, , ⎛ N, Kg, ⎜, or, ⎜, 2, cm 2, ⎝ mm, , ⎛ N, Kg, ⎜, or, ⎜, cm2, ⎝ cm2, , 1 Compressive stress on connecting rod on the first part, of power stroke, 2 Compressive stress on push rod during valve opening, 3 Clutch lining when the clutch is engaged, , Area of original cross section, , Load (or) Force, Area, , Eg., , ⎞, ⎟, ⎟, ⎠, , ⎞, ⎟, ⎟, ⎠, , 3 Shear stress: When a material is subjected to two, equal and opposite forces acting tangentially across, the resisting section, the body tends to be sheared off, across the cross section. The stress included is called, shear stress. It is represented by  . The corresponding, strain is called shear strain. (Fig 6), , Types of Stress, 1 Tensile stress, 2 Compressive stress, 3 Shear stress, 4 Torsional Stress, 1 Tensile stress: When a material is subjected to two, equal and opposite axial pulls, the material tends to, increase in length. The resistance offered against this, increase in length is called tensile stress. The, corresponding strain is called tensile strain. (Fig 4), , Shear stress (τ ) =, , F, A, , ⎛ N, Kg, ⎜, or, ⎜, 2, cm2, ⎝ cm, , ⎞, ⎟, ⎟, ⎠, , Eg., 1 Rivets, 2 Gudgeon Pin, 3 Spring shackle pin, 4 Brake rod rivets, E.g.:, , 5 Chassis rivets, , 1 When brake is applied the brake rod is under tensile, stress., , 6 Fly wheel holding bolts, , 2 During tightening of bolt or nut., 3 Belt driving the fan., 4 Crane rope (When rope is pulling), , 7 Swivel pins, 8 Gear box shaft, 9 Axle shaft, , Workshop Calculation & Science : (NSQF) Exercise 2.5.10, , Copyright free, under CC BY Licence, , 45

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4 Torsional stress: When a shaft is subjected to the, action of two equal and opposite couples acting in, parallel planes, then the shaft is said to in torsion. The, stress set up by the torsion is known as torsional shear, stress., , 3 A load of 600 kg is placed on a hollow cast iron, cylinder of 200 mm outer diameter and 100 mm, internal diameter. Find the stress on the cylinder., Hollow cylinder, , Eg., , Outer diameter (D) = 200 mm = 20 cm, , 1 Rear axle, , Outer radius (R), , 2 Crank shaft, , Internal diameter (d) = 100 mm = 10 cm, , 3 Coil springs, , Inner radius (r), , = 5 cm, , 4 Propeller shaft, , Weight, , = 600 kg, , 5 Starter motor armature shaft, Examples, 1 A steel wire 3 mm dia. is loaded in tension with a, weight of 50 kg. Find out the stress developed., Diameter of the steel wire, , = 1.5 mm, , Weight, , = 50 kg, , Stress, , =, , =, , Area, , = π (R + r) (R − r), 22, × (10 + 5) × (10 − 5), 7, 22, =, × 15 × 5, 7, 1650, =, = 235.7cm 2, 7, , Area(A), , Stress, , 22, × 1.5 × 1.5, 7, 49.5, =, = 7.07 mm2, 7, , =, , =, , 50, 7.07, = 7.072 Kg/mm2, , Stress, , =, , 2 A force of 500 N is applied on a metallic wire of, 5mm diameter. Find the stress., Diameter of the wire, , = 2.5 mm, , Force, , = 500 Newton, , 600, kg/cm2, 235.7, , = 2.546 kg/cm2, 4 Calculate the minimum cross section area of a, M.S. bar to withstand a load 6720 kg. Take the, maximum stress of the material as 698.2 kg/cm2., Weight, , = 6720 kg, , Maximum stress, , = 698.2 kg/cm2, , Stress, , =, , Area(A), , =, , = 5 mm, , Radius, , Area(A), , =, , Force(F), , =  r2, , Force(F), , Stress, , = 3 mm, , Radius, , Area of circular wire (A), , = 10 cm, , Force(F), Area(A), , 6720, 698.2, , = 9.625 cm2, Stress, , =, , Area of circular wire (A), , Force(F), Area(A), , =  r2, , To calculate diameter, , Area, , 22, × 2.5 × 2.5, 7, 137.5, =, = 19.64mm2, 7, , =, , πd 2, 4, , =, , Stress, , =, , 500, 19.64, , = 25.46 N/mm2, 46, , d2, , = 4 x 9.625 x, , =, , 134.75, 11, , d2, , = 12.25, , d, , = 3.5 cm, , Workshop Calculation & Science : (NSQF) Exercise 2.5.10, , Copyright free, under CC BY Licence, , 7, 22

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5 A load of 300 kg hanging from a rod of 3 metre, length and 5 mm diameter extends it by 4 mm., Find the stress in the material and the strain it, causes., , Thickness of the plate = 5 mm = 0.5 cm, Diameter of the punch = 30 mm = 3 cm, Shear stress, , = 400 kg/cm2, , Length of the rod, , = 3 m = 3000 mm, , Force, , = Shear stress x area, , Increased length, , = 4 mm, , Shear area, , = Circumference x thickness, , Diameter, , = 5 mm;, , Radius, , = 2.5 mm, , Weight, , = 300 kg, , Strain, , =, , =, , =, , Change in length, Original length, , ==, Stress, , = Dt, , 4, = 0.00133, 3000, , Force(F), Area(A), , Area of circular rod (A) = r2, , 22, =, × 2.5 × 2.5, 7, 137.5, =, 7, , =, Required force, , 22, 7, , x 3 x 0.5, , 33, 7, , = 4.71 mm2, , = 400 x 4.71, = 1885.71 kg, , 8 What force will be required to shear off a bar of, 30 mm dia. if the ultimate shear stress of the, material is 35 kg/mm2., Diameter of the bar, , = 30 mm, , Shear stress, , = 35 kg/mm2, , Stress(), , Force(F), , =, , Area(A), , = 19.643 mm2, 35, , 300, =, 19.643, , Stress, , Thickness of the plate= 1 mm, , = 24750 kg, 9 A Hole of 2 cm dia is to be punched out of a plate, of 1.4 cm thick. If the force applied to the punching, die is 12 KN. Calculate the shear stress., Dia. of the hole, , = 2cm, , Thickness, , = 1.4 cm, , Force, , = 12 KN, , Shear stress, , =?, , Dia. of the punch, , = 10 mm, , Shear stress, , = 50 Newton/mm2, , Force, , = Shear stress x area, , Shear area, , = Circumference x thickness, , Punched out area, , =, =, , =, , dt, 22, 7, , F, π × 15 × 15, , = 35 x  x 15 x 15 kg, , F, , = 15.273 kg/mm2, 6 Find the force required to punch a hole of 10 mm, dia. in a 1 mm thick plate, if the allowable shear, stress is 50 N/mm2., , =, , = Circumference of the hole,  Thickness, = 2 r  t unit2, = 2 1  1.4, , x 10 x 1, , = 2.8 cm2, , 220, , = 31.43 mm2, 7, Force = 50 x 31.43, , Shear stress ( ) =, , = 1571.5 Newton, 7 A hole of 30 mm diameter is punched in a plate of, 5 mm thickness. If the shear stress is 400 kg/cm2., Find the force required to punch the hole., , =, , F, A, 12KN, 2.8 π cm2, , = 1.364 KN/cm2, , Workshop Calculation & Science : (NSQF) Exercise 2.5.10, , Copyright free, under CC BY Licence, , 47

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10 A square rod of 10 mm side is tested for a tensile, load of 1016 kg. Calculate the tensile stress?, Side of square rod a, , = 10 mm, , Tensile force F, , = 1016 kg, , Tensile stress , , =?, , Stress(), , =, , =, , =, , Force(F), Area(A), , 2, , 1016, 10x10, , = 10.16 Kg/mm2, 11 A M.S. tie bar 3.5 cm dia. is under a state of stress, which carries a load of 6720 kg. Find the intensity, of stress in the material., , If the external force is so large that the stress exceeds the, limit, the material loses to some extent its property of, elasticity. If now the force is removed, the material will not, return to its original shape and size and there will be a, residual deformation in material., , d = 3.5 cm, , Yield point, , r = 1.75 cm, , The yeild point of a material is the point at which there is, a marked increase in elongation without increase in load., , F = 6720 kg, Stress(), , When an external force acts on a body, the body tends to, under go some deformation. If the external force is removed, and the body comes back to its original shape and size, (Which means the deformation disappers completely)., The body is known as elastic body. This property by virtue, of which certains materials return back to their original, position after the removal of the external force is called, elasticity., The body will regain its previous shape and size only when, the deformation caused by the external force is with in a, certain limit. Thus there is a limiting value of force up to and, within which the deformation completely disapperas on the, removal of the force. The value of stress corresponding to, this limiting force is known as elastic limit of the material., , Force, a, , Elasticity and Elastic limit, , =, , =, , Hooke’s law, , Force(F), , Robert Hooke discovered a relationship between stress, and strain. According to Hooke’s law stress is proportional, to strain within elastic limit., , Area(A), Force, , πr, , Young’s Modulus or Modulus of Elasticity, , 2, , 6720, , =, , 3.14 x 1.75 x 1.75, , =, , 6720, , The ratio of stress to strain within elastic limit is known as, young’s modulus or modulus of elasticity. This is expressed by a symbol “E”. The unit of Young’s modulus is, same that of stress., , , 9.616, , = 698.8 Kg/cm, , 2, , 12 A rivet of 10 mm dia. is subjected to a double shear, force of 1.5 KN. Find the shear stress in the rivet., , Double shear force is acting on the rivet, consider, the area as double., Stress, , =, , F, 2Area, , =, , 1.5, , Modulus of Rigidity (N) , , Shear stress, Shear strain, , Bulk Modulus, When a body is subjected to three mutually perpendicular, forces of the same intensity, the ratio of volumetric stress, to the volumetric strain is known as Bulk Modulus. It is, usually represented by the letter K., , , Bulk Modulus (K) , , Volumetric stress, , 2x3.14x5x5, = 0.00955 KN/mm2, , 48, , Strain, , The ratio of shear stress to shear strain is known as, “modulus of rigidity” represented by symbol “N”., , , Shear stress = ?, , Stress, , Modulus of Rigidity, , dia. of the rivet = 10 mm, r = 5 mm, , Young' s modulus (E) , , Workshop Calculation & Science : (NSQF) Exercise 2.5.10, , Copyright free, under CC BY Licence, , Volumetric strain

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Relationship between three moduli for a given, material, , ==, , The relationship between three moduli for a given material, is as follows :, , ⎛ 2⎞, ⎛ 1⎞, E = 2N⎜1+ ⎟ = 3 K ⎜1− ⎟., m, ⎝ m⎠, ⎝, ⎠, , =, , , , 2, π, × 2.5, 4, , π × 6.25, , cm2, 4, Force applied, Stress =, Area of original cross section, , Where, , 4500, π × 6.25, =, 4, , E = Young’s modulus of elasticity, N = Modulus of rigidity, K = Bulk modulus, , 4500× 4, = π, × 6.25, , 1, = Poisson’s ratio, m, , =, , Example, 1 A steel rod of 10 mm diameter and 175 mm long, is subjected to a tensile load of 15 kN. If E = 2 x105, N/mm2, calculate the change in length., Tensile load, , = 15 kN = 15000 N, 22, , Area of cross section = (r2) =, , , , Stress =, , 15000N, , , , Strain, , =, , Change in length =, , 2880, Kg/cm2, π, , , , Strain, , =, , Change in length, Original length, , =, , 0.008 8 1000, =, 20, 20, , =, , 8, 4, =, 20×1000 10000, , =, , 4, 10000, , Stress, , , , Strain, , E = 2  105 N/mm2 =, , Stress =, , = 191 N/mm2, , 0.785×100mm2, , Young's modulus E =, , , ,  5  5 mm2 = 78.57, , 7, , 191N/mm2, Strain, , Strain, ,  Young’s modulus, , 2 ×105, , 2 ×105, , mm, , 2 A bar of steel 2.5 cm diameter was subjected to, compressive load of 4500 kg. The compression in, a length of 20 cm was found to be 0.008 cm. Find, the Young’s modulus of elasticity of bar., Solution, = 2.5 cm, , Force applied i.e. compressive load = 4500 kg, , 2880, 4, ÷, π, 10000, , =, , 2880 10000, ×, π, 4, , =, , 7200000, π, , = 2292000 Kg/cm2, = 2.292 x 106 Kg/cm2, 3 A force of 10 tonnes is applied axially on a rod of, 12 cm dia. the original length is 100 mm.If modulus, of elasticity is 2 x 1012 kg/cm2. Calculate stress, and strain developed in the rod., Force applied = 10 tonnes, , = 0.008 cm, ,  Area of original cross-section =, , =, , Solution, , Original length of bar = 20 cm, Change in length, , Stress, Strain, , = 0.167 mm., , Diameter of bar d, , =, , 191, , 175 ×191, , 2880, π, , π, 4, , = 10 x 1000 kg, = 104 kg, , d2, , Diameter (d) = 12 mm, , = 1.2 cm, , Young’s modulus (E), , = 2 x 1012 kg/cm2, , Workshop Calculation & Science : (NSQF) Exercise 2.5.10, , Copyright free, under CC BY Licence, , 49

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Stress =, , Force applied, Area of original cross section, , =, , =, , Strain, , 104, π 12 12, × ×, 4 10 10, , 104 × 4 × 10 × 10, π × 12× 12, , =, , = 0.0136, , = 8841 kg/cm2, , We know, Stress, , = Young’s modulus, , Strain, , Strain x Young’s modulus, Strain, , = Stress, =, , =, , Stress, Young's Modulus, , 8841, , E, , =, , Stress, , = 8841 kg/cm2, , Strain, , = 4420.5 x 10-12, , 4 A bar of 100 cm elongates to 101.36 cm when a, load of 15000 kg is applied to it. Take the area of, cross section of bar as 10 cm2. Find the stress,, strain and youngs modulus., L2, , = 101.36 cm, , = 110300 kg/cm2, 5 What force is required to stretch a steel wire of 10, mm long and 10 mm dia. to double its length. E, of steel is 205 KN/cm2., d, , = 10 mm = 1 cm, , r, , = 0.5 cm, , L1, , = 1 cm, , L2, , = 2 cm, , , , = L2 - L1 = 2 - 1 = 1 cm, , E, , = 205 KN/cm2, =, , E, , =, , 205, , =, , Stress, , = 1 x 205 = 205 KN/cm2, , Stress, , =, , 205, , =, , A, , = 10 cm2, , Stress, , =, , =, , Strain, , Stress, , 1, , Force(F), Area(A), Force, , Force, , = 205 x 3.14 x 0.5 x 0.5, = 161 KN, , 6 A wire of 1.6 cm diameter is subjected to a tensile, load of 2000 Kg. Find the stress and strain if, youngs modulus = 2 x 106 kg/cm2., , Force(F), Area(A), , F, , = 2000 kg, , 15000, , d, , = 1.6 cm, , 10, , r, , = 0.8 cm, , E, , = 2 x 106 Kg/cm2, , = 1500 kg/cm2, 50, , Stress, , 3.14 x 0.5 x 0.5, , = 101.36 - 100 = 1.36 cm, = 15000 kg, , =, , 1, =1, 1, , Strain, , = L2 - L1, F, , 1500, , 2 ×1012, , = 4420.5 x 10-12, , = 100 cm, , Strain, , 0.0136, , =4420.5, 12, 10, , L1, , Stress, , Youngs modulus =, , = 8841 kg/cm2, Stress, , 1.36, 100, , 106, =, 36 π, , , , =, , Workshop Calculation & Science : (NSQF) Exercise 2.5.10, , Copyright free, under CC BY Licence

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Stress, , =, , E, , F, , Rectangular rod length = 2 cm, , A, , =, , = 2 x 106 kg/cm2, , Breadth = 1 cm, , 2000, Stress() =, , πr 2, 2000, , =, , =, , 3.14 x 0.8 x 0.8, , =, , E, , = 995.2 kg/cm2, , 2 x 106, , =, , lxb, , 2000, , =, , Stress, Strain, 1000, Strain, , 995.2, , Strain, , Strain, , Strain, , =, , Stress, Strain, , =, , Force, , = 1000 kg/cm2, , 2.0096, , 2 x 106, , Area(A), , =, , 2 x1, , 2000, , Youngs modulus =, , Force(F), , =, , 1000, 6, 2 x 10, , = 0.0005, , 995.2, 6, 2 x 10, , = Strain, , = 0.0005, 7 A tensile load of 2000 kg is applied on a, rectangular rod of 2 cm x 1 cm whose length is 2, metres. Calculate the elongation in length as E =, 2 x 106 Kg/cm2., F, , = 2000 Kg., , L1, , = 2 m = 200 cm, , , 200, , = 0.0005, ,   = 200 x 0.0005, , = 0.1 cm, ,  Elongated length = 0.1 cm, , Assignment, Strain, 1 Find the compressive strain if a metal bar is 150 cm, long. When 2.5 KN is applied, its length becomes, 148.6 cm., , 2 Calculate the intensity of stress if a mild steel rod having, a cross sectional area of 40 mm2 is subjected to the, load of 1000 kg., , 2 Calculate the strain if a metallic bar is 150 cm long., When 2500 kg is applied its length becomes 150.5, cm., , 3 Calculate the tensile stress if a square rod of 10 mm, side is tested for a tensile load of 1000 kg., , 3 Find the strain it causes if a load of 300 kg hanging, from a rod of 3 metres length and 5 mm diameter, extends it by 4 mm., 4 A tensile force of 10 kg is applied on a copper wire of, diameter 1 cm. So that the length of wire increases by, 5 mm. If the original length of wire was 2 metres, findout, the strain., , 4 Calculate the maximum stress if a bar of 9 cm2 cross, sectional area 300 cm long carries a tensile load of, 3500 kg., 5 Find out the stress on the rod. if a load of 500 kg is, placed on a M.S.rod of dia. 35 mm., 6 A metallic bar of 8 cm diameter is under stress carrying, a load of 8620 N. Calculate the intensity of stress., 7 A steel wire 2 mm diameter is loaded in tension with a, weight of 20 kg. Find out the stress developed., , Stress, 1 Calculate the intensity of stress in the material if a, copper rod of 40 mm diameter is subjected by tensile, load of 4000 Newtons., , 8 A rod having a cross sectional area of 25 mm2 is, subjected to a load of 1500 kg. Find out stress on the, rod., , Workshop Calculation & Science : (NSQF) Exercise 2.5.10, , Copyright free, under CC BY Licence, , 51

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9 A square rod of 10 mm side is tested for a tensile load, of 2500 kg. Calculate the tensile stress of the rod., Youngs modulus, 1 A piece of wire 2 m long, 0.8 mm2 in cross section, increases its length by 1.6 mm on suspension of 8 kg, weight from it. Calculate the stress, strain and youngs, modulus., 2 A wire of 16 mm dia. is subjected to a tensile load of, 2000 kg. Find the stress and strain if young’s modulus, E = 2 x 1016 kg/cm2., 3 A wire is of 2 metres long and its area of cross section, is 0.78 mm2. If 78 kg weight is suspended on this, wire, then the length of the wire is increased by, 1.4 mm. Find out stress, strain and youngs modulus, of elasticity., , 52, , 4 A wire 2800 mm long is stretched by 0.5 mm, when a, weight of 9 kg is hung on it, its diameter is 2 mm., Calculate stress and youngs modulus for the substance, of the wire., 5 A force of 1000 kg is applied axially on rod of 12 mm, diameter the original length is 100 mm. If modulus of, elasticity is 2 x 1012 kg/cm2. Calculate the stress and, strain developed in the rod., 6 A steel wire 3.2 mm diameter and 3.65 metre long, stretches by 2.03 mm under the load of 115 kg., Calculate the stress and youngs modulus of elasticity., 7 A mass of 10 kg is hung from a vertical wire 300.25 cm, long and 0.0005 sq. cm cross section. When the load, is removed the wire is found to be 300 cm long. Find, the modulus of elasticity for the wire material., , Workshop Calculation & Science : (NSQF) Exercise 2.5.10, , Copyright free, under CC BY Licence

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Elasticity - Ultimate stress and working stress, Ultimate stress and Working stress, The minimium load at which a material develops failure is, called as ultimate load or breaking load. The stress, produced in a material at ultimate load is called as ultimate, stress or breaking stress.,  Ultimate stress =, , Ultimate load, Area of original cross section, , The load which is considered safe for the machine element, is known as safe load or working load and the, corresponding stress at this load is called as safe stress, or working stress.,  Safe stress =, , Safe load, Area of original cross section, , Factor of safety (Fig 1), , Exercise 2.5.11, , A metal (say mild steel) is subjected to increasing load and, the extensions are measured with an extensometer. On, plotting a graph between the loads and elongations produced,, in the beginning, there is a straight-line relationship. It, continues up to ‘a’ which is called the limit of proportionality,, i.e. up to ‘a’ in Fig 2 ‘Stress is proportional to strain’., Point b denotes the elastic limit. Below this point, the body, regains its original shape, if the load is removed. Beyond, this point the body does not recover its original shape, completely, even if the load is removed., Upto a point beyond the elastic limit, a considerable, amount of elongation takes place even with a slight, increase in load. The point C where it occurs, is called the, yield point., At ‘d’ the maximum or the ultimate load is reached. After, this, a waist or local contraction is formed in the specimen,, and fracture occurs as illustrated in Figure., Example, A standard steel bar of 30 mm square cross section is, subjected to tensile stress. If the factor of safety is 4 and, ultimate stress is 370 N/mm2 determine the load to which, the bar is subjected. (Fig 3), , The ratio of ultimate stress to working stress (i.e. safe, stress) is known as factor of safety. The ratio of ultimate, load to the safe load may also be termed as factor of, safety. It has no unit. Hence it is expressed in a number., , Factory of safety =, , Ultimate stress, Working stress, , Ultimate stress, = Factor of safety = 4, Working stress, , or, Ultimate load, Factor of safety =, Safe load, , Working stress =, , N/mm2, 4, Area of cross section = 900 mm2, (a2 = 302 = 900, a =, , Stress-Strain graph, Load-extension graph (Fig 2), , 370, , =30), , Load = Working tensile stress x area, = 900 mm2 x, , 370, , N/mm2, , 4, , = 83250 N, Example, 1 A rod of  60 mm is subjected to a maximum, tensile load of 1600 kg. Calculate the stress and, strength of the material. If the factory of safety is, 5., Dia. of rod, , = 60 mm, , Tensile load F, , = 1600 kg, , Copyright free, under CC BY Licence, , 53

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i, , Stress, , Factor of safety (FS) = 4, , F, 1600, =, A π × 30 × 30, , =, , = 0.5658 kg/mm2 Ans., ii Factor of safety, Factor of safety, , 5=, Ultimate stress, , =, , Ultimate stress, Working stress, , 4, , =, , 25 kg/mm 2, W.S, , WS, , =, , 25, kg/mm2, 4, , =5, =, , Ultimate stress, Working stress, , Ultimate stress, , = 6.25 kg/mm2, , 0.5685 kg/mm 2, = 5  0.5658 kg/mm2, , Stress, , Strength of the material = 2.829 kg/mm2, 2 Find the safe load which can be suspended from, a 4.2 mm dia. wire. If the ultimate stress is 25 kg/, mm2 and the factor of safety is 4., Dia. of wire d, , FS., , =, , 6.25 kg/mm2, F, , = 4.2 mm, , Safe load F, , Ultimate stress U.S = 25 kg/mm2, , =, , F, A, F kg, π × 2.1 × 2.1mm2, , = 6.25   2.1  2.1 kg, = 86.6 kg, , Assignment, 1, , A = 60 x 15 mm2, Rm = 370 N/mm, , F = 19000 N, Rm = 420 N/mm2, , F = ______ N, , Factor of safety = 5, , Rm = Ultimate, stress, , d = ______ mm, , F = Breaking Force, 2, , 4, , 2, , 5, , F = 35000 N, , A = 25 x 6 mm2, , Working shear, stress = 110 N/mm2, , F = 63000 N, , b = 10 mm, , Rm = ______ N/mm2, , l = ______ mm, , 6, , Ultimate tensile, stress = 420 N/mm2, d = 10 mm, , 3, , A = 490.87 mm, , 2, , Shear force Max, = ______ N, , F = 206.22 kN, d = ______ mm, Rm = ______ N/mm2, , 7, , F = 140 kN (Tensile), Factor of safety = 4, Ultimate tensile, stress = 500 N/mm2, Ultimate shear, stress = 400 N/mm2, d = ______ mm, , 54, , Workshop Calculation & Science : (NSQF) Exercise 2.5.11, , Copyright free, under CC BY Licence

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8, , 9, , Ultimate tensile, strength, = 370 N/mm2, Shear force, = 380 kN, R = 30 mm, Thickness of plate, punched, = ______ mm, Ultimate tensile, strength = 330 N/mm2, s = 3 mm, F = 60 kN, (Shear force), , 10, , 15 A steel rod of 1.5 cm diameter and 8 metres long pulled, by a forced of 80 kg at both ends. Find out the, expansion and strain on the rod. The coefficient of, elasticityE=2.10106 kg/cm2., 16 Find out the load which can be suspended from a 3.2, mm dia wire taking the factor of safety as 2. Ultimate, stress is 25 kg/mm2., 17 A rod of 60 mm dia is subjected to a maximum tensile, load of 1600kg. Calculate the stress and strength of the, material if the factor of safety is 5., , F = 43180 N, , 18 A wire of length 3.5 m and diameter 0.35 mm is, stretched by a force of 2kg weight. If the elongation is, 4 mm. Calculate the young's modulus of the material of, wire., , Ultimate tensile, strength, = 220 N/mm2, d = ______ mm, , d = 12 mm, F = 36.2 kN, Ultimate tensile, strength = ____ N/mm2, , 12, , 14 A steel rod 1.5 metres long and of 30 mm diameter is, pulled at both ends by a force of 1500 kg. If modulus of, elasticity of steel is 2.4 106 kg/cm2, determine increase, in length of rod and strain produced in it., , d = ______ mm, , s = 2.5 mm, , 11, , 13 A steel rod whose diameter is 1 cm and 60 cm in length., This rod is pulled at both ends by a force of 700 kg. If, modulus of elasticity of steel is 2.1 106 kg/cm2, find out, increase in length of rod and strain produced in it., , Tensile load, = 15000 N, Ultimate tensile, strength, = 9.5 N/mm2, , 19 A mass of 1kg is suspended from a metal wire 100 cm, long and 0.5 mm diameter. An increase in length of wire, equal to 2 mm is observed. Calculate the young's, modulus of wire., 20 A 4 metre long copper wire of diameter 3 mm is used to, support a mass of 50kg. What will be the elongation of, the wire. Young's modulus of elasticity for copper is, 7x1010 N/mm2., 21 Calculate the change in length of a rod of dia 16 mm and, 160 mm long when it carries a load of 40KN. Take E =, 200000 N/mm2., 22 A hollow C.I. column with a wall thickness of 2 cm is, subjected to an axial compressive load of 80 tonnes. If, the maximum stress is not to be exceeded 1 tonne per, cm2, determine the internal diameter of column. Calculate, compressive strain, if E = 950 tonnes per cm2., , s = 20 mm, Width `b' = ______ mm, , Workshop Calculation & Science : (NSQF) Exercise 2.5.11, , Copyright free, under CC BY Licence, , 55

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Heat treatment and advantages, , Exercise 2.6.12, , Heat treatment and its purpose, , Lower critical temperature, , The properties of steel depend upon its composition and, its structure. These properties can be changed to a, considerable extent, by changing either its composition or, its structure. The structure of steel can be changed by, heating it to a particular temperature, and then, allowing, it to cool at a definite rate. The process of changing the, structure and thus changing the properties of steel, by, heating and cooling, is known as ‘heat treatment of steel’., , The temperature, at which the change of structure to, austenite starts - 723°C, is called the lower critical, temperature for all plain carbon steels., , Structure of steel when heated (Fig 1), If steel is heated, a change in its structure commences, from 723°C. The new structure formed is called ‘Austenite’., Austenite is non-magnetic. If the hot steel is cooled, slowly, the old structure is retained and it will have fine, grains which makes it easily machinable., , Upper critical temperature, The temperature at which the structure of steel completely, changes to AUSTENITE is called the upper critical, temperature. This varies depending on the percentage of, carbon in the steel. (Fig 2), Example, 0.57% and 1.15% carbon steel: In these cases the lower, critical temperature is 723°C and the upper critical, temperature is 800°C., For 0.84% carbon steel, both LCT and UCT are 723°C., This steel is called eutectoid steel., Three stages of heat treatment, •, •, •, , Heating, Soaking, Quenching, , When the steel on being heated reaches the required, temperature, it is held in the same temperature for a period, of time. This allows the heating to take place throughout, the section uniformly. This process is called soaking., Soaking time, , If the hot steel is cooled rapidly the austenite changes into, a new structure called ‘Martensite’. This structure is very, fine grained, very hard and magnetic. It is extremely wearresistant and can cut other metals., Critical temperatures (Fig 2), , This depends upon the cross-section of the steel, its, chemical composition, the volume of the charge in the, furnace and the arrangement of the charge in the furnace., A good general guide for soaking time in normal conditions, is five minutes per 10 mm of thickness for carbon and low, alloy steels, and 10 minutes per 10 mm of thickness for, high alloy steels., Heating steel, This depends on the selection of the furnace, the fuel used, for heating, the time interval and the regulation in bringing, the part up to the required temperature. The heating rate, and the heating time also depend on the composition of, the steel, its structure, the shape and size of the part to, be heat-treated etc., Preheating, Steel should be preheated at low temperatures up to, 600°C as slowly as possible., Quenching, Depending on the severity of the cooling required, different, quenching media are used., The most widely used quenching media are:, • brine solution, • water, • oil, • air, , 56, , Copyright free, under CC BY Licence

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Brine solution gives a faster rate of cooling while air cooling, has the slowest rate of cooling., , Advantages, , Brine solution (Sodium chloride) gives severe quenching, because it has a higher boiling point than pure water, and, the salt content removes the scales formed on the metal, surfaces due to heating. This provides a better contact, with the quenching medium and the metal being heattreated., , •, , The hardening devices are brought to the workpiece., , •, , It is advantageous for large workpieces., , •, , Short hardening time., , •, , Great depth of hardening., , •, , Small distortion., , •, , Low fuel consumption., , Water is very commonly used for plain carbon steels., While using water as a quenching medium, the work, should be agitated. This can increase the rate of cooling., The quenching oil used should be of a low viscosity., Ordinary lubricating oils should not be used for this, purpose. Special quenching oils, which can give rapid and, uniform cooling with less fuming and reduced fire risks, are, commercially available. Oil is widely used for alloy steels, where the cooling rate is slower than plain carbon steels., , The following are the advantages of this type of hardening., , Disadvantages, The following are the disadvantages., •, , Not suitable for small workpieces because of the, danger of hardening throughout., , •, , The workpieces must be stress-relieved before, hardening., , Cold air is used for hardening some special alloy steels., , Workshop Calculation & Science : (NSQF) Exercise 2.6.12, , Copyright free, under CC BY Licence, , 57

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Heat treatment - Different heat treatment process – Hardening, tempering,, annealing, normalising, case hardening, Exercise 2.6.13, Heat treatment processes and purpose, Because steel undergoes changes in structure on heating, and cooling, its properties may be greatly altered by, suitable heat treatment., The following are the various heat treatments and their, purposes., Hardening:, , To add cutting ability., To increase wear resistance., , Tempering:, , Annealing:, , Normalising:, , To remove extreme brittleness, caused by hardening to an extent., , For annealing, the hypoeutectoid steel is heated to, 30°C to 50°C above the upper critical temperature, and, it is 50°C above the lower critical temperature for, hypereutectoid steel. (Fig 3), The soaking time at this temperature is 5 mts/10 mm, of thickness for carbon steel., The cooling rate for carbon steel is 100°C to 150°C/hour., The cooling is done in the furnace itself by switching off, the furnace or the steel is covered either in sand or dry, lime and dry ash., Purpose of annealing, , To induce toughness and shock, , Annealing is done:, , resistance., , •, , to obtain softness, , To relieve strain and stress., , •, , to improve machinability, , To eliminate strain/hardness., , •, , to increase ductility, , To improve machinability., , •, , to relieve internal stresses, , To soften the steel., , •, , to reduce or eliminate structural in homogeneity, , To refine the grain structure of, the steel., , •, , to refine grain size and to prepare the steel for subsequent heat treatment processes., Annealing temperature, , Annealing and normalising, The treatments that produce equilibrium conditions are, annealing and normalising., , Carbon content %, , Temperature °C, , <0.12, , 875 to 925, , 0.12 to 0.25, , 840 to 970, , 0.25 to 0.50, , 815 to 840, , Annealing, , 0.50 to 0.90, , 780 to 810, , In this process, steel is heated to a suitable temperature, depending upon its carbon content (Fig 3), and is held at, that temperature for sufficient time, and then slowly, cooled to room temperature., , 0.90 to 1.3, , 760 to 780, , The treatments that produce non-equilibrium conditions, are hardening and tempering (usually done in conjunction, with each other)., , Normalising, Due to continuous hammering or uneven cooling, strains, and stresses are formed in the internal structure of, steel. These should be removed from forgings or castings;, as otherwise, they may fail at any time while in use., Normalising is done to produce a fine grain for uniformity, of the structure and for improved mechanical properties., The normalising process, In this process, the steel is heated to a suitable temperature, depending upon its carbon content, (Fig 4) and held at, that temperature, and then, cooled freely in the air., Normalising is usually done, before machining and, before hardening, to put the steel in the best condition for, these operations., , The heating, soaking (holding at that temperature) and, slow cooling cause the grains to become large, and in, the process, produce softness and ductility., , The steel is heated to a temperature (30 to 40°C above, the upper critical temperature) at which only austenite is, present even in the case of high carbon steel. This is, because this process is the first step towards producing, the final properties, and it is necessary to start with, austenite to ensure uniformity., , 58, , Copyright free, under CC BY Licence

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The increase in carbon content will result in an increase, in the hardness produced by the treatment., Steel with less than about 0.15% carbon will not respond, to this treatment., Process of hardening, In order to produce the desired effect, sufficient carbon, must be put into the solid solution to cause internal, distortion when it is trapped in the iron by quenching., When the carbon content is less than 0.83%, the steel is, heated to only just above its upper critical point (heating)., When its carbon content is more than 0.83% the steel is, heated only to just above its lower critical point (heating)., The heated piece for normalising should not, be kept at any wet place, in wet air or kept, in forced air as they will induce some hardness., , Fig 6 illustrates the temperatures to which steels are, heated before quenching, and the quenching temperature, of steels with different carbon content., , Hardening & tempering, If a piece of steel is heated to a sufficiently high temperature,, all the carbon will be dissolved in the solid iron to form a, solid solution, called austenite of the steel. When it is, slowly cooled, the change in the arrangement of the iron, atoms will cause a solid solution called ferrite to be, produced. The solid solution can only contain up to, 0.006% carbon, and so the excess carbon will be forced, to leave the solid solution, and produce cementite. This, will, with ferrite, form a laminated structure called pearlite., The principle of hardening, If steel is cooled rapidly (quenched) the excess carbon, will not have sufficient time to leave the solid solution, with the result that it will be trapped in the iron, and so, cause an internal distortion. This internal distortion is the, cause for the increase in the hardness of steel with, a corresponding reduction in its strength and ductility., This is the basis of the hardening process., The mechanical properties produced as a result of this, treatment will depend upon:, •, , the carbon content of the steel, , •, , the temperature to which it is heated, , •, , the duration of heating, , •, , the temperature of the steel at the start of quenching, , •, , the cooling rate produced by quenching., , The effect of carbon content upon the hardness produced, by the hardening process is illustrated in Fig 5., , Soaking time, After heating, the steel is held at that temperature for, some time. Normally 5 mts are allowed as soaking time, for 10 mm thickness of steel., Cooling, Then the steel is cooled in a suitable quenching medium, at a certain minimum rate called the critical cooling rate., The critical cooling rate depends upon the composition, of the steel. This cooling transforms all the austenite into, a fine, needle-like structure called martensite, (the, appearance of which is shown in Fig 7)., The structure of steel treated this way is very hard and, strong, but very brittle., The quenching medium, The quenching medium controls the rate of cooling., For a rapid quenching a solution of salt or caustic soda, in water is used., For very slow quenching a blast of air is sufficient., Oil gives an intermediate quenching., Water and oil are the most common quenching media, used., , Air quenching is suitable only for certain special alloy, steels., Workshop Calculation & Science : (NSQF) Exercise 2.6.13, 59, , Copyright free, under CC BY Licence

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This method is not, however, suitable for accurate, temperature assessment., In a manufacturing plant, when heat treating is done on, a production basis, modern methods are used. Tempering, is done in controlled-atmosphere furnaces with the, temperatures controlled by modern instruments. Under, such conditions, it is possible to obtain accurate and, uniform results in any number of pieces., Table 1, Tempering temperature, Temper colour, Pale straw, Dark straw, Brown, Brownish purple, Purple, Dark purple, Blue, , Tempering, After hardening, steel is usually reheated to a suitable, temperature below the lower critical point (heating) to, improve its toughness and ductility but it is done at the, expense of hardness and strength. It is done in order to, make the steel more suitable for service requirements., , Temperature in °C, 230, 240, 250, 260, 270, 280, 300, , Fig 9 illustrates the appearance of the microstructure of, hardened, and then tempered steel., , Purpose of tempering the steel, Steel, in its hardened condition, is generally too brittle and, too severely strained. In this condition, steel cannot be, used, and hence it has to be tempered., The aims of tempering are:, •, , to relieve the steel from the internal stresses and, strains, , •, , to regulate the hardness and toughness, , •, , to reduce the brittleness, , •, , to restore some ductility., , Process of tempering, The tempering temperature depends upon the properties, required, but it is between 180°C and 650°C. (Fig 8) The, duration of heating depends upon the thickness of the, material. Tools are usually tempered at a low temperature., The temperature itself is judged by the colour of the oxide, film produced on the surface during heating. (Table 1), , Generally, tempering in the lower temperature range for, an increased time provides greater control in securing, the desirable mechanical properties. Such heat treatment, may not be feasible under all conditions. For precision, work, where results justify the method, and for certain, combination of mechanical properties, tempering for long, periods of time in a lower temperature range provides, a reliable method of getting the desired results., Fig 10 illustrates how the mechanical properties of hardened, steel can be modified by tempering., Surface Hardening of Steels, Most of the components must have a hard, wear-resisting, surface supported by a tough, shock-resisting core for, better service condition and longer life. This combination, of different properties can be obtained in a single piece of, steel by surface hardening. (Fig 11), Types of surface hardening, , 60, , •, , Case hardening, , •, , Nitriding, , •, , Flame hardening, , •, , Induction hardening, , Workshop Calculation & Science : (NSQF) Exercise 2.6.13, , Copyright free, under CC BY Licence

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Case hardening, Parts to be hardened by this process are made from a, steel with a carbon content of 0.15% so that they will not, respond to direct hardening., The steel is subjected to treatment in which the carbon, content of the surface layer is increased to about 0.9%., When the carburised steel is heated and quenched, only, the surface layer will respond, and the core will remain soft, and tough as required. (Fig 12), The surface which must remain soft can be insulated, against carburising by coating it with suitable paste or by, plating it with copper., Case hardening takes place in two stages., •, , Carburising in which the carbon content of the surface, is increased., , •, , Heat treatment in which the core is refined and the, surface hardened., , Carburising, In this operation, the steel is heated to a suitable, temperature in a carbonaceous atmosphere, and kept at, that temperature until the carbon has penetrated to the, depth required., The carbon can be supplied as a solid, liquid or gas., In all cases, the carbonaceous gases coming from these, materials penetrate (diffuse) into the surface of the workpiece, at a temperature between 880° and 930°C., Pack carburising (Fig 13), The parts are packed into a suitable metal box in which, they are surrounded by the carburising medium., , The lid is fitted to the box and sealed with fireclay and, tied with a piece of wire so that no carbon gas can escape, and no air can enter the box to cause decarburisation., The carburising medium can be wood, bone, leather or, charcoal, but an energiser, such as barium carbonate, is, added to speed up the process., Liquid carburising, Carburising can be done in a heated salt-bath. (Sodium, carbonate, sodium cyanide and barium chloride are typical, carburising salts.) For a constant time and temperature of, carburisation, the depth of the case depends on the, cyanide content., Salt-bath carburising is very rapid, and is not always, suitable because it produces an abrupt change in the, carbon content from the surface to the core. This produces, a tendency for the case to flake., This is suitable for a thin case, about 0.25 mm deep. Its, advantage is that heating is rapid and distortion is minimised., , Workshop Calculation & Science : (NSQF) Exercise 2.6.13, , Copyright free, under CC BY Licence, , 61

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Gas carburising, The work is placed in a gas-tight container which can be, heated in a suitable furnace, or the furnace itself may be the, container., The carburising gas is admitted to the container, and the, exit gas in vented., , If the part is not required to resist shock, it is unnecessary, to carry out the core refining operation; in these conditions,, a coarse martensite at the surface may not cause trouble,, and so this part may be quenched directly after carburising., Fig 15 illustrates the appearance of the structure, across its section produced by case hardening., , The gas such as methane or propane may be fed directly, into the container in which the work is placed., In a continuous gas carburising furnace, the carburising,, quenching and tempering processes are carried out in, sequence in the same closed furnace as they progress on, a conveyer from one operation to the next., Fig 14 illustrates the appearance of the structure across its, section produced by carburising., , Nitriding, In the nitriding process, the surface is enriched not with, carbon, but with nitrogen. There are two systems in, common use, gas nitriding and salt bath nitriding., Gas nitriding, The gas nitriding process consists of heating the parts at, 500°C in a constant circulation of ammonia gas for up to, 100 hours., Heat treatment, After the carburising has been done, the case will contain, about 0.9% carbon, and the core will still contain about, 0.15% carbon. There will be a gradual transition of carbon, content between the case and the core., Owing to the prolonged heating, the core will be coarse,, and in order to produce a reasonable toughness, it must be, refined., To refine the core, the carburised steel is reheated to about, 870°C and held at that temperature long enough to produce, a uniformity of structure, and is then cooled rapidly to, prevent grain growth during cooling., The temperature of this heating is much higher than that, suitable for the case, and, therefore, an extremely brittle, martensite will be produced., The case and the outer layers of the core must now, be refined., The refining is done by reheating the steel to about, 760°C, to suit the case, and quenching it., Tempering, Finally the case is tempered at about 300°C to relieve the, quenching stresses., , 62, , During the gas nitriding process the parts are in an, externally heated gas-tight box, fitted with inlet and, outlet bores for the ammonia gas which supplies the, nitrogen. At the completion of the 'soaking' the, ammonia is still circulated until the temperature of the, steel has fallen to about 150°C, when the box is opened,, and the cooling completed in air. Nitriding causes a, film to be produced on the surface but this can be, removed by light buffing operation., Nitriding in salt-bath, Special nitriding baths are used for salt-bath nitriding., This process is suitable for all alloyed and unalloyed, types of steel, annealed or not annealed, and also for cast, iron., Process, The completely stress-relieved workpieces are preheated, (about 400°C) before being put in the salt-bath (about, 520°C-570°C). A layer 0.01 to 0.02 mm thick is formed, on the surface which consists of a carbon and nitrogen, compound. The duration of nitriding (half an hour to, three, hours) depends on the cross-section of the, workpiece . (It is much shorter than for gas nitriding.), After being taken out of the bath, the workpieces are, quenched and washed in water, and dried., , Workshop Calculation & Science : (NSQF) Exercise 2.6.13, , Copyright free, under CC BY Licence

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Advantages, , Induction hardening, , The parts can be finish-machined before nitriding because, no quenching is done after nitriding, and, therefore,, they will not suffer from quenching distortion., , This is a production method of surface hardening in which, the part to be surface-hardened is placed within an, induction coil through which a high frequency current is, passed. (Fig 18) The depth of penetration by the heating, becomes less, as the frequency increases., , In this process the parts are not heated above the critical, temperature, and, hence warping or distortion does not, occur., The hardness and wear- resistance are exceptional. There, is a slight improvement in corrosion-resistance as well., Since the alloy steels used are inherently strong when, properly heat treated, remarkable combinations of strength, and wear-resistance are obtained., Flame hardening, In this type of hardening, the heat is applied to the surface, of the workpiece by specially constructed burners. The, heat is applied to the surface very rapidly, and the work, is quenched immediately by spraying it with water., (Figs16 and 17) The hardening temperature is generally, about 50° C higher than that for full hardening., , The depth of hardening for high frequency current is 0.7 to, 1.0mm. The depth of hardening for medium frequency, current is 1.5 to 2.0mm. Special steels and unalloyed, steels with a carbon content of 0.35 to 0.7% are hardened., After induction hardening of workpieces, stress-relieving, is necessary., Advantages, The following are the advantages of this type of hardening., •, , The depth of hardening, distribution in width and the, temperature are easily controllable., , •, , The time required and the distortion due to hardening, are very small., , •, , The surface remains free from scale., , •, , This type of hardening can easily be incorporated in, mass production., , Heat treatment of high speed steel, High speed steels get their name from the fact that they, may be operated as cutting tools at much higher, speeds than is possible with plain carbon tool steels., Since the maximum hardness of high speed steel is, obtained on tempering at high temperatures, it can be, operated as a cutting tool in the same temperature range, without loss of hardness. That is, the rise in temperature, due to friction will not reduce the temper of the tool point, as it cuts into the steel. Tool steels are often annealed, for softening before machining or forming, and also for, obtaining grain refinement., The workpiece is maintained at the hardening temperature, for a very short period only, so that the heat is not, conducted to more than the necessary depth into the, workpiece., , Annealing, Soak at 900°C for about four hours. Then cool slowly at, not more than 20°C per hour to 600°C. It may then be, cooled to room temperature in still air., , Steels used for surface hardening by the flame hardening, method will have a carbon content of 0.35% to 0.7%., Workshop Calculation & Science : (NSQF) Exercise 2.6.13, , Copyright free, under CC BY Licence, , 63

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Hardening, Preheat to 850° C slowly to prevent cracking. Then heat, rapidly in a salt-bath furnace approximately to 1250°C, depending upon the alloy used. This rapid heating in a, salt-bath furnace reduces grain growth and prevents, oxidation of steel during heating. Quench in an air blast, or oil depending on the mass of the component. When a, salt bath furnace is not available, this can be minimised, in a double chamber muffle furnace (Fig 19) by using the, excess fuel to give a carburising atmosphere. However,, this reduces the combustion efficiency, and there may, be some difficulty in reaching the hardening temperature., With high temperature salts available, modern practice, favours the use of the salt-bath furnace. (Figs 20 and 21), , Secondary hardening (Tempering ), This is sometimes called tempering. However, this term is, not strictly true. Not only does secondary hardening, increase the toughness of the steel, it also increases, the hardness whereas tempering increases the toughness, at the expense of hardness. The effect of secondary, hardening helps these steels to work effectively at higher, temperatures than plain carbon tool steels., Heat treatment of non ferrous metals, Annealing, Like steel, non-ferrous metals can be softened by heating, and allowing to cool. However, since they cannot be, hardened by rapid cooling, the rate of cooling is, comparatively unimportant. In fact, copper components, are often quenched in water. This not only saves time, but also cleans the black oxide film from the surface of, the components because of the rapid contraction., Hardening, Only a very few non-ferrous alloys can be hardened by, heating and quenching like steel, and most non-ferrous, metals are hardened by cold working. As anyone, who, has worked as a coppersmith will know, the metal starts, to work-harden as it is beaten to shape and it must be, annealed from time to time to prevent it from cracking., , 64, , Workshop Calculation & Science : (NSQF) Exercise 2.6.13, , Copyright free, under CC BY Licence

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Assignment A, 1 a What is two purpose of Hardening., , iv What is the heating method for normal hardening?, , b What is the purpose of normalising., , a Temperature controlled furnace, , c Write four annealing purposes., , b Electric induction coil heat, , d What is the purpose of tempering., , c By use of Oxy-Acetylene torche, , e What is the purpose of annealing., , d By Ammonia gas, , f, , Which cooling process is suitable for annealing., , g Which cooling process is suitable for normalising., , v Which quenching medium for hardening of H.S, Steel?, , h What type of structure in the normalising process., , a Water spray, , i, , What reason the steel is reheated (after hardening), to a suitable temperature (180°C to 650°C)., , b Still air, , j, , Why steel is reheated., , d Air blast or oil, , 2 i, , What is the heating method for nitriding?, a Temperature controlled furnace, , c Bath of water, , vi Which quenching medium for hardening of carbon, steel?, , b Electric induction coil heat, , a Water spray, , c By use of Oxy-Acetylene torche, , b Still air, , d By Ammonia gas, , c Bath of water, , ii What is the heating method for flame hardening?, a Temperature controlled furnace, b Electric induction coil heat, c By use of Oxy-Acetylene torche, d By Ammonia gas1 Hardening Process, , d Air blast or oil, vii Which quenching medium for normalising low carbon, or medium carbon steel?, a Water spray, b Still air, c Bath of water, , iii What is the heating method for induction hardening?, , d Air blast or oil, , a Temperature controlled furnace, b Electric induction coil heat, c By use of Oxy-Acetylene torche, , viii Which quenching medium for flame hardening?, a Water spray, b Still air, , d By Ammonia gas, , c Bath of water, d Air blast or oil, , B MCQ, 1 The process of changing the structure of a metal by, heating and cooling is, A Heat treatment, , B Machining, , C Hot rolling, , D Melting, , 2 The three stages of heat treatment is Heating, Soaking,, and, A Smelting, , B Quenching, , C Hardening, , D Tempering, , 3 The quenching oil viscosity should be., A normal, , B medium, , C low, , D high, , 4 Which relieves stress and strain., A Hardening, , B Normalising, , C Tempering, , D Annealing, , Workshop Calculation & Science : (NSQF) Exercise 2.6.13, , Copyright free, under CC BY Licence, , 65

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5 Which process produce equilibrium conditions., , 8 What is the purpose of Nitriding process., , A Annealing and Normalising, , A partial, , B whole, , B Hardening and Tempering, , C case, , D surface, , C Annealing and Hardening, , 9 Which temperature change structure to austenite., , D Normalising and Tempering, 6 Which quenching medium is commonly used., A Water & Lead, , B Water & Oil, , C Water & Brine solution D Lead & Brine solution, 7 Which process is used to reduce the brittleness., A Annealing, , B Normalising, , C Tempering, , D Case hardening, , A lower, , B upper, , C higher, , D larger, , 10 Which temperature change structure of steel to, austenite, A lower, , B upper, , C smaller, , D lesser, , Key Answers, A, 1 a i, , f, , To add cutting ability, , g Air, , ii To increase water resistance, b To refine grain structure of the steel, c, , i To relieve strain and stress, ii To eliminate strain/hardness, iii To improve machinability, , h Grain, i, , Ductility, , j, , Tempering, , 2 i d, , ii c, , iii b, , iv a, , To remove extreme brittleness caused by hardening, , ii To reduce toughness and shock resistance, , vi c, , vii b viii a, , 1 A, , 2 B, , 3 C, , 4 D, , 5 A, , 6 B, , 7 C, , 8 D, , 9 A, , 10 B, , e Ductility, , 66, , v d, , B MCQ, , iv To soften the steel, d i, , Room temperature, , Workshop Calculation & Science : (NSQF) Exercise 2.6.13, , Copyright free, under CC BY Licence

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Profit and loss - Simple problems on profit & loss, Definition of 'profit and loss statement (P&L), A profit and loss statement (P&L) is a financial statement, that summarizes the revenues, costs and expenses, incurred during a specific period of time, usually a year., These records provide information about a company's, ability - to generate profit by increasing revenue, reducing, costs, or both. The P&L statement is also referred to as, "statement of profit and loss", "income statement",, "statement of operations", "statement of financial results", and "income and expense statement"., Profit and loss, , Exercise 2.7.14, , 8 Cost price: (C.P), , ⎛, , C.P = ⎜, , ⎞, , 100, , ⎝ (100 + Gain %), , x S.P ⎟, , ⎠, , 9 Cost price: (C.P), , ⎛, , C.P = ⎜, , ⎞, , 100, , ⎝ (100 − Loss %), , x S.P ⎟, , ⎠, , 10 If an article is sold at a gain of say 35%, then S.P.=135%, of C.P., , Important facts, , 11 If an article is sold at a loss of say, 35% then S.P=65%, of C.P., , Cost price, , Example, , The price, at which an article is purchased is called its cost, price, abbreviated as C.P., , 1 A dealer bought a television set for Rs.10,000 and, sold it for Rs.12,000. Find the profit/loss made by, him for 1 television set. If he had sold 5 television, sets, find the total profit / loss, , Selling price, The price at which an article is sold, is called its selling, prices, abbreviated as S.P., , Solution, , Profit or gain, , Selling price of the television set = Rs.12,000, , If S.P. is greater than C.P., the seller is said to have a profit, or gain., , Cost price of the television set, , = Rs.10,000, , S.P. > C.P., there is a profit, , Loss, , Profit, , = S.P. - C.P, = 12000-10000, , If S.P. is less than C.P., the seller is said to have incurred, a loss., , Profit, , = Rs.2,000, , Discount, , Profit on 1 television set, , = Rs.2000, , The reduction given to the selling price of a product is the, discount., , Profit on 5 television sets, , = 2000 x 5, , Important formulae, , = Rs.10,000, , 2 Loss=(C.P)-(S.P), , 2 Sanjay bought a bicycle for Rs.5000. He sold it for, Rs.600 less after two years. Find the selling price, and the loss percent., , 3 Loss or gain always depends on C.P., , Solution, , 1 Profit or Gain=(S.P)-(C.P), , 4 Profit/gain is always expressed in %., , Cost price of the bicycle, , ⎛ Gain x 100 ⎞, Gain% = ⎜, ⎟, C.P., ⎝, ⎠, , Loss, , = Rs.600, , Selling price, , = Cost price - loss, = 5000 - 600, , 5 Loss percentage: (Loss %), ⎛ Loss x 100 ⎞, Loss % = ⎜, ⎟, C.P. ⎠, ⎝, , Selling price of the bicycle, Loss %, , 6 Selling price: (S.P), , ⎛ 100 + Gain%, ⎞, x C.P ⎟, 100, ⎝, ⎠, ⎛ (100 − loss %), ⎞, x C.P ⎟, 100, ⎝, ⎠, , SP = ⎜, , = Rs.4400, =, , Loss, x 100, C.P., , = 600 x 100, 5000, , SP = ⎜, , 7 Selling price: (S.P), , = Rs.5000, , Loss, , = 12%, , 3 A man bought an old bicycle for Rs.1250. he spent, Rs.250 on its repairs. He then sold it for Rs.1400., Find his gain or loss %., Solution, Cost price of the bicycle = Rs.1250, , Copyright free, under CC BY Licence, , 67

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Repair Charges, , = Rs.250, , Loss, , Total cost price, , = 1250+250 = Rs.1500, , S.P. = C.P. - Loss, , Selling price, , = Rs.1400, , = 50 - 5 = 45, , C.P > S.P, there is a loss, Loss, , = Rs.5, , Selling price of the pen, , = Cost price - Selling price, , 1500 - 1400 = 100, , 6 Find the total amount if 12% of it is ₹ 1080, Let the total amount be 'x ', , Loss, , = Rs.100, , Loss %, , =, , Loss, x 100, C.P., , 12, xx, 100, , =, , 100, x 100, 1500, , x, , =, , 20, 2, = 6 % (or) 6.67%, 3, 3, , Loss, , = Rs.45, , Given: 12% of the total amount = Rs.1080, , = 6.67%, , = 1080, =, , = ₹ 9000, , The total amount, , = Rs.9000, , Applications of profit and loss, , Profit percentage or loss percentage is always, calculated on the cost price of the article., 4 A fruit seller bought 8 boxes of grapes at Rs.150, each. One box was damaged. He sold the, remaining boxes at Rs.190 each. Find the profit/, loss percent., , In this section, we learn to solve problems on applications, of profit and loss., i, , Illustration of the formula for S.P., , Consider the following situation, Rajesh buys a pen for Rs.80 and sells it to his friend., If he wants to make a profit of 5%, can you say the price, for which he would have sold?, , Solution, Cost price of 1 box of grapes, , = Rs.150, , Cost price of 8 boxes of grapes, , = 150 x 8, = Rs.1200, , Number of boxes damaged, , =1, , Number of boxes sold, , =8-1=7, , Selling price of 1 box of grapes, , = Rs.190, , (Rajesh bought the pen for Rs.80 which is the cost price, (C.P.). When he sold, he makes a profit of 5% which is, calculated on the C.P.),  Profit = 5% of C.P. =, , S.P. = C.P. + Profit, = 80 + 4 = Rs.84,  The price for which Rajesh would have sold = Rs.84, , Profit = Selling price - Cost price, , The same problem can be done using the formula., , = 1330-1200, = 130, Profit, Percentage of profit, , =, , Selling price (S.P), , =, , (100 + Profit %) x C.P, 100, , = Rs.130, Profit, , =, , x 100, , C.P, , =, , Profit, , 130, 1200, , =, x 100, , (100 + 5), 100, , x 80, , 105, x 80 = Rs. 84, 100, , ii Illustration of the formula for C.P, , = 10.83, , Consider the following situation, , = 10.83%, , Suppose a shopkeeper sells a wrist watch for Rs. 540, making a profit of 5%, then what would have been the, cost of the watch?, , 5 Ram, the shopkeeper bought a pen for Rs.50 and, then sold it at a loss of Rs.5. Find his selling price., Solution, Cost price of the pen, , x 80 = Rs.4, , Since there is a profit, S.P > C.P., , = Rs.1330, S.P.>C.P, there is a profit, , 5, 100, , Selling price of 7 boxes of grapes = 190 x 7, , 68, , 1080x100, 12, , = Rs.50, Workshop Calculation & Science : (NSQF) Exercise 2.7.14, , Copyright free, under CC BY Licence

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(The shopkeeper had sold the watch at a profit of 5% on, the C.P. Since C.P. is not known, let us take it as Rs., 100), , The watch would have cost Rs. 514.29 for the shopkeeper., The above problem can also be solved by using the formula, method, , Profit of 5% is made on the C.P., ,  Profit = 5% of C.P., =, , = Rs. 514.29, , 5, x 100 = Rs. 5, 100, , We know S.P. = C.P + Profit, , ⎛, ⎞, 100, ⎟⎟ x S.P, C.P. = ⎜⎜, ⎝ 100 +Profit % ⎠, , 100, ⎞, = ⎛⎜, x 540 ⎟, ⎝ 100 + 5, ⎠, , = 100 + 5, = Rs. 105, Here, if S.P. is Rs.105, then C.P. is Rs. 100, When S.P. of the watch is Rs. 540, C.P =, , =, , 100, x 540 = Rs. 514.29, 105, , 540 x 100, 105, , We now summarize the formulae to calculate S.P. and C.P. as follows., 1 When there is a profit, , 1 When there is a loss, , 100, ⎛, ⎞, ⎟ x S.P., C.P = ⎜, ⎝ 100 + Profit% ⎠, 2 When there is a profit, , 2 When there is a loss, , ⎛ 100 + Profit% ⎞, S.P = ⎜, ⎟ x C.P., 100, ⎝, ⎠, Example, 1 Hameed buys a colour T.V. set for Rs. 15,200 and, sells it at a loss of 20%. What is the selling price, of the T.V. set?, Method - I, , = 80 x15200, 100, = Rs. 12,160, , Method - I, , Loss is 20% of the C.P., , Loss of 15% means,, , 20 x 15200 = Rs. 3040, 100, S.P, 15200 - 3040, , If C.P. is Rs. 100, Loss = Rs. 15, Therefore, S.P. would be, , = C.P - Loss, = Rs.12160, , Method - II, , (100-15) = Rs. 85, If S.P. is Rs. 85, C.P. is Rs. 100, When S.P is Rs. 13600 then, , C.P = Rs 15,200, , C.P = 100 x13600 = Rs. 16000, 85, , Loss = 20%, , =, , ⎛ 100 − Loss% ⎞, S.P = ⎜, ⎟ X C.P., 100, ⎝, ⎠, , 2 A scooty is sold for Rs. 13600 and fetches a loss of, 15%. Find the cost price of the scooty., , Solution, , S.P =, , 100, ⎛, ⎞, C.P = ⎜⎝ 100 − Loss% ⎟⎠ X S.P., , 100 − Loss%, xC.P., 100, , 100 − 20, x15200, 100, , Method - II, Loss, , = 15%, , S.P., , = Rs. 13600, , C.P., , 100, ⎛, ⎞, = ⎜, ⎟ x S.P, ⎝ 100 − Loss% ⎠, , Workshop Calculation & Science : (NSQF) Exercise 2.7.14, , Copyright free, under CC BY Licence, , 69

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=, , 100, x 13600, 100 − 15, , =, , 100, x 13600, 85, , Method - I, The discount is given in percentage, Hence, the M.P. is taken as Rs. 100, , = Rs. 16000, , Rate of discount, , Hence the cost price of the scotty is Rs. 16000, Discount, Discount is the reduction in value on the marked price or, list price of the article., The market price of a product is Rs.550, Amount paid by pooja to the shop keeper is Rs. 440, Discount, , Solution, , 5, x 100, 100, , Amount of discount =, Selling price, , = M.P - Discount, = 100 - 5 = Rs. 95, , If S.P. is Rs. 95, then M.P. is Rs.100, When S.P. is Rs. 5225, , = Rs. 550 - Rs. 440, M.P., , = Rs. 110, = Marked price - Selling price, Hence we conclude the following, Discount, , = 5%, , =, , 100, x 5225, 95, , M.P of the almirah = Rs. 5500, Method - II, , = Marked price - Selling price, , Selling price = Marked price - Discount, , S.P, Discount, , Marked price= Selling price + Discount, Example, , = Rs. 5225, = 5%, , M.P, , =?, , M.P, , 100, ⎞, = ⎛⎜, ⎟ x S.P., ⎝ 100 − Discount% ⎠, , 1 A bicycle marked at Rs. 1500 is sold for Rs. 1350., What is the percentage of discount?, Marked price = Rs. 1500, , ⎛ 100 ⎞, =⎜, ⎟ x 5225, ⎝ 100 − 5 ⎠, , Selling price = Rs. 1350, , = Rs. 5500, , = Rs. 150, , 4 A shopkeeper allows a discount of 10% to his, customers and still gains 20%. Find the marked, price of an article which costs Rs.450 to the, shopkeeper., , Discount for Rs. 1500, , = Rs. 150, , Solution, , Discount for Rs. 100, , =, , Amount of discount = Marked price - Selling price, = 1500 - 1350, , Method - I, , 150, x 100, 1500, , Let M.P be Rs. 100, Discount, , Percentage of discount = 10%, , = 10 of M.P = 10 x 100, 100, 100, , 2 The list price of a Frock is Rs.220. A discount of, 20% on sales is announced. What is the amount, of discount on it and its selling price?, Amount of discount =, , Discount, x M.P., 100%, , Amount of discount = 20 x 220 = Rs. 44, 100, , = 10% of M.P, , = Rs. 10, S.P, , = M.P - Discount, = 100 - 10, = Rs. 90, , Gain, , = 20% of C.P., , Selling price of the frock = Marked price - Discount, =, , 220 - 44= Rs. 176, 3 An almirah is sold at Rs. 5225 after allowing a, discount of 5%. Find its marked price., , 70, , S.P, , 20, x 450 = Rs. 90, 100, , = C.P + Gain, = 450 + 90 = Rs. 540, , Workshop Calculation & Science : (NSQF) Exercise 2.7.14, , Copyright free, under CC BY Licence

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If S.P. is Rs. 90, then M.P is Rs. 100, M.P., , =, , 540 x 100, = Rs. 600, 90, , The M.P. of an article = Rs. 600, Method - II, Discount = 10%,, , Gain = 20%, , Method - II, Discount, , = 10%, , Gain, , = 10%, , M.P., , = Rs. 220, , C.P., , =, , 100 - Discount%, x M.P, 100 +Gain %, , =, , 100 - 10, x 220, 100 +10, , =, , 90, x 220, 110, , C.P. = Rs. 450 , M.P. = ?, M.P, , 100 + Gain%, x C.P, =, 100 − Discount%, , (100 + 20)x450, =, (100 − 10), =, , 120, x 450, 90, , = Rs. 600, 5 A dealer allows a discount of 10% and still gains, 10%. What is the cost price of the book which is, marked at Rs. 220?, Solution, , = Rs. 180, 6 A trader buys an article for Rs. 1200 and marks it, 30% above the C.P. He then sells it after allowing, a discount of 20%. Find the S.P. and profit, percent., Solution, Let C.P. of the article be Rs. 100, M.P. =30% above C.P. = Rs. 130, , Method - I, , If C.P is Rs. 100, then M.P. is Rs. 130, M.P, , Discount, , = Rs.220, = 10% of M.P, , When C.P. is Rs. 1200,, M.P., , =, , 10, =, x 220, 100, , = Rs. 22, S.P, , 1200 X 130, 100, , Discount = 20% of 1560 =, , 20, x1560, 100, , Discount = 20% of 1560 =, , 20, x1560, 100, , = M.P - Discount, = 220 - 22, = Rs. 198, , Let, C.P. be Rs. 100, Gain, , = 10% of C.P., 10, =, x 100, 100, , = Rs. 10, S.P., , = Rs. 1560, , = Rs. 312, S.P, , = 1560 - 312, = Rs. 1248, Profit, , = Rs. 48, , = 100 + 10, , Profit % =, , If S.P. is Rs. 110, then C.P is Rs. 100, , Profit, x 100, C.P., , = 48 x 100, 1200, , When S.P. is Rs. 198,, =, , = S.P - C.P, = 1248 - 1200, , = C.P + Gain, = Rs. 110, , = M.P. - Discount, , 198 x 100, 110, , = 4%, , = Rs. 180, , Workshop Calculation & Science : (NSQF) Exercise 2.7.14, , Copyright free, under CC BY Licence, , 71

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Summary, Percent means per hundred. A fraction with its denominator 100 is called a percent., In case of profit, we have Profit = S.P - C.P., Profit % =, , In case of loss, we have Loss = C.P - S.P., , Profit, x 100, C.P., , Loss % =, , Loss, x 100, C.P., , S.P, , ⎛ 100 + Pr ofit% ⎞, ⎟ x C.P., = ⎜, 100, ⎝, ⎠, , S.P, , ⎛ 100 − Loss% ⎞, ⎟ x C.P., = ⎜, 100, ⎝, ⎠, , C.P, , 100, ⎛, ⎞, ⎟ x S.P., = ⎜, 100, +, Pr, ofit, %, ⎝, ⎠, , C.P, , 100, ⎛, ⎞, ⎟ x S.P., = ⎜, ⎝ 100 − Loss% ⎠, , M.P., , =, , 100, x S.P, 100 − Discount %, , = 100 − Discount % x M.P, 100, C.P.= 100 − Discount % x M.P, 100 + Profit %, S.P., , M.P.=, , 100 + Profit %, x C.P, 100 − Discount %, , Discount percent =, , Discount, x 100, M.P., , Discount is the reduction given on the Marked price., Selling price is the price payable after reducing the discount from the marked price., Discount = M.P. - S.P., , Assignment, 1 Find the cost price if the product is sold at Rs. 572, with a profit of Rs. 72., , 6 Find out the profit amount if the C.P. and S.P. are Rs., 2500 and Rs. 2700 respectively., , 2 Find the C.P if the product is sold at Rs.1973 with a, profit of Rs. 273, , 7 Calculate the percentage of loss if the C.P. and S.P, are Rs. 40 and Rs. 38 respectively., , 3 Find the selling price if the cost price is Rs. 7282 with, a profit of Rs. 208, , 8 A computer table bought at Rs. 1150 with Rs. 50 as a, transport charge. Calculate the S.P. if the profit is of, 5%, , 4 Find out the S.P. if the C.P. is Rs. 9684 with a loss of, Rs. 684, 5 Find out the profit percentage if the C.P is Rs. 320 and, S.P is Rs. 384., , 9 By selling a table for Rs. 1320 with a gain of 10%., Find the C.P., 10 The C.P. of 16 bolts is equal to the S.P. of 12 bolts., Find the gain percent., , Key Answers, 1 Rs. 500, , 6 Rs. 200, , 2 Rs. 2246, , 7 5%, , 3 Rs. 7490, , 8 Rs. 1260, , 4 Rs. 9000, , 9 Rs. 1200, , 5 Rs. 20%, , 10 33 13 %, , 72, , Workshop Calculation & Science : (NSQF) Exercise 2.7.14, , Copyright free, under CC BY Licence

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Profit and loss - Simple and compound interest, Interest, When we borrow (or lent) money we pay (or receive) some, additional amount in addition to the original amount. This, additional amount that we receive is termed as Interest. It, is denoted as 'I'. Money can be borrowed/lent deposited, in banks to get Interest. The amount borrowed//lent is, called the principal. (P), The principal added to the Interest is called the Amount(A)., Amount = Principal + Interest, A=P+I, Interest depends on principal and duration of time. But it, also depends on one more factor called the rate of interest., Rate of interest is the amount calculated annually for `100., (ie) if rate of interest is 10% per annum, then interest is, ₹ 10 for ₹ 100 for 1 year., So,, , r=, , 100I, Pn, , n=, , 100 I, Pr, , P=, , 100I, rn, , 'n' is always calculated in years. When 'n' is given in, months or days, convert it into years., Example :, 12 Months = 1 year, 6 Months =, , 6, 1, year =, year, 12, 2, , 3 Months =, , 3, 1, year = year, 12, 4, , 146 days =, , 2, 146, year = year, 5, 365, , Interest depends on, Amount deposited or borrowed/lent - Principal - P, Period of time - mostly expressed in years - n, Example, , Rate of interest - r, This interest is termed as Simple interest., When the interest is paid on the principal only, it is called, simple interest., Calculation of interest, If 'r' is the rate of interest, Principal is 100,, r, The interest for 1 year = 100 x 1 x, 100, r, for 2 years = 100 x 2 x, 100, , for 3 years = 100 x 3 x, , 1 Vimal invested ₹ 3000 for 1 year at 7% per annum., Find the simple interest and the amount received, by him at the end of one year., Solution, Principal (P), , = ₹ 3,000, , Number of years (n), , =1, , Rate of interest (r), , = 7%, , Interest(I), , =, , Pnr, 100, , =, , 3000 x 1 x 7, 100, , r, 100, , I = 210, , r, for n years = 100 x n x, 100, So,, , I=, , Amount(A), , Amount received by him (A) = ₹ 3,210, 2 Ramani invested ₹ 5000 for 2 years at 11% per, annum. Find the simple interest and the amount, received by him at the end of 2 years., , A=P+I, Pnr, 100, , Solution, , nr ⎞, ⎛, ⎟, A = P ⎜1 +, ⎝ 100 ⎠, Interest = Amount - Principal, , The other formulae derived from, I=, , =P+I, = 3000 + 210, , Pnr, 100, , A=P+, , Exercise 2.7.15, , Principal (P), , = ₹ 5,000, , Number of years (n), , = 2 yrs, , Rate of interest (r), , = 11%, , Interest(I), , =, , Pnr, 100, , Pnr, 100, , 73, , Copyright free, under CC BY Licence

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=, , 3x8 ⎞, ⎛, ⎟, = 7500⎜1 +, 2 x 100 ⎠, ⎝, , 5000 x 2 x 11, 100, , ⎛ 28 ⎞, = 7500⎜ ⎟, ⎝ 25 ⎠, = 300 x 28, , = 1100, I = ₹ 1100, Amount(A), , =P+I, , = 8400, , = 5000 + 1100, , A = ₹ 8400, , Amount received by him (A) = ₹ 6,100, 3 Find the simple interest and the amount due, on ₹ 7,500 at 8% per annum for 1 year 6 months., Solution, Principal (P), , = ₹ 7,500, , Number of years (n), , = 1 yr. 6 months, 6, =1, yrs, 12, , =1, r, , Interest (I), , = 8400 - 7500, Interest(II) = ₹ 900, Amount = ₹ 8,400, 4 Find the simple interest and the amount due, on ₹ 6,750 for 219 days at 10% per annum., Solution, Principal (P), , = ₹ 6,750, , Number of years (n), , = 219 days, , 1, 3, yrs =, yrs., 2, 2, , =, , = 8%, , Interest(I), , =, , =, , =, , Pnr, 100, , = 10%, , I, , =, , Pnr, 100, , =, , 6750 x 3 x 10, 5 x 100, , 3, x8, 2, 100, , = 405, , 7500 x 3 x 8, 2 x 100, , I = ₹ 405, =P+I, , Amount, , = 6750 + 405, , I = ₹ 900, , Amount due on, , = ₹ 8,400, , Interest = ₹ 900, Amount = ₹ 8,400, Alternative method, Principal (P), Number of years (n), Rate of interest (r), A, , = ₹ 7,500, 3, =, yrs, 2, = 8%, , nr ⎞, ⎛, ⎟, = P ⎜1 +, ⎝ 100 ⎠, 3, ⎞, ⎛, x8⎟, ⎜, 2, ⎟, ⎜, = 7500 1 +, ⎜, 100 ⎟, ⎟, ⎜, ⎠, ⎝, , 74, , Amount due on = ₹ 7,155, , =P+I, = 7500 + 900, , 219, 3, year = year, 365, 5, , r, , 7500 x, , = 900, , Amount, , =A-P, , Interest(II) = ₹ 405, Amount = ₹ 7,155, 5 Ravi borrowed ₹ 4000 on 7 th june 2006 and, returned it on 19th August 2006. Find the amount, he paid, if the interest is calculated at 5% per, annum., Solution, Principal (P), , = ₹ 4,000, r, , = 5%, , Number of days, June, , = 24(30 - 6), , July, , = 31, , August, , = 18, , Total number of days, , = 73, , n, , Workshop Calculation & Science : (NSQF) Exercise 2.7.15, , Copyright free, under CC BY Licence, , = 73 days, =, , 73, year, 365

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=, , nr ⎞, ⎛, ⎟, = P ⎜1 +, ⎝ 100 ⎠, , Amount, , =, , Number of years, Alternative method, A, , nr ⎞, ⎛, ⎟, = P ⎜1 +, 100, ⎝, ⎠, , 1 ⎞, ⎛, ⎟, = 4000⎜1 +, ⎝ 100 ⎠, , 11000, , ⎛ nx5⎞, ⎟, = 10000⎜1 +, 100 ⎠, ⎝, , ⎛ 101 ⎞, ⎟, = 4000⎜, ⎝ 100 ⎠, , 11, 10, , =, , 1x 5 ⎞, ⎛, = 4000⎜1 +, ⎟, ⎝ 5 x 100 ⎠, , = 4,040, Amount, , 100 x 1000, 10000 x 5, = 2 years., , 1, year, 5, , = ₹ 4,040, , 6 Find the rate percent per annum when a principal, of ₹ 7,000 earns a S.I. of ₹ 1,680 in 16 months., , 20 + n, 20, , 11, x 20, 10, , = 20 + n, , 22, , = 20 + n, , 22 - 20 = n, Number of years = 2, , Solution, Principal (P), , = ₹ 7,000, n, , = 16 months, 16, 4, =, yr = yr, 12, 3, , I, , = ₹ 1,680, , r, , =?, , r, , 100I, =, Pn, , 8 A sum of money triples itself at 8% per annum, over a certain time. Find the number of years., Solution, Let principal be ₹ P, Amount, , 100 x 1680, =, 4, 7000 x, 3, , (r), , = 18%, , r, , = ₹ 3P, = 8%, , n, , =?, , I, , =A-P, = 300 - 100, , 100 x 1680 x 3, =, 7000 x 4, = 18, , Rate of interest, , = triple the principal, , I, , = ₹ 200, , n, , =, , 100 I, 100 x 200, =, Pr, 100 x 8, , n, , =, , 200, = 25, 8, , Number of years = 25, , 7 Vijayan invested `10,000 at the rate of 5% simple, interest per annum. He received ₹ 11,000 after, some years. Find the number of years., , 9 A certain sum of money amounts to ₹ 10,080 in 5, years at 8%. Find the principal., , Solution, , Solution, A, , = ₹ 11,000, , P, , = ₹ 10,000, , r, , = 5%, , I, , =A-P, = 11,000 - 10,000, = 1,000, , I, , = ₹ 1,000, , n, , =, , 100 I, Pr, , A, n, , = ₹ 10,080, = 5 years, , r, , = 8%, , P, , =?, , Amount (A), , ₹ 10080, , nr ⎞, ⎛, ⎟, = P ⎜1 +, ⎝ 100 ⎠, ⎛ 5x8⎞, ⎟, = P ⎜1 +, 100 ⎠, ⎝, , Workshop Calculation & Science : (NSQF) Exercise 2.7.15, , Copyright free, under CC BY Licence, , 75

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₹ 10080, , ⎛ 5x8⎞, ⎟, = P ⎜1 +, 100 ⎠, ⎝, , ₹ 10080, , ⎛7⎞, = P⎜ ⎟, ⎝5⎠, , 5, ₹ 10080 x 7, , =P, , Suppose he fails to pay the simple interest ₹ 2,000 at the, end of first year, then the interest ₹ 2,000 is added to the, old principal ₹ 50,000 and now the sum = P + I = ₹ 52,000, becomes the new principal for the second year for which, the interest is calculated., , =P, , Now in the second year he will have to pay an interest of, , 7,200, Principal, , =, , = ₹ 7,200, , Simple interest =, , 10 A certain sum of money amounts to ₹ 8,880 in 6, years and ₹ 7,920 in 4 years respectively. Find, the principal and rate percent., Solution, Amount at the end of 6 years, , = Principal +, interest for 6 years, = P + I6 = 8880, , Amount at the end of 4 years, , = Principal +, interest for 4 years, , 50000 x 1 x 4, = ₹ 2,000, 100, , =, , Pxnxr, 100, , 52000 x 1 x 4, = `2,080, 100, , Therefore Rajesh will have to pay more interest for the, second year., This way of calculating interest is called compound Interest., If the interest is paid on the principal as well as on the, accrued interest, it is called compound interest., , = 8880 - 7920, , Generally in banks, insurance companies, post offices, and in other companies which lend money and accept, deposits, compound interest is followed to find the interest., , = 960, , Example, , Interest at the end of 2 years, , = `960, , Interest at the end of 1st years, , =, , Ram deposited ₹ 8,000 with a finance company for 3, years at an interest of 15% per annum. What is the, compound interest that Ram gets after 3 years?, , = P + I4 = 7920, I2, , 960, 2, , = 480, Interest at the end of 4 years, , Solution, Step 1 :, , = 480 x 4, , Principal for the first year, , = ₹ 8,000, , Interest for the first year, , =, , = 1,920, P + I4, P + 1920, , = 7920, P, , = 7920 - 1920, , P, , = 6,000, , Principal, , = `6,000, r, , Rate of interest, , = 7920, , =, , 100I, Pn, , =, , 100 x1920, 6000 x 4, , 80000 x 1 x 15, 100, = ₹ 1,200, , =, , Amount at the end of first year, , = 8,000 + 1,200, = ₹ 9,200, Step 2 :, Principal for the 2nd year, , = ₹ 9,200, , Interest for the 2nd year, , =, , Rajesh borrowed ₹ 50,000 from a bank for a fixed time, period of 2 years. at the rate of 4% per annum., Rajesh has to pay for the first year., Simple interest =, 76, , =P+I, , (r) = 8%, , Compound Interest, , Pxnxr, 100, , Pxnxr, 100, , Workshop Calculation & Science : (NSQF) Exercise 2.7.15, , Copyright free, under CC BY Licence, , Pxnxr, 100, , 9200 x 1 x 15, 100, = ₹ 1,380, , =

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Amount at the end of 2nd year, , =P+I, = 9,200 + 1,380, , r ⎞, ⎛, P⎜1 +, ⎟ x 1x r, 100, ⎠, = ⎝, 100, , Interest for the 2 year, nd, , = ₹ 10,580, , (using the S.I. formula), , Step 3 :, Principal for the 3rd year, , = ₹ 10,580, , Interest for the 3rd year, , =, , Pxnxr, 100, , r ⎞, r, ⎛, ⎟x, = P⎜1 +, 100, 100, ⎝, ⎠, Amount at the end of 2nd year = P + I, , r ⎞, r ⎞, r, ⎛, ⎛, ⎟ + P⎜1 +, ⎟x, = P⎜1 +, ⎝ 100 ⎠, ⎝ 100 ⎠ 100, , 10580 x 1 x 15, =, 100, = ₹ 1,587, , Amount at the end of 3rd year, , r ⎞⎛, r ⎞, ⎛, ⎟ ⎜1 +, ⎟, = P⎜1 +, 100, 100, ⎝, ⎠⎝, ⎠, , =P+I, = 10,580 + 1,587, , r ⎞, ⎛, ⎟, = P⎜1 +, ⎝ 100 ⎠, , = ₹ 12,167, Hence, the compound interest that Ram gets after 3 years, is, A - P = 12,167 - 8,000 = ₹ 4,167, , 2, , Step 3 :, 2, , Deduction of formula for compound interest, , r ⎞, ⎛, ⎟, Principal for the 3 year = P⎜1 +, ⎝ 100 ⎠, , The above method which we have used for the calculation, of compound interest is quite lengthy and cumbersome,, especially when the period of time very large. Hence we, shall obtain a formula for the computation of amount and, compound interest., , r ⎞, ⎛, P⎜1 +, ⎟ x 1x r, rd, Interest for the 3 year = ⎝ 100 ⎠, 100, , rd, , 2, , (using the S.I. formula), , Example, If the principal is P, Rate of interest per annum is r%, and the period of time or the number of years is n,, then we deduce the compound interest formula as, follows:, , Amount at the end of 3rd year = P + I, 2, , Step 1:, Principal for the first year, , =P, , Interest for the first year, , Amount at the end of first year, , =, , Pxnxr, 100, , =, , P x 1x r, Pr, =, 100, 100, , =P+I, =P+, , Pr, 100, , r ⎞, = P⎛⎜1 +, ⎟, ⎝ 100 ⎠, , 2, , r ⎞, r ⎞, r, ⎛, ⎛, ⎟ + P⎜1 +, ⎟ x, = P⎜1 +, 100, ⎝ 100 ⎠, ⎝ 100 ⎠, 2, , r ⎞ ⎛, r ⎞, ⎛, ⎟ ⎜1 +, ⎟, = P⎜1 +, 100, 100, ⎝, ⎠ ⎝, ⎠, 3, , r ⎞, ⎛, ⎟, = P⎜1 +, 100, ⎝, ⎠, Similarly, Amount at the end of n th year is, n, , r ⎞, ⎛, ⎟ and C.I. at the end of 'n' years is given, A = P⎜1 +, ⎝ 100 ⎠, by, Compound Interest (C.I) = A – P, n, , Step 2 :, Principal for the 2nd year, , 2, , r ⎞, r, ⎛, ⎟ x, = P⎜1 +, 100, 100, ⎝, ⎠, , r ⎞, = P⎛⎜1 +, ⎟, 100, ⎝, ⎠, , r ⎞, ⎛, ⎟ -P, (ie.) Compound Interest (C.I) = P⎜1 +, 100, ⎝, ⎠, , Workshop Calculation & Science : (NSQF) Exercise 2.7.15, , Copyright free, under CC BY Licence, , 77

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To compute compound interest, , A is given by, , Case 1 : Compounded Annually, When the interest is added to the principal at the end of, each year, we say that the interest is compounded, annually., Here,, , r ⎞, ⎛, ⎟, A = P⎜1 +, 100, ⎝, ⎠, , r ⎞, ⎛, ⎟, A = P⎜1 +, ⎝ 100 ⎠, , 5, , ⎡ 1 ⎛ r ⎞⎤, ⎟⎥ and C.I = A - P, ⎢1 + ⎜, ⎣ 4 ⎝ 100 ⎠⎦, , for 5 years, , for, , 1, years, 4, , Example, , n, , and C.I = A - P, , Find the C.I. on ₹ 15,625 at 8% p.a. for 3 years, compounded annually., , Case 2 : Compounded half-yearly (semi-annually), When the interest is compounded half-yearly, there are, two conversion periods in a year each after 6 months. In, such situations, the half-yearly rate will be half of the annual, rate, that is ()., , Solution, We know,, Amount after 3 years, , In this case,, ⎛, 1 ⎛ r ⎞⎞, ⎟⎟, A = P⎜⎜1 + ⎜, 2 ⎝ 100 ⎠ ⎟⎠, ⎝, , 3, , 8 ⎞, ⎛, ⎟, = 15625 ⎜1 +, ⎝ 100 ⎠, , 2n, , and C.I = A - P, , 2 ⎞, ⎛, ⎟, = 15625 ⎜1 +, 25 ⎠, ⎝, , Case 3 : Compounded quarterly, When the interest is compounded quarterly, there are four, conversion periods in a year and the quarterly rate will be, one-fourth of the annual rate, that is ()., , 3, , 3, , ⎛ 27 ⎞, ⎟, = 15625 ⎜, ⎝ 25 ⎠, , In this case,, ⎛, 1 ⎛ r ⎞⎞, ⎟ ⎟⎟, A = P⎜⎜1 + ⎜, 4, ⎝ 100 ⎠ ⎠, ⎝, , r ⎞, ⎛, = P⎜1 +, ⎟, ⎝ 100 ⎠, , =15625 x, 4n, , and C.I = A - P, , 27 27 27, x, x, 25 25 25, , = ₹ 19,683, , Case 4 : Compounded when time being fraction of a, year, When interest is compounded annually but time, being a fraction., , Now, compound interest= A - P, = 19,683 - 15,625, = ₹ 4,058, , In this case, when interest is compounded annually but, , To find the C.I. when the interest is compounded, annually or half-yearly., , 1, years, then amount, 4, , Let us see what happens to `100 over a period of one year, if an interest is compounded annually or half-yearly., , time being a fraction of a year, say 5, S. No., , Annually, , Half yearly, , 1, , P = ₹ 100 at 10% per annum compounded, annually., , P = ₹ 100 at 10% per annum compounded, half-yearly., , 2, , The time period taken is 1 year., , The time period is 6 months or 1/2 year., , 3, , I=, , 4, , A = 100 + 10 = ₹ 110, , 100 x 10 x 1, = ₹ 10, 100, , I=, , 100 x 10 x 1, 2 = 10, ₹, 100, , A = 100 + 5 = ₹ 105, For the next 6 months, P = ₹ 105, So, I =, , 105 x 10 x 1, 2 = 5.25 and, ₹, 100, , A = 105 + 5.25 = ₹ 110.25, 5, 78, , A = ₹ 110, , A = ₹ 110.25, , Workshop Calculation & Science : (NSQF) Exercise 2.7.15, , Copyright free, under CC BY Licence

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Thus, if interest is compounded half-yearly, we compute, the interest two times and rate is taken as half of the, annual rate., , 23 23 21, x, x, 20 20 20, = ₹ 27,772.50, = 20000 x, , Example, 1 Find the compound interest on ₹ 1000 at the rate, of 10% per annum for 18 months when interest is, compounded half-yearly., Solution, Here, P = ₹ 1000, r = 10% per annum., 3, 18, and n = 18 months =, years = years = 1 12 years, 2, 12,  Amount after 18 months, , ⎡ 1 ⎛ r ⎞⎤, ⎟⎥, = P ⎢1 + ⎜, ⎣ 2 ⎝ 100 ⎠⎦, , ⎡ 1 ⎛ 10 ⎞⎤, = 1000 ⎢1 + ⎜, ⎟⎥, ⎣ 2 ⎝ 100 ⎠⎦, , 10 ⎤, ⎡, = 1000 ⎢1 +, ⎥, ⎣ 200 ⎦, , 2x, , 2n, , 3, 2, , = 27,772.50 - 20,000, Compound Interest= ₹ 7,772.50, Inverse problems on compound interest, , r ⎞, ⎛, ⎟, We have already learnt the formula, A = P⎜1 +, 100, ⎝, ⎠, , 1 At what rate per annum will ₹ 640 amount, to ₹ 774.40 in 2 years, interest being compounded, annually?, , 3, , Solution, Given : P = ₹ 640, A = ₹ 774.40, n = 2 years , r = ?, We know,, , 21 21 21, x, x, 20 20 20, , r ⎞, ⎛, ⎟, A = P⎜1 +, 100, ⎝, ⎠, , = ₹ 1157.625, C.I = A - P, = 1157.63 - 1000, = ₹ 157.63, , 2 Find the compound interest on ₹ 20,000 at 15% per, 1, , annum for 2 years., 3, , Solution, 1, , 2, , 15 ⎞, ⎛, ⎟, = 20000 ⎜1 +, ⎝ 100 ⎠, , 2, , r ⎞, 774.40 ⎛, ⎟, = ⎜1 +, 640, ⎝ 100 ⎠, , 2, , r ⎞, 77440 ⎛, ⎟, = ⎜1 +, 64000 ⎝ 100 ⎠, , 2, , r ⎞, ⎛, ⎟, = ⎜1 +, 100, ⎝, ⎠, , 2, , 121, 100, , Here, P = ₹ 20,000, r = 15% per annum. and n = 2, 3, years., , r ⎞, ⎛, ⎟, Amount after 2 years A = P⎜1 +, 100, 3, ⎝, ⎠, , n, , r ⎞, ⎛, ⎟, 774.40 = 640 ⎜1 +, 100, ⎝, ⎠, , = ₹ 1157.63, , 1, , n, , Example, , ⎛ 21 ⎞, ⎟, = 1000 ⎜, ⎝ 20 ⎠, , Compound Interest, , =A-P, , Where four variable A, P, r and n are involved. Out of these, four variables, if any three variables are known, then we, can calculate the fourth variable., , 3, , = 1000 x, , C.I, , ⎛ 11 ⎞, ⎜ ⎟, ⎝ 10 ⎠, , ⎛, 1 ⎛ r ⎞⎞, ⎜⎜1 + ⎜, ⎟ ⎟⎟, ⎝ 3 ⎝ 100 ⎠ ⎠, , 11, 10, , ⎛, 1 ⎛ 15 ⎞ ⎞, ⎜⎜1 + ⎜, ⎟ ⎟⎟, ⎝ 3 ⎝ 100 ⎠ ⎠, , r, 100, , 2, , 3 ⎞ ⎛, 1 ⎞, ⎛, ⎟ ⎜1 +, ⎟, = 20000 ⎜1 +, 20, 20, ⎝, ⎠ ⎝, ⎠, , 2, , r ⎞, ⎛, ⎟, = ⎜1 +, ⎝ 100 ⎠, , = 1+, =, , 2, , 2, , r, 100, , 11 - 10, 10, , r, 1, =, 100 10, , 2, , ⎛ 23 ⎞ ⎛ 21 ⎞, ⎟, ⎟ ⎜, = 20000 ⎜, ⎝ 20 ⎠ ⎝ 20 ⎠, , r, , =, , 100, 10, , Rate r = 10% per annum, Workshop Calculation & Science : (NSQF) Exercise 2.7.15, , Copyright free, under CC BY Licence, , 79

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2 In how much time will a sum of ₹ 1600 amount, to ₹ 1852.20 at 5% per annum compound interest., , 2, , 4 ⎞, ⎛, ⎟ - 1632, = P ⎜1 +, ⎝ 100 ⎠, , Solution, Given : P = `1600, A = ₹ 1852.20, r = 5% per annum,, n=?, , =Px, P, , We know,, A, , 1852.20, , 5 ⎞, ⎛, ⎟, = 1600 ⎜1 +, 100, ⎝, ⎠, , 1852.20, 1600, , ⎛ 105 ⎞, ⎟, = ⎜, ⎝ 100 ⎠, , = 1.0816P - 1632, , 1.0816P - 1632 = P, , n, , r ⎞, ⎛, ⎟, = P⎜1 +, 100, ⎝, ⎠, , 104 104, x, - 1632, 100 100, , n, , 1.0816P - P, , = 1632, , 0.0816P, , = 1632, P, , n, , = 1632, 0.0816, = 20,000, , Principal, , = ₹ 20,000, , 185220, 160000, , ⎛ 21 ⎞, ⎟, = ⎜, ⎝ 20 ⎠, , n, , 9261, 8000, , ⎛ 21 ⎞, ⎟, = ⎜, ⎝ 20 ⎠, , n, , When P is the Principal, n = 2 years and r is the rate of, interest., , ⎛ 21 ⎞, ⎟, = ⎜, ⎝ 20 ⎠, , n, , ⎛ r ⎞, ⎟, Difference between C.I and S.I for 2 years = P⎜, ⎝ 100 ⎠, Example, , ⎛ 21 ⎞, ⎜, ⎟, ⎝ 20 ⎠, n, , 3, , Difference between simple interest and compound, interest, , = 3 years, , 3 Find the principal that will yield a compound, interest of ₹ 1632 in 2 years at 4% rate of interest, per annum., , 2, , Find the difference between simple interest and, compund interest for a sum of ₹ 8,000 lent at 10% p.a., in 2 years., Solution, , Solution, , Here, P = ₹ 8000, n = 2 years, r = 10% p.a., , Given : C.I = ₹ 1632, n = 2 years, r = 4% p.a, , Difference between compound interest and simple interest, , P=?, for 2 years, , We know,, Amount - Principal = Compound interest, A-P, -P, , ⎛ r ⎞, ⎟, = P⎜, ⎝ 100 ⎠, , 2, , ⎛ 10 ⎞, ⎟, = 8000 ⎜, ⎝ 100 ⎠, , = C.I, = C.I - A, , ⎛ 1⎞, = 8000 ⎜ ⎟, ⎝ 10 ⎠, , + P = A - C.I, , 2, , 2, , n, , P, , r ⎞ - C.I, = P⎛⎜1 +, ⎟, 100, ⎝, ⎠, , = 8000 x, , 1, 1, x, 10 10, , = ₹ 80, , Assignment A, 1 If principal = Rs. 5000, Interest = Rs. 500. Find the, amount., , 4 If principal = Rs. 8450, Interest is 750. Calculate the, amount., , 2 If principal = Rs. 12500, Amount= Rs. 17500. Find the, Interest., , 5 If principal = Rs. 12000, Amount= Rs. 15600. Find the, Interest., , 3 If the amount is Rs. 25000, its interest is 6000,, calcaulate its principal., 80, , Workshop Calculation & Science : (NSQF) Exercise 2.7.15, , Copyright free, under CC BY Licence

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Assignment B, Convert the following, 1 6 Months = _______ year., , 6 292 days into year., , 2 10 Months = _______ year., , 7 The month of July and August = ____ days, , 3 259 days into week., , 8 2 year 6 months = ____ years, , 4 22 weeks into days., , 9 15 years = ____ months, , 5 170 days into year., , 10 144 Months = _____ years., , Assignment C, 1 Ramani invested Rs. 1000 for 2 years at 10% per, annum. Find the simple interest., , 9 In how many years will a sum of ₹ 5,000 amount to, ₹ 5,800 at the rate of 8% per annum., , 2 Find the S.I. and the amount on ₹ 5,000 at 10% per, annum for 5 years., , 10 A sum of money doubles itself in 10 years. Find the, rate of interest., , 3 Find the S.I. and the amount on ₹ 1,200 at 12½% per, annum for 3 years., , 11 A sum of money doubles itself in 12½ per annum over, a certain period of time. Find the number of years., , 4 Kamesh invested ₹ 10,000 in a bank that pays an, interest of 10% per annum. He withdraws the amount, after 2 years and 3 months. Find the interest, he, receives., , 12 A certain sum of money amounts to ₹ 6,372 in 3 years, at 6%. Find the principal., , 5 Find the amount when ₹ 2,500 is invested for 146 days, at 13% per annum., 6 Find the S.I. and the amount on ₹ 12,000 from May, 21" 1999 to August 2nd 1999 at 9% per annum., 7 Shanthi deposited ₹ 6,000 in a bank and received 7500, at the end of 5 years. Find the rate of interest., , 13 A certain sum of money amounts to ₹ 6,500 in 3 years, and ₹ 5,750 in 1½ years respectively. Find the principal, and the rate percent., 14 Find the S.I. and the amount on ₹ 3,600 at 15% per, annum for 3 years and 9 months., 15 Find the principal that earns ₹ 2,080 as S.I. in 3¼ years, at 16% p.a., , 8 Find the principal that earns ₹ 250 as S.I. in 2½ years, at 10% per annum., , Assignment D, 1 Find the amount and compound interest in the following cases:, Sl. No., , Principal in Rs., , Rate % per annum, , Time in years, , a, , 1000, , 5%, , 3, , b, , 4000, , 10%, , 2, , c, , 18000, , 10%, , 2½, , 2 Sankari borrowed Rs. 8,000 from Alex for 2 years at, 12½% per annum. What interest did sankari pay to, Alex if the interest is compounded annually., , 6 Find the amount that Manimegalai would receive if she, invests Rs. 80,000 for 18 months at 10% per annum,, the interest being compounded half-yearly., , 3 Find the compound interest on Rs. 24000 compounded, semi annually (half yearly) for 1½ years at the rate of, 10% per annum., , 7 Find the compound interest on Rs. 15625 for 9 months, at 16% per annum compounded quarterly., , 4 Find the amount that Divakar would receive if he invests, Rs. 8192 for 18 months at 12½% per annum, the, interest being compounded half-yearly., 5 Anbu took a loan of Rs.80,000 from a bank for 1½, years at 10% per annum. What interest did Anbu pay, to bank if the interest is compounded annually., , 8 Raju took a loan of Rs. 80,000 from a bank. If the rate, of interest is 10% p.a. Find the difference in amounts, he would be paying after 1½ years if the interest, compounded annually is Rs. 92400, compounded half, yearly is Rs. 92610, , Workshop Calculation & Science : (NSQF) Exercise 2.7.15, , Copyright free, under CC BY Licence, , 81

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9 Guna borrowed Rs. 26400 from a bank to buy a scooter, at the rate of 15% p.a. compounded yearly. What, amount will he pay at the end of 2 years and 4 months, to clear the loan., 10 Find the difference between simple interest and, compound interest on ₹ 2400 at 2 years at 5% per, annum compounded annually., , 11 Find the difference between simple interest and, compound interest on ₹ 6400 for 2 years at 6¼% p.a., compounded annually., 12 Find the difference between compound interest and, simple interest for 2 years on a sum of money lent at, 5% p.a. is ₹ 5. Find the sum of money lent., , Assignment E, I MCQ, 1 Reduction from original selling price is called ____, A loss, , B list price, , C profit, , D markdown, , 2 A man buys an article for Rs. 27.50 and sells it for Rs., 28.60. Find his gain percent, A 1%, , B 2%, , C 3%, , D 4%, , 3 A TV is purchased at Rs.5000 and sold at Rs. 4000,, find the lost percent., A 10%, , B 20%, , C 25%, , D 28%, , 4 A person incurs a loss of 5% be selling a watch for Rs., 1140. At what price should the watch be sold to earn, 5% profit., A Rs. 1200, , B Rs. 1230, , C Rs. 1260, , D Rs. 1290, , 5 A book was sold for Rs. 27.50 with a profit of 10%. If it, were sold for Rs.25.75, What would have been, percentage of profit and loss?, , A 660, , B 760, , C 860, , D 960, , 10 A plot is sold for Rs. 18,700 with a loss of 15%. At, what price it should be sold to get profit of 15%., A Rs. 25300, , B Rs. 22300, , C Rs. 24300, , D Rs. 21300, , 11 A man gains 20% by selling an article for a certain, price. If he sells it at double the price, the percentage, of profit will be, A 130%, , B 140%, , C 150%, , D 160%, , 12 If the cost price of 12 pens is equal to the selling price, of 8 pens, the gain percent is?, A 12%, , B 30%, , C 50%, , D 60%, , 13 Ryan buys a clock for Rs. 75 and sells it for Rs. 100., His gain percent is _______, , A 2% profit, , B 3% profit, , A 25%, , B 33 13 %, , C 2% loss, , D 3% loss, , C 20%, , D 37 12 %, , 6 Alfred buys an old scooter for Rs. 4700 and spends, Rs. 800 on its repairs. If he sells the scooter for, Rs. 5800 his gain percent is ________, A 6.19%, , B 6.17%, , C 5.4545%, , D 3.5111%, , 7 If the cost price is 25% of selling price. Then what is, the profit percent?, , 14 A bat is bought for Rs. 120 and sold for Rs. 105, the, loss percent is _____, A 15 13 %, , B 14 15 %, , C 15%, , D 16 2 3 %, , 15 A man bought apples at the rate of 8 for Rs.34 and, sold them at the rate of 12 for Rs. 57. How many, apples should be sold to earn a net profit of Rs. 45?, , A 150%, , B 200%, , A 90, , B 100, , C 300%, , D 350%, , C 135, , D 150, , 8 The cost price of 20 articles is the same as the selling, price of x articles. If the profit is 25%, find out the value, of x., , 82, , 9 A man buys an item at Rs. 1200 and sells it all the, loss of 20 percent. Then what is the selling price of, that item., , 16 A tradesman sold an article at a loss of 20%. Had he, sold it for Rs. 100 more , he should have gained 5%., The cost price of the article was ________, , A 13, , B 14, , A Rs. 360, , B Rs. 400, , C 15, , D 16, , C Rs. 425, , D Rs. 450, , Workshop Calculation & Science : (NSQF) Exercise 2.7.15, , Copyright free, under CC BY Licence

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17 At what percentage above the cost price must an article, be marked so as to gain 33% after allowing a customer, a discount of 5%?, A 35%, , B 38%, , C 40%, , D 42%, , 18 A shopkeeper earns a profit of 12% on selling a book, at 10% discount on the printed price. The ratio of the, cost price and the printed price of the book is, A 45:56, , B 45:51, , C 47:56, , D 47:51, , 19 By selling a bicycle for Rs. 2,850 a shopkeeper gains, 14%. If the profit is reduced to 8%, then the selling, price will be, A Rs. 2600, , B Rs. 2700, , C Rs. 2800, , D Rs. 3000, , 20 A person sold a horse at a gain of 15%. Head he, bought it for 25% less and sold it for Rs. 600 less, he, would have made a profit of 32%. The cost price of the, horse was:, A Rs. 3750, , B Rs. 3250, , C Rs. 2750, , D Rs. 2250, , 21 If a man were to sell his chair for Rs. 720, he would, lose 25%. To gain 25% he should sell it for:, A Rs. 1200, , B Rs. 1000, , C Rs. 960, , D Rs. 900, , 22 If harsh sold a match ticket for Rs. 340 at a profit of, 25%, at what price did he purchase the ticket?, A 280, , B 255, , C 300, , D 272, , 23 Eleven bags are bought for Rs. 1000 and sold at 10 for, Rs. 1100. What is the gain or loss in percentage?, , 2 Rs. 1200 is lent out at 5% per annum simple interest, for 3 years. Find the amount after 3 years, A Rs. 1380, , B Rs. 1290, , C Rs. 1470, , D Rs. 1200, , 3 Interest obtained on a sum of Rs. 5000 for 3 years is, Rs. 1500. Find the rate percent., A Rs. 8%, , B Rs. 9%, , C Rs. 10%, , D Rs. 11%, , 4 Rs. 2100 is lent at compound interest of 5% per annum, for 2 years. Find the amount after two years., A Rs. 2300, , B Rs. 2315.25, , C Rs. 2310, , D Rs. 2320, , 5 Find the difference between the simple interest and, the compound interest at 5% per annum for 2 years, on principal of Rs. 2000?, A Rs. 5, , B Rs. 10.5, , C Rs. 4.5, , D Rs. 5.5, , 6 A bank offers 5% compound interest calculated on half, yearly basis. A customer deposits Rs. 1600 each on, 1st January and 1st july of a year. At the end of the year,, the amount he would have gained by way of interest is:, A Rs. 120, , B Rs. 121, , C Rs. 122, , D Rs. 123, , 7 There is 60% increase in an amount in 6 years at simple, interest. What will be the compound interest of, Rs. 12,000 after 3 years at the same rate?, A Rs. 2160, , B Rs. 3120, , C Rs. 3972, , D Rs. 6240, , 8 What is the difference between the compound interest, on Rs. 5000 for 1 12 years at 4% per annum, compounded yearly and half-yearly?, , A 10%, , B 21%, , A Rs. 2.04, , B Rs. 3.06, , C 25%, , D 20%, , C Rs. 4.80, , D Rs. 8.30, , 24 A man buys an article for Rs. 27.50 and sells it for rs., 28.60. Find its gain percent?, A 1%, , B 2%, , C 3%, , D 4%, , II MCQ, 1 Find the simple interest on Rs. 5200 for 2 years at 6%, per annum., A Rs. 450, , B Rs. 524, , C Rs. 600, , D Rs. 624, , 9 The compound interest on Rs. 30,000 at 7% per annum, is Rs. 4347. Their period (in years) is, A Rs.2, , B Rs.2 12, , C Rs.3, , D Rs.4, , 10 What will be the compound interest on a sum of Rs., 25000 after 3 years at the rate of 12 p.c.p.a?, A Rs. 9000.30, , B Rs. 9720, , C Rs. 10123.20, , D Rs. 10483.20, , Workshop Calculation & Science : (NSQF) Exercise 2.7.15, , Copyright free, under CC BY Licence, , 83

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11 At what rate of compound interest per annum will a, sum of Rs. 1200 become Rs. 1348.32 in 2 years?, A 6%, , B 6.5%, , C 7%, , D 7.5%, , 12 Albert invested an amount of Rs. 8000 in a fixed deposit, scheme for 2 years at compound interest rate 5 P.C.P.A., How much amount will Albert get on maturity of the, fixed deposit?, A Rs. 8600, , B Rs. 8620, , C Rs. 8820, , D Rs. 8940, , Key Answers, A, , 3 Rs. 3783/-, , 8 Rs. 210/-, , 1 Rs. 5500/-, , 4 Rs. 9826/-, , 9 Rs. 36659.7/-, , 2 Rs. 5000/-, , 5 Rs. 92400/-, , 10 Rs. 6/-, , 3 Rs. 19000/-, , 6 Rs. 92610/-, , 11 Rs. 25/-, , 4 Rs. 9200/-, , 7 Rs. 1951/-, , 12. Rs. 2000/-, , 5 Rs. 3600/-, , E MCQ, I, , B, 1, , 1, 2, , 6, , 2, , 5, , 7 62, , 6, , 292, , 365, , II, , 1, , A, , 1, , D, , 2, , D, , 2, , A, , 3, , B, , 3, , C, , 3 37, , 8 2 12, , 4, , C, , 4, , B, , 4 154, , 9 180, , 5, , B, , 5, , A, , 5, , 10 12, , 6, , C, , 6, , B, , 7, , C, , 7, , C, , 8, , D, , 8, , A, , 9, , D, , 9, , A, , 34, , 73, , C, 1 Rs. 200/-, , 9 2 years, , 2 Rs. 2500/-, Rs. 7500/-, , 10 10%, , 3 Rs. 450/-, Rs. 1650/-, , 11 8 years, , 4 Rs. 2250/-, , 12 Rs. 5400/-, , 5 Rs. 2630/-, , 13 Rs. 5000/- ,10%, , 6 Rs. 216/-, Rs. 12216/-, , 14 Rs. 2025/-,, Rs. 5625/-, , 7 5%, , 15 Rs. 4000/-, , 8 Rs. 1000/-, , 10 A, , 10 C, , 11 B, , 11 A, , 12 C, , 12 B, , 13 B, 14 A, 15 A, 16 B, 17 C, , D, , 18 A, , 1 a Rs. 1157.63, Rs. 157.63, , 19 B, , b, , Rs. 4840/-, Rs. 840/-, , 20 A, , c, , Rs. 22869/-, Rs. 4869/-, , 21 A, , 2 Rs. 2125/-, , 22 D, 23 B, 24 D, , 84, , Workshop Calculation & Science : (NSQF) Exercise 2.7.15, , Copyright free, under CC BY Licence

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Estimation and costing - Simple estimation of the requirement of material etc.,, as applicable to the trade, Exercise 2.8.16, Estimation is the method of calculating the various, quantities and the expenditure to be incurred on a particular, job or process., In case the funds available are less than the estimated, cost the work is done in part or by reducing it or, specifications are altered,, The following essential details are required for preparing, an estimate., Drawings like plan, elevation and sections of important, parts., Detailed specifications about workmanship & properties, of materials, etc., , Reference table, A refereance table may mean a set of references that are, author may have cited (or) gained inspiration from whilst, writing an article, similar to a bibliography., It can also mean an information table that is used as a, quick and easy reference for things that are difficult to, remember such as comparing imperial with metric, measurements. This kind of data is known as reference, data., Estimation of Bush, Fig 1, , Standard schedule of rates of the current year., Estimating is the process of preparing an approximation, of quantities which is a value used as input data and it is, derived from the best information available., An estimate that turns out to be incorrect will be an, overestimate if the estimate exceeded the actual result,, and an underestimate if the estimate fell short of the actual, result., A cost estimate contains approximate cost of a product, process or operation. The cost estimate has a single total, value and it is inclusive of identifiable component, values., , Raw Material Cost, Size of Raw material =  90 x 80 mm, 90 mm = 9 cm, 80 mm = 8 cm, Volume of Raw material =  r2 h cm3, , Hand box and reference table, A hand book is a type of reference work, or other collection, of instruction. That is intended to provide ready reference., The term originally applied to a small portable book, containing information useful for its owner, but the oxford, english disctionary defines as “any book .... givng, information such as facts on a particular subject, guidence, in some art or occupation, instruction for operating a, machine etc. A handbook is sometimes referred to as a, pocket reference., Hand book may deal with any topic, and arc generally, having compact information in a particular field (or), technique. They are designed to be easily consulted and, provides quick answer in a certain area., , =, , 22, 7, , x 4.52 x 8 cm3, , = 509.14 cm3, Density of material = 7.8 gm/cm3, Welght of Raw material = Volume of raw material x, density of raw material, = 509.14 x 7.8 gm/cm3, = 3971.3 gms say 3.971 kg, Raw material cost = Rs. 140/kg, Total raw material cost = 3.971 x 140, , Example of engineering hand book include parry’s, cheorikal engineers hand book, mark standard hand book, for machine engineer and the CRC hand book of chemistry, and physics., , = Rs. 555.94 say Rs.556, , 85, , Copyright free, under CC BY Licence

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ESTIMATION SHEET, Operation, No., , Operation, description, , Lathe, , Estimated, time, , Rate Rs.100, per hr., , 01, , Setting and aligning job on lathe, , -, , 10 min, , 16.66, , 02, , Set speed and feed, , -, , 2 min, , 3.32, , 03, , Align cutting tool in position, , -, , 2 min, , 3.32, , 04, , Turn the job, , -, , 50 min, , 83.00, , 05, , Chamfer 45º angle corner, , -, , 8 min, , 13.28, , 06, , Reverse the job on Lathe, , -, , 10 min, , 16.66, , 07, , Turn the job, , -, , 20 min, , 33.34, , 08, , Chamfer 45º on other side, , -, , 20 min, , 33.34, , 09, , Centre drilling, , -, , 10 min, , 16.66, , 10, , Mount drill chuck and drill, using tail stock, , Drilling, , 3 min, , 4.98, , 11, , Set drill rpm, , Drilling, , 2 min, , 3.32, , 12, , Drill holes, , Drilling, , 20 min, , 33.34, , 13, , Set the boring tool, , Drilling, , 15 min, , 24.90, , 14, , Bore to the required diameter, , 8 min, , 13.28, , 15, , Check the bore dia, , 10 min, , 16.66, , 16, , Deburr the job and clean the, machine, , 10 min, , 16.66, , 18, , Total estimation, , 200 min, , Rs. 332.72, Rounded off, Rs. 333, , Machining cost =, , 100, 60, , -, ,  200 = Rs. 333.33 say Rs. 333, , Total Cost = Total raw material cost + Machining cost, = 556 + 333, Total Estimation Cost = Rs. 889/-, , 86, , Workshop Calculation & Science : (NSQF) Exercise 2.8.16, , Copyright free, under CC BY Licence

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1 Estimating, Estimating means to determine the quantities of various, items required to execute a job and to asses the cost of, the execution. The estimator keeping in view the, requirements arrived at during initial planning, chalk out a, list of items and quantities. The cost is determined by him, by consulting the price catalogue and schedule of labour, rates. The various steps to form an estimate are:, i, , Chalk out a list of items and quantities required., , ii Consult the rate catalogues for pricing the various, items., iii Assess the exact number of worker required to complete, the job and after consulting the schedule of labour rates, add the labour cost to the estimate under preparation., It should be noted that number of workmen required is, dependent upon the time limit fixed to complete the, services., iv Add supervision charges and executor's profit., v In case of Govt. organization, where the work is to be, executed by the contractor, the tenders are floated, only after correctly specifying the description of each, item, to avoid any misunderstanding while execution., For preparing an estimate the following are necessary., 2 Drawing : The necessary drawings of the building with, dimension are required for execution of the work and, also for estimation. (Fig 1) For the electrical installation, the layout of the wiring should be shown on a separate, drawing along with the position of the points etc., , Determination of Conductor size, , 3 Specifications of materials : Specification gives the, details of materials, brand name, grade of quality,, rating of current and voltage, quality certification like, BIS or ISO etc. Specification helps both the wireman, and the consumer to select the material according to, commercial practice, cost and the requirement., , a Current carrying capacity, , Price Catalogue, It is in the form of a booklet in which rates of various terms, are indicated. The price catalogue is required to be, amended as and when there is variation in the market., Schedule of labour rates, It is also in the form of booklet indicating the labour rates., Schedule of rates and estimating data, Almost all the Govt. departments have published schedule, of rates or estimating data to facilitate the process of, estimating. In these schedules or data, the estimated cost, including labour charges for per meter run of wiring,, overhead lines etc. With various sizes of wire, aluminium, conductor, pole etc. are given. With the help of such a, document, only overhead charges viz. supervision,, departmental charges, etc. are required to be added after, working one cost of the any services., , Before making an estimate it is necessary to find out the, size of wire, cable or aluminium conductor. The following, essential points are to be considered while calculating the, size:, b Voltage drop, c Minimum permissible size, There have been discussed separately in the following, articles., Current carrying capacity, In any circuit the value of the current will be more as, compared to sub - circuits. In sub circuits as the load, decreases, the current is also reduced. Thus it is very, necessary to divide the services into groups in accordance, with the amount of current which will flow through them., Afterwards, size of wire in each group is determined., Voltage drop, As in known to the readers that the voltage drop is there, as when the current flows through the wiring and the same, should be as low as permissible and economical. The, voltage drop can be determined by Ohm's law. As the, resistance is inversely proportional to area, so the voltage, drop will be less if the area of wire is more., , Workshop Calculation & Science : (NSQF) Exercise 2.8.16, , Copyright free, under CC BY Licence, , 87

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Minimum permission size, Due to mechanical reason the minimum permissible size, of wire, U/G cables and conductors should be as follows:, , It should be noted that the maximum voltage drop should, not be more than as given below., a Lighting circuit. In any circuit, , a Wire : The area of aluminium wire should not be less, than 1.5 sq.mm and its single strand should not be less, than 1.40 mm diameter., , i, , ii at 210 volts, supply voltage drop should not be more, than 5.1 volts;, , b U/G Cable : The area of conductor for two core cable, should not be less than 6 sq.mm and for 25 sq.mm. The, area of conductor for three and half cores cable should, be 50 sq.mm or more., , iii at 220 volts supply, voltage drop should not be more, than 5.4 volts., iv at 30 volts supply, voltage drop should not be more, than 5.6 volts, , A.C.S.R. The size of A.C.S.R should not be less than, 6, 0.83, , inch or, , 6, 1 21.08, , mm having total area of cross section, , v at 250 volts supply, voltage drop should not be more, than 6.0 volts., , as 20-71 sq.mm., Conductor size calculation for internal domestic, wiring, The important point to be considered is the current, carrying capacity, the voltage drop is usually of very small, magnitude and will not have much effect for small domestic, wiring. For multi - storeyed building, factories and industries,, the voltage drop is required to be ascertained. If the voltage, drop is much the house - hold appliance and motors will, not work., , at 200 volt supply, voltage drop should not be more, than, 5 V., , From, the above it will be seen that the permissible voltage, drop in a lighting circuit is 2% of the supply voltage plus, one volt., b Industrial loads : The maximum voltage drop at the, extreme and equipment or motor should not be more, than 5% of the declared supply voltage., In tables 1 various size of wires, current rating and voltage, drop if loaded fully are given. Considering the load in, amperes and voltage drop, suitable size of wire with, required insulation is selected., , Table 1, Current ratings and voltage drop for vulcanised rubber PVC or polythene insulated or tough rubber PVC, lead sheathed singles core aluminium wires or tables, Size of, Conductor, , 2 Cables 1c. or, Single -phase a.c, Current, rating in, amperes, , Approx., length of, run for 1, volt - drop, in meters, , 3 or 4 cables of, balanced 3 - phase, Current, rating in, amperes, , Approx., length of, run for 1, volt - drop, in meters, , 4 Cables d.c, , Normal, area, sq.mm, , Number, and, diameter, of wire in, mm, , 1.5, , 1/1.40, , 10, , 2.3, , 9, , 2.9, , 9, , 2.5, , 2.5, , 1/1.80, , 15, , 2.5, , 12, , 3.6, , 11, , 3.4, , 4.0, , 1/2.24, , 20, , 2.9, , 17, , 3.9, , 15, , 4.1, , 6.0, , 1/2.80, , 27, , 3.4, , 24, , 4.3, , 21, , 4.3, , 10.0, , 1/3.55, , 34, , 4.3, , 31, , 5.4, , 27, , 5.4, , 16.0, , 7/1.70, , 43, , 5.4, , 38, , 7.0, , 35, , 6.8, , 25.0, , 7/2.24, , 59, , 6.8, , 54, , 8.5, , 48, , 8.5, , 35.0, , 7/2.50, , 69, , 7.2, , 62, , 9.3, , 55, , 9.0, , 50.0, , 7/.3.0, , 91, , 7.9, , 82, , 10.1, , 69, , 10.0, , 19/1.80, , 88, , Workshop Calculation & Science : (NSQF) Exercise 2.8.16, , Copyright free, under CC BY Licence, , Current, rating in, amperes, , Approx., length of, run for 1, volt - drop, in meters

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After deciding the type of wiring system the electrician has, to prepare the following details as per the consumer’s, interest, durability of wiring accessories, cost and available, work force., 1 Position of electrical points in each area of the house, based on consumer’s requirement. (Fig 2), 2 Prepare a layout of wiring on the building plan. (Fig 3), 3 Calculate total connected load and type of load to, decide whether three phase or single phase supply is, required., 4 Prepare a circuit diagram showing the connections and, number of lighting and power circuits. Fig 4 shows the, cubical wiring diagram model., 5 Prepare a list of electrical accessories to be procured., 6 Calculate the length of different sizes of cables required, based on the load, voltage drop and number of cables, in each circuit run., 7 Calculate the length of different sizes of batten/PVC, casing and capping/PVC/metal conduit required based, on the load, voltage drop and number of cables in each, circuit run., 8 Prepare a list of hardwares like screw, nails, etc., required to execute the job., , 10 Calculate the cost of accessories, base materials like, batten/conduit/PVC casing and capping, cables,, hardware and labour charges., 11 Calculate the total cost of wiring including 5%, contingencies for the unforeseen items and variation in, prices., In the above list, items 1 to 5,8 and 11 were, dealt in various stages of skill development., Let us consider item No.6 i.e. calculation of, length of different sizes of cables in the wiring., Before selection of sizes of cable and their, length we have to calculate the total load of, the house and also to decide whether three, phase or single phase supply to be requisitioned, from local electricity board., Total connected load : As this exercise is a part of, cognitive skill development in 1st Year. Let us assume the, total connected load to be 2400 watts., Assuming the total connected load is 2400w., then current will be 2400w/240V = 10amp, (Assumption PF is unity and single phase supply), Hence copper cable 3/0.036 could be selected for main, board connection which can safety carry 10 amp., , 9 Calculate the labour charges for entire wiring., , Workshop Calculation & Science : (NSQF) Exercise 2.8.16, , Copyright free, under CC BY Licence, , 89

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a one tube light point, b one light point, , circuit - 1, , c two fan points, d one 6 amps three-pin socket, Room No.2 should be provided with, a one tube light point, b one light point, , circuit - 2, , c one fan point, d two 6 amp three-pin socket, Total connected load of Room No.1 & 2 will be, Circuit 1, Sl. Description, No., , Quantity Wattage Total, rating, wattage, , 1, , Tube light, (1200mm), , 1, , 50, , 50, , 2, , Light point, , 2, , 60, , 120, , 3, , Fan points, , 2, , 60, , 120, , 4, , 6A 3 pin socket, , 1, , 100, , 100, 390 watts, , Circuit 2, Sl. Description, No., , Quantity Wattage Total, rating, wattage, , 1, , Tube light, (1200mm), , 1, , 50, , 50, , 2, , Light point, , 1, , 60, , 60, , 3, , Fan point, , 1, , 60, , 60, , 4, , 6A 3 pin socket, , 2, , 100, , 200, 370 watts, , Total wattage = 390 + 370 = 760 watt of circuit 1 & 2, As this is well within 800 watt rating of a branch circuit we, can wire up the hall and sitout in one branch circuit., Remember that each branch circuit termination should, have one phase cable and one neutral cable brought to the, DB., Length of cables : Main cable say from meter board to the, DB, 2 x 1.5 = 3 m of size 3/0.36 copper cable, (1.5m is taken as a rough value), At this stage we have to analyse the house plan as well as, the consumer’s requirement of electrical points. Only, Room No.1 & 2 are taken in the following example., According to the consumer the requirements are:, Room No.1 should be provided with, , 90, , Draw the position of switches, sockets, lamps and fans in, the house plan (position of accessories shown in Fig 2), Then draw the layout diagram for the hall and sitout, keeping in view of the position of accessories. (Refer, layout diagram Fig 3), After drawing the layout diagram, draw the circuit diagram, for the hall and sitout., At this stage you may have to modify the layout diagram, in most of the cases considering economical use of cables, and the conduit and also considering aesthetic sense of, wiring layout., , Workshop Calculation & Science : (NSQF) Exercise 2.8.16, , Copyright free, under CC BY Licence

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After finalising the layout and the circuit diagram, analyse, the cable runs, indicate in the layout diagram the number, of cables to be drawn in each section of the conduit., In this case understudy, the total load is of lighting, fan and, 3A pin sockets amounting to 760 watt which is less than, 5 amp as such cable selected is 1 sq.mm (1/1.12) or (1/, 18) copper cable or say aluminium cable of 1.5 sq.mm, (1/1.4)., Horizontal run, J, , As thumb rule length of cable = length of conduit x 3 = 25m, x 3 = 75m. This length is more or less the same as the, one calculated., Required length of copper wire 14 SWG (For earthing), Router K, H, F, E, D and A, 0.8 + 0.9 + 1.8 + 0.9 + 0.8 + 1 + 0.8 + 0.9 + 1.5 + 1.5 +, 1.5 + 0.8 + 0.9 + 1 = 15.1 m, Say 20m or 600 grams, , -150cm 1.5m, , Total of 2 to 5, , Say 90m of cable required, , 25m, , say 25m, , 1/1.2 copper cable required for, Conduit Run J = (2.5 + 0.9 + 0.8)2, , =, , 8.4m, , Now prepare a complete list of accessories required,, length of conduit required, length of the cable required and, the required copper wire 14 SWG. Calculate the cost of, the above materials. Say in this case Rs.2500/-., , -do-, , A = (1.5 + 0.9 + 0.8)7, , = 22.4m, , Method of calculating labour charges, , -do-, , B = 1.1 x 2, , =, , 2.2m, , -do-, , C = 1.05 x 4 + 1.85 x 2, , =, , 7.9m, , Work can be completed in two days by one electrician and, one helper., , -do-, , D = (1.5 + 0.8 + 0.9)3, , =, , 9.6m, , -do-, , F = (1.8 + 0.9 + 0.8)5, , = 17.5m, , -do-, , G = 2.2 x 2, , =, , 4.4m, , -do-, , H = (2.2 + 0.8)2, , =, , 6.0m, , -do-, , K = (0.9 + 0.2)2, , =, , 2.2m, , Electrician @ Rs.150/-, , 2 days = Rs.300, , Helper @ Rs.90/-, , 2 days = Rs.180, Rs.480, , On the other hand if the consumer wishes to give wiring on, contract, based on number of points, the calculation of, contract cost according to points., , 80.6m, , Workshop Calculation & Science : (NSQF) Exercise 2.8.16, , Copyright free, under CC BY Licence, , 91

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Estimation and Costing - Problems on estimation and costing, Exercise 2.8.17, 1 What is the total estimation cost of M.S rod dia, 50mm and length 500 mm, if cost of material, Rs.80/kg and the density of the M.S rod is, 7.8gm/cm3. Lathe time taken to complete the, component is 1hr 30minutes and lathe charge is, Rs. 80/hr., , Area of square job = 60 x 60 mm = 6 x 6 = 36 cm2, Area of cut sizes, , = 1.5 x 1.5 = 2.25 cm2, = 1.5 x 3 = 4.5 cm2, = 1.5 x 4.5 = 6.75 cm2, , Total area of cut size = 2.25 + 4.5 + 6.75 = 13.5 cm2, Volume = Area x Thickness, Volume of cut size, , = 13.5 x 4 = 54 cm3, , Volume of square job = 36 x 4 = 144 cm3, Volume of job, , M.S rod dia(d) = 50 mm or 5 cm, , = 144 - 54 = 90 cm3, , r = 2.5 cm, l = 500 mm = 50 cm, , Weight of job, , Volume of M.S rod = r x l unit, 2, , =, , 3, , = 90 cm3 x 7.8 gm/cm3, = 702 gms = 0.702 kg, , 22, x 2.5 x 2.5 x 50, 7, , = 981.875 cm3, Weight of M.S rod = Volume x density of M.S rod, Weight = 981.875 x 7.8, , Material cost, , = 0.702 x 60 = Rs.42.12, , Machining cost, , = Rs.200/hr, , Machining cost for 2hrs = 2 x Rs.200/hr = Rs.400, Total cost for job, , Weight of M.S rod = 7.659 kg, , = Material cost + Machining, cost, = Rs. 42.12 + Rs.400, = Rs.442.12, , Cost of material = Rs.80/kg, Total cost for job, , cost of M.S. rod = 7.659 x 80, = Rs.612.72, Weight of M.S rod = 7.659 kg, Cost of material M.S.rod = Rs.612.72, Machining cost in lathe =, , = Total volume - cut out volume, , = Rs.442.12, , 3 What is the estimation of milling cost of a, rectangular block size 100 x 80 x 60 mm, if cost of, the milling is Rs.2/sq.cm., , 80,  90 = Rs. 120, 60, , (1hr 30min = 90 mins), Total Estimation of cost = Rs 120 + 612.72, = Rs 732.72 say Rs. 733, 2 What is the total estimation cost of job as shown, in figure the density of the material is 7.8 gm/cm3, time taken in milling machine to complete the job, is 2 hours, charge for the milling machine is, Rs.200/hr and material cost is Rs.60/kg., , Rectangular block size, , = 100 x 80 x 60 mm, = 10 x 8 x 6 cm, , Surface area of 6 x 8 x 2, , = 96 cm2, , Surface area of 6 x 10 x 2 = 120 cm2, Surface area of 8 x 10 x 2 = 160 cm2, Total surface area of, rectangular block, , = 96 + 120 + 160, = 376 cm2, , Milling cost, , 92, , Copyright free, under CC BY Licence, , = Rs.10/cm2

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Milling cost of rectangular, block, = 376 cm2 x Rs.2/cm2, = Rs.752, Milling cost of rectangular block = Rs.752., , 5 What is the estimation cost of the job shown in, figure if the cost of material is Rs.140/kg and the, density of the material is 7.8 gm/cm3. Machining, time taken for milling machine 3hr 25 mins and, machinig charge is Rs 180/hr., , 4 What is the total estimation cost of component if, the filing & scraping cost is Rs.20/cm2 in the block, as shown in figure, the density of the material is, 7.8 gm/cm 3 and the cost of the material is, Rs.130/kg., , Volume of blank, , = 78 x 48x 10 mm, = 7.8 x 4.8 x 1 cm, = 37.44 cm3, , Volume of cut-out pieces = 2 no.of triangular pieces +, 1 rectangular pieces, M.S block size, , = 90 x 48 x 18 mm, , Volume of triangular pieces =, , = 9 x 4.8 x 1.8 cm, Volume of block, , =, , = 9 x 4.8 x 1.8 cm3, , Area of filing & scraping =, =, , 2r, 2, , x 1.8, , 2  3.142  20, 2, , x 1.8 = 113.112 cm3, , = 1.96 cm3, Volume of rectangular piece= 2.8 x 2 x 1, = 5.6 cm3, Volume of job as per fig = volume of blank size of, job - (volume of 2 triangular, pieces + volume of 1, rectangular piece), , Cost of filing & scraping = 113.112 x Rs.20 = Rs.226.22, , = 37.44 - (1.96 + 5.6) cm3, , Volume of the block = Volume of whole size - Volume, of cur piece, , = 37.44 - 7.56 cm3, = 29.88 cm3, , Volume of block = 77.76 - 2.83 = 74.93 cm3, Weight of block, , = Volume x density, = 74.93 x 7.8 = 584.45 gms, , Machining charge, , = 0.58445 kg x Rs.130/kg, = Rs.75.98, , =, , 180,  205, 60, , = Rs.615, , = 0.58445 kg, Material cost, , 1, x 1.4 x 1.4 x 1, 2, , = 0.98 x 2 pieces, , = 77.76 cm3, r 2, Volume of half circle =, x 1.8 cm3 = 2.83 cm3, 2, , 1, x b x h x thickness, 2, , Cost of material, Weight of material, , = Rs.140 /kg, = 29.88 x 7.8 gm/cm3, , Total cost, , = Material cost + Filling & scraping cost, , = 233.064 gms, , Total cost, , = 75.98 + 226.22, , = 0.233 kg, Material cost of job, , = Rs.302.20, , = Rs.32.62, , Material cost of M.S plate as per fig = Rs.75.98, Filing & scraping cost = Rs.226.22, , = 0.233 kg x Rs.140, , Estimation cost of job = Rs.32.62 + 615, , Total cost of material = Rs.302.20, , = Rs.647.62, , Workshop Calculation & Science : (NSQF) Exercise 2.8.17, , Copyright free, under CC BY Licence, , 93

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6 What is the estimation of labour charge for making, inside square of size 30 x 30 mm, if making charge, Rs.500/10cm2., , Volume of smaller dia of mandrel = r2 x l, = 3.142 x 1 x 1 x 1.8 x 2 pieces, = 11.311 cm3, Total volume of mandrel, , = 84.834 + 11.311 cm3, = 96.145 cm3, , Density of mandrel, , = 7.8 gm/cm3, , Weight of mandrel, , = 96.145 x 7.8 gm/cm3, = 749.931 gms = 0.750 kg, , Weight of mandrel, , = 0.75 kg, , Estimation of mandrel cost = 0.75 x 240 = Rs. 180, Size of square hole = 30 x 30 mm, = 3 x 3 cm = 9 cm2, Making charge, , = Rs.500 per 10 cm2, , Labour charge for making, square of 9 cm2, , 9 What is the material cost of component as shown, in figure, if density of material is 7.8 gm/cm3., Material cost Rs. 180/kg., , , , =, , 500,  9 = Rs.450, 10, , Labour charge for making square hole = Rs.450, 7 What is the total estimation cost for making the, component of 8 drilled hole dia 10mm and 4Nos, of M6 taps in the plate, if Rs.8 per drilled holes and, Rs.12 per drill & tap., No.of drilled holes, , = 8 Nos, , No.of tapped hloes, , = 4 Nos, , Charge for making drilled hole = Rs.8 per hole, Charge for making tapped hole = Rs.12 per tapped hole, , Size of blank, , = 8 x 8 x 1 cm = 64 cm3, , Cost for making 8 drilled holes = 8 x Rs.8 = Rs.64, Cost for making 4 tapped holes = 4 x Rs.12 = Rs.48, Total estimation cost for making, 8 drilled holes and 4 tapped, holes, , = 64 + 48, , Area of triangle cut =, =, , = Rs.112, , 8 What is the total estimation cost for mandrel, as, shown in figure, if density is 7.8 gm/cm3 and, material cost is Rs. 240/kg., , = 80 x 80 x 10 mm, , 1, xbxh, 2, 1, x 4 x 4 cm2 = 8 cm2, 2, , For 2 triangle pieces = 2 x 8 cm2 = 16 cm2, Volume, , = area x thickness, = 16 x 1 = 16 cm3, , Volume of material, , = volume of blank size –, volume of 2 triangle cut-out, pieces, = 64 – 16 = 48 cm3, , Size of larger dia of mandrel = dia 30 mm, radius = 1.5 cm, , Density of material, , Weight of component = volume x density, = 48 x 7.8, , Size of smaller dia of mandrel = dia 20 mm, , = 374.4 gms, , radius = 1 cm, Volume of larger dia of mandrel = r2 x l, = 3.142 x 1.5 x 1.5 x 12 cm, , = 7.8 gm/cm3, , Weight of the component = 0.374 kg, Cost of material, , = 84.834 cm3, 94, , Workshop Calculation & Science : (NSQF) Exercise 2.8.17, , Copyright free, under CC BY Licence, , = 0.374 x 180, = Rs. 67.32

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Assignment, 1 What is the material cost if cost is Rs.120/kg and, density of the material is 7.8gm/cm3., , 4 What is the material cost and machining cost for, complete the job. Material density is 7.8 gm/cm3 and, material cost is Rs.90/kg. Labour time taken for complete, 2 What is the total manufacturing cost of bush for turning, 1 hour 20 minutes and drilling the bush 25 minutes if, lathe charged for Rs.60/hr., , the job 6, , 1, hours he charged is Rs.300 per day of 8hrs., 2, , 5 What is the estimation of labour charge for making 10, numbers centre gauge if each time taken for complete, 2, 3 Find the estimation of manufacturing cost of the, component, if density is 7.8 gm/cm3, and machining, cost Rs 380/hr and machining time 4hr 15mins., , 1, hours, and labour charge is Rs.800 per 6 hrs., 2, , Workshop Calculation & Science : (NSQF) Exercise 2.8.17, , Copyright free, under CC BY Licence, , 95