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2, , www.aktutor.in, PUBLISHED BY :, , Apram Singh, , Quantum Publications, (A Unit of Quantum Page Pvt. Ltd.), Plot No. 59/2/7, Site - 4, Industrial Area,, , Sahibabad, Ghaziabad-201 010, Phone : 0120 - 4160479, Email : pagequantum@gmail.com, , Website: www.quantumpage.co.in, , Delhi Office : 1/6590, East Rohtas Nagar, Shahdara, Delhi-110032, © ALL RIGHTS RESERVED, No part of this publication may be reproduced or transmitted,, in any form or by any means, without permission., Information contained in this work is derived from sources, believed to be reliable. Every effort has been made to ensure, accuracy, however neither the publisher nor the authors, guarantee the accuracy or completeness of any information, published herein, and neither the publisher nor the authors, shall be responsible for any errors, omissions, or damages, arising out of use of this information., , Engineering Mechanics (CE : Sem-3), 1st Edition : 2019-20, 2nd Edition : 2020-21, , Price: Rs. 90/- only, Printed Version : e-Book.
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3, , www.aktutor.in, , CONTENTS, KCE 301 : Engineering Mechanics, UNIT-1 : INTRODUCTION TO ENGINEERING MECHANICS, (1–1 C to 1–44 C), Force Systems, Basic concepts, Rigid Body equilibrium; System of, Forces, Coplanar Concurrent Forces, Components in Space –, Resultant Moment of Forces and its Applications; Couples and, Resultant of Force System, Equilibrium of System of Forces, Free, body diagrams, Equations of Equilibrium of Coplanar Systems., Friction: Types of friction, Limiting friction, Laws of Friction, Static, & Dynamic Friction; Motion of Bodies, wedge friction, screw jack, & differential screw jack., , UNIT-2 : CENTROID & CENTRE OF GRAVITY, , (2–1 C to 2–45 C), , Centroid and Centre of Gravity, Centroid of simple figures from first, principle, centroid of composite sections; Centre of Gravity and its, implications; Area moment of inertia Definition, Moment of inertia of, plane sections from first principles, Theorems of moment of inertia,, Moment of inertia of standard sections and composite sections; Mass, moment inertia of circular plate, Cylinder, Cone, Sphere, Hook., , UNIT-3 : BASIC STRUCTURAL ANALYSIS, , (3–1 C to 3–28 C), , Basic Structural Analysis, Equilibrium in three dimensions;, Analysis of simple trusses by method of sections & method of, joints, Zero force members, Simple beams and support reactions., , UNIT-4 : REVIEW OF PARTICLE DYNAMICS, , (4–1 C to 4–38 C), , Review of particle dynamics- Rectilinear motion; Plane curvilinear motion, (rectangular, path, and polar coordinates). Work-kinetic energy, power,, potential energy. Impulse-momentum (linear, angular); Impact (Direct, and oblique)., , UNIT-5 : KINETICS OF RIGID BODIES, , (5–1 C to 5–34 C), , Introduction to Kinetics of Rigid Bodies, Basic terms, general, principles in dynamics; Types of motion, Instantaneous centre of, rotation in plane motion and simple problems; D’Alembert’s, principle and its applications in plane motion and connected bodies;, Work energy principle and its application in plane motion of, connected bodies; Kinetics of rigid body rotation., Virtual Work and Energy Method : Virtual displacements, principle, of virtual work for particle and ideal system of rigid bodies,, Applications of energy method for equilibrium, Stability of, equilibrium., , SHORT QUESTIONS, SOLVED PAPERS (2019-20), , (SQ–1 C to SQ–16 C), (SP-1C to SP-16C)
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www.aktutor.in, , ENGINEERING MECHANICS, , (L-T-P 3-1-0), , Credit – 4, , Course Outcomes: At the end of this course the student will be able to1. Use scalar and vector analytical techniques for analyzing forces in statically determinate, structures, 2. Apply fundamental concepts of kinematics and kinetics of particles to the analysis of simple,, practical problems., 3. Apply basic knowledge of mathematics and physics to solve real-world problems., 4. Understand basic dynamics concepts – force, momentum, work and energy;, 5. Understand and be able to apply Newton’s laws of motion;, UNIT - I Introduction to Engineering Mechanics: Force Systems, Basic concepts, Rigid Body, equilibrium; System of Forces, Coplanar Concurrent Forces, Components in Space – ResultantMoment of Forces and its Applications; Couples and Resultant of Force System, Equilibrium of, System of Forces, Free body diagrams, Equations of Equilibrium of Coplanar Systems., Friction: Types of friction, Limiting friction, Laws of Friction, Static and Dynamic Friction;, Motion of Bodies, wedge friction, screw jack & differential screw jack; [8 Hours], UNIT- II Centroid and Centre of Gravity, Centroid of simple figures from first principle,, centroid of composite sections; Centre of Gravity and its implications; Area moment of inertiaDefinition, Moment of inertia of plane sections from first principles, Theorems of moment of, inertia, Moment of inertia of standard sections and composite sections; Mass moment inertia of, circular plate, Cylinder, Cone, Sphere, Hook. [8 Hours], UNIT - III Basic Structural Analysis, Equilibrium in three dimensions; Analysis of simple, trusses by method of sections & method of joints, Zero force members, Simple beams and, support reactions. [8 Hours], UNIT - IV Review of particle dynamics- Rectilinear motion; Plane curvilinear motion, (rectangular, path, and polar coordinates). Work-kinetic energy, power, potential energy., Impulse-momentum (linear, angular); Impact (Direct and oblique). [8 Hours], UNIT - V Introduction to Kinetics of Rigid Bodies, Basic terms, general principles in dynamics;, Types of motion, Instantaneous centre of rotation in plane motion and simple problems;, D’Alembert’s principle and its applications in plane motion and connected bodies; Work energy, principle and its application in plane motion of connected bodies; Kinetics of rigid body rotation, Virtual Work and Energy Method- Virtual displacements, principle of virtual work for particle, and ideal system of rigid bodies, Applications of energy method for equilibrium, Stability of, equilibrium. [8 Hours], Books and References, 1. Irving H. Shames (2006), Engineering Mechanics, 4th Edition, Prentice Hall
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www.aktutor.in, , 2. F. P. Beer and E. R. Johnston (2011), Vector Mechanics for Engineers, Vol I - Statics, Vol II,, – Dynamics, 9th Ed, Tata McGraw Hill, 3. R. C. Hibbler (2006), Engineering Mechanics: Principles of Statics and Dynamics, Pearson, Press., 4. Andy Ruina and Rudra Pratap (2011), Introduction to Statics and Dynamics, Oxford, University Press, 5. Shanes and Rao (2006), Engineering Mechanics, Pearson Education,, 6. Hibler and Gupta (2010),Engineering Mechanics (Statics, Dynamics) by Pearson Education, 7. Reddy Vijaykumar K. and K. Suresh Kumar(2010), Singer’s Engineering Mechanics, 8. Bansal R.K.(2010), A Text Book of Engineering Mechanics, Laxmi Publications, 9. Khurmi R.S. (2010), Engineering Mechanics, S. Chand & Co., 10. Tayal A.K. (2010), Engineering Mechanics, Umesh Publications, 11. Strength of Materials by Timoshenko and Yσungs, East West Press., 12. Textbook of Applied Mechanics-Dynamics and Statics by Prasad I.B, Khanna Publications.
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Engineering Mechanics, , 1, , www.aktutor.in, , 1–1 C (CE-Sem-3), , Introduction to, Engineering, Mechanics, , CONTENTS, Part-1, , :, , Introduction to Engineering ................... 1–2C to 1–7C, Mechanics, Force Systems,, Basic Concepts, , Part-2, , :, , Rigid Body Equilibrium, ......................... 1–7C to 1–12C, System of Forces, Coplanar, Concurrent Forces, Components, in Space, Resultant, , Part-3, , :, , Moment of Forces .................................. 1–12C to 1–15C, and its Applications, , Part-4, , :, , Couples and Resultant .......................... 1–16C to 1–22C, of Force System, , Part-5, , :, , Equilibrium of System .......................... 1–22C to 1–28C, of Forces, Free Body Diagrams, , Part-6, , :, , Equations of Equilibrium ..................... 1–28C to 1–30C, of Coplanar Systems, , Part-7, , :, , Friction,, Limiting, Friction,, Friction,, , Part-8, , :, , Wedge Friction ....................................... 1–35C to 1–37C, , Part-9, , :, , Screw Jack and Differential ............... 1–37C to 1–44C, Screw Jack, , Types of Friction, ................... 1–30C to 1–35C, Friction, Laws of, Static and Dynamic, Motion of Bodies
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1–2 C (CE-Sem-3), , Introduction to Engineering Mechanics, , www.aktutor.in, , Introduction to Engineering Mechanics,, Force Systems, Basic Concepts., , CONCEPT OUTLINE, Engineering Mechanics : It is that branch of science which deals, with the behaviour of a body when the body is at rest or in motion., Branches of Mechanics :, i., , Statics : Branch of mechanics which deals with the study of, body when the body is at rest is known as statics., , ii., , Dynamics : Branch of mechanics which deals with the study of, body when the body is in motion is known as dynamics. It is, further divided into kinematics (force not considered) and, kinetics (force considered)., , Scalar Quantity : A quantity which is completely specified by, magnitude only is known as scalar quantity., Example : Mass, length, time, etc., Vector Quantity : A quantity which is specified by both magnitude, and direction is known as vector quantity., Example : Velocity, force, displacement, etc., , Questions-Answers, Long Answer Type and Medium Answer Type Questions, , Que 1.1., , Define free, fixed and forced vectors., , Answer, i., , ii., , Free Vector : A vector which can be moved parallel to its position, anywhere in space provided its magnitude, direction and sense remain, the same is known as free vector. Fig. 1.1.1(a) shows free vector., Fixed Vector : A vector whose initial point is fixed, is known as fixed, vector. Fig. 1.1.1(b) shows fixed vector., , iii. Forced Vector : A vector which can be applied anywhere along its line, of action is known as forced vector. Fig. 1.1.1(c) shows a forced vector.
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Engineering Mechanics, , 1–3 C (CE-Sem-3), , www.aktutor.in, , A, , A, , B, A, , (a) Free vector A, , (b) Fixed vector, , (c) Forced vector, , Fig. 1.1.1., , Que 1.2., , State and prove parallelogram law of forces., , Answer, A., , Statement : Parallelogram law states that if two forces, acting at a, point be represented in magnitude and direction by the two adjacent, sides of a parallelogram, then their resultant is represented in magnitude, and direction by the diagonal of the parallelogram passing through that, point., B, , C, , B, , Q, , R, , Q, , , , O, , , (a), , P, , A, , O, , , (b ), , P, , A, , D, , Fig. 1.2.1., , B., , Proof :, , 1., , Let two forces P and Q act at a point O as shown in Fig. 1.2.1(a). The, force P is represented in magnitude and direction by OA whereas the, force Q is represented in magnitude and direction by OB., , 2., , Let the angle between the two forces be ‘’. The resultant of these two, forces will be obtained in magnitude and direction by the diagonal (passing, through O) of the parallelogram of which OA and OB are two adjacent, sides. Hence draw the parallelogram with OA and OB as adjacent sides, as shown in Fig. 1.2.1(b)., , 3., , The resultant R is represented by OC in magnitude and direction., , 4., , From C draw CD perpendicular to OA produced., , 5., , Let,, , = Angle between two forces P and Q = AOB, = Angle made by resultant with OA., , 6., , In parallelogram OACB, AC is parallel and equal to OB.
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1–4 C (CE-Sem-3), , AC = Q, , , 7., , In triangle ACD, AD = AC cos = Q cos , and, , 8., , Introduction to Engineering Mechanics, , www.aktutor.in, CD = AC sin = Q sin , , In triangle OCD, OC2 = OD2 + DC2, But, , OC = R, OD = OA + AD = P + Q cos , DC = Q sin , R2 = (P + Q cos )2 + (Q sin )2, = P2 + Q2 cos2 + 2PQ cos + Q2 sin2 , = P2 + Q2 (cos2 + sin2 ) + 2PQ cos , , , , = P2 + Q2 + 2PQ cos , R=, , , , ( cos2 + sin2 = 1), , P 2 Q2 2PQ cos , , ...(1.2.1), , Eq. (1.2.1) gives the magnitude of resultant force R., 9., , Now from triangle OCD,, tan =, , CD, Q sin , , OD P Q cos , , Q sin , = tan – 1 , , P Q cos , , , , ...(1.2.2), , Eq. (1.2.2) gives the direction of resultant (R)., , Que 1.3., , Discuss the law of parallelogram of forces. Two forces, , equal to P and 2P act on a rigid body. When the first force is increased, by 100 N and the second force is doubled, the direction of the, resultant remains unchanged. Determine the value of P., , AKTU 2013-14, (I) Marks 05, , Answer, A., , Parallelogram Law of Forces : Refer Q. 1.2, Page, Unit-1., , B., , Numerical :, Given : F1 = P, F2 = 2P, F1 = P + 100, F2 = 2P, To Find : Value of P., , , , , 2P, R, P, , Fig. 1.3.1., , 1., , We know that,
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Engineering Mechanics, , www.aktutor.in, , tan =, 2., , 2 P sin , P 2 P cos , , ...(1.3.1), , According to question if P is now changed to P + 100 and 2P is now, changed to 4P then again direction of resultant remains same i.e.,, tan =, , 3., , 1–5 C (CE-Sem-3), , 4 P sin , ( P 100) 4 P cos , , ...(1.3.2), , From eq. (1.3.1) and eq. (1.3.2), we have, , 2P sin , 4 P sin , =, P 2P cos , (P 100) 4 P cos , sin [P + 100 + 4P cos ] = 2 sin [P + 2P cos ], sin [P + 100 + 4P cos – 2P – 4P cos ] = 0, Either, , sin = 0, , or, , P + 100 – 2P = 0, , P = 100 N, So, the value of P = 100 N, , Que 1.4., , Two forces P and Q are inclined at an angle of 75°,, , magnitude of their resultant is 100 N. The angle between the, resultant and the force P is 45°. Determine the magnitude of P and Q., , AKTU 2016-17, (II) Marks 10, , Answer, Given : = 75°, = 45°, R = 100 N, To Find : Magnitude of P and Q., 1., , The resultant R of P and Q is given by,, R=, 100 =, , P 2 Q2 2PQ cos , P 2 Q2 2PQ cos 75, , (100)2 = P2 + Q2 + 0.517 PQ, 2., , ...(1.4.1), , The inclination of R to the direction of the force P is given by,, tan =, , tan 45° =, , Q sin , P Q cos , , Q sin 75, P Q cos 75, , P + 0.259 Q = 0.966 Q, P = 0.707 Q, , ...(1.4.2)
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1–6 C (CE-Sem-3), 3., , Introduction to Engineering Mechanics, , www.aktutor.in, , Putting value of P from eq. (1.4.2) in eq. (1.4.1), we get, (100)2 = (0.707 Q)2 + Q2 + 0.517 (0.707)Q2, (100)2 = 1.865Q2, Q = 73.22 N, , 4., , From eq. (1.4.2), we have, P = 0.707 × 73.22 = 51.76 N, , Que 1.5., , What are the basic laws of mechanics ?, , Answer, Following are the basic laws of mechanics :, i., , Newton’s First Law of Motion : It states that every body continues in, a state of rest or uniform motion in a straight line unless it is compelled, to change that state by some external force acting on it., , ii., , Newton’s Second Law of Motion : It states that, the net external, force acting on a body in the direction of motion is directly proportional, to the rate of change of momentum in that direction., , iii. Newton’s Third Law of Motion : It states that to every action there, is always equal and opposite reaction., iv. Gravitational Law of Attraction : It states that two bodies will be, attracted towards each other along their connecting line with a force, which is directly proportional to the product of their masses and inversely, proportional to the square of the distance between their centres., , m1 m2, r2, G = Universal gravitational constant of proportionality., , Mathematically, F = G, where,, , Que 1.6., , What do you understand by resolution of force ?, , Answer, 1., , Resolution of a force means finding the components of a given force in, two given directions., , 2., , Let a given force be R which makes an angle with X-axis as shown in, Fig. 1.6.1. It is required to find the components of the force R along, X-axis and Y-axis., , 3., , Components of R along X-axis = R cos , Components of R along Y-axis = R sin , Hence, the resolution of force is the process of finding components of, forces in specified directions.
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Engineering Mechanics, , 1–7 C (CE-Sem-3), , www.aktutor.in, , Y, C, , B, R sin , , R, , , O, , R cos , , A, , X, , Fig. 1.6.1., , Rigid Body Equilibrium, System of Forces, Coplanar Concurrent, Forces, Components in Space, Resultant., , CONCEPT OUTLINE, Rigid Body : A body which does not deform under the action of, external forces is known as rigid body., System of Forces : When several forces act on a body then, they are, said to form a system of forces., Coplanar Force System : If in a system, all the forces lie in the same, plane, then the force system is known as coplanar., Non-Coplanar Force System : If in a system, all the forces lie in, different planes, then the force system is known as non-coplanar., , Questions-Answers, Long Answer Type and Medium Answer Type Questions, , Que 1.7., , Discuss in short about rigid body equilibrium., , Answer, 1., , The external forces acting on a rigid body can be reduced to a forcecouple system at some arbitrary point., , 2., , When the force and the couple are both equal to zero, the external, forces form a system equivalent to zero, and the rigid body is said to be, in equilibrium.
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1–8 C (CE-Sem-3), 3., , 4., 5., , Introduction to Engineering Mechanics, , www.aktutor.in, , The necessary and sufficient conditions for the equilibrium of a rigid, body are :, F = 0, MO = (r × F) = 0, In general, the point O should be fixed with respect to an inertial, reference frame., Resolving each force and each moment into its rectangular components,, we can express the necessary and sufficient conditions for the equilibrium, of a rigid body with following six scalar equations :, Fx = 0, Fy = 0, Fz = 0, Mx = 0, My = 0, Mz = 0, , Que 1.8., , Give the classification of system of forces and also, , explain the systems involved., , Answer, A., , Classification of System of Forces :, Force System, , Non-coplanar, , Coplanar, , Collinear, , Concurrent, , Parallel, , Non-concurrent, Non-parallel, , Concurrent, , Parallel, , Non-concurrent, Non-parallel, , Fig. 1.8.1., , B., a., , Explanation :, Coplanar Collinear System of Forces : Fig. 1.8.2 shows three forces, F1, F2 and F3 acting in the same plane. These three forces are in the, same line, i.e., these three forces are having a common line of action., This system of forces is known as coplanar collinear force system., Plane, , F1, , F2, , F3, , Fig. 1.8.2. Coplanar collinear forces., , b., , Coplanar Concurrent System of Forces : Fig. 1.8.3 shows three, forces F1, F2 and F3 acting in the same plane and these forces intersect
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Engineering Mechanics, , 1–9 C (CE-Sem-3), , www.aktutor.in, , or meet at a common point O. This system of forces is known as coplanar, concurrent force system., F2, Plane, O, , F1, , F3, Fig. 1.8.3. Concurrent coplanar forces., , c., , Coplanar Parallel System of Forces : Fig 1.8.4 shows three forces, F1, F2 and F3 acting in the same plane and these forces are parallel. This, system of forces is known as coplanar parallel force system., , Plane, F2, F3, F1, Fig. 1.8.4. Coplanar parallel forces., , d., , Coplanar Non-concurrent Non-parallel System of Forces :, Fig. 1.8.5 shows four forces F1, F2, F3 and F4 acting in a plane. The lines, of action of these forces lie in the same plane but they are neither, parallel nor meet or intersect at a common point. This system of forces, is known as coplanar non-concurrent non-parallel force system., F3, , F2, , F1, Plane, , F4, Fig. 1.8.5. Non-concurrent non-parallel forces.
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1–10 C (CE-Sem-3), , Que 1.9., , Introduction to Engineering Mechanics, , www.aktutor.in, , Define the principle of transmissibility of forces., , AKTU 2011-12, Marks 02, , Answer, 1., , Principle of transmissibility of forces states that if force acting at a point, on a rigid body is shifted to any other point which is on the line of action, of the force, the external effect of the force on the body remains, unchanged., F1 = F, , F1 = F, O, , O, , O, , F2 = F, O, , O, F, , O, , F, (a), , (b ), , (c ), , Fig. 1.9.1., , 2., , For example, consider a force F acting at a point O on a rigid body as, shown in Fig. 1.9.1(a)., , 3., , On this rigid body, “there is another point O in the line of action of the, force F., , 4., , Suppose at this point O, two equal and opposite forces F1 and F2 (each, equal to F and collinear with F) are applied as shown in Fig. 1.9.1(b)., , 5., , The force F and F2 being equal and opposite will cancel each other, leaving a force F1 at point O as shown in Fig. 1.9.1(c). But force F1 is, equal to force F., , 6., , The original force F acting at point O has been transferred to point O, which is along the line of action of F without changing the effect of the, force on the rigid body., , 7., , Hence any force acting at a point on a rigid body can be transmitted to, act at any other point along its line of action without changing its effect, on the rigid body. This proves the principle of transmissibility of a force., , Que 1.10. Describe the component of forces in space and also give, the formula for resultant., , Answer, 1., , Consider a force F acting at the origin O of the system of rectangular, coordinates X, Y and Z.
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Engineering Mechanics, , 1–11 C (CE-Sem-3), , www.aktutor.in, , 2., , To define the direction of F, we draw the vertical plane OBAC containing, F [Fig. 1.10.1 (a)]. This plane passes through the vertical Y-axis; its, orientation is defined by the angle it forms with the XY plane., , 3., , The direction of F within the plane is defined by the angle y that F, forms with Y-axis. The force F may be resolved into a vertical component, Fy and a horizontal component Fh, this operation is shown in Fig. 1.10.1(b),, is carried out in plane OBAC., Y, , Y, , B, , B, y F, , A, , Fy, , O, , X, Fh, , C, Z, , A, , O, , X, , , , y F, , C, Z, , (a), , (b), , Y, B, Fy, O, Fz, , Fx, D, , , , X, , Fh, C, , E, Z, , (c ), Fig. 1.10.1., , 4., , The corresponding scalar components are :, , 5., , But Fh may be resolved into two rectangular components Fx and Fz, along the X and Z axes, respectively. This operation shown in, Fig. 1.10.1(c) is carried out in the XZ plane., , 6., , We obtain the following expression for the corresponding scalar, components :, , Fy = F cos y, , Fh = F sin y, , ...(1.10.1), , Fx = Fh cos = F sin y cos , Fz = Fh sin = F sin y sin , 7., , ...(1.10.2), , The given force F has thus been resolved into three rectangular vector, components Fx, Fy, Fz which are directed along the three coordinate, axes.
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1–12 C (CE-Sem-3), 8., , Introduction to Engineering Mechanics, , www.aktutor.in, , Applying the Pythagorean Theorem to the triangles OAB and OCD of, Fig. 1.10.1, we write, F2 = (OA)2 = (OB)2 + (BA)2 = Fy2 + Fh2, Fh2 = (OC)2 = (OD)2 + (DC)2 = Fx2 + Fz2, , 9., , Fh2, , Eliminating, from these two equations and solving for F, we obtain, the following relation between the magnitude of F and its rectangular, scalar components,, F=, , Fx2 Fy2 Fz2, , 10. We also have,, Fx = F cos x, Fy = F cos y and Fz = F cos z, where,, , x, y, z, = Angle made of F with X-axis, Y-axis and, Z-axis, respectively., , Que 1.11., , A force F has the components Fx = 100 N, Fy = – 150 N,, , Fz = 300 N. Determine its magnitude F and the angles x, y, z it, forms with the coordinates axes., , Answer, Given : Fx = 100 N, Fy = – 150 N, Fz = 300 N, To Find : F, x, y and z, 1., , We know that,, F=, , 2., , Fx2 Fy2 Fz2, , =, , (100)2 ( 150)2 (300)2, , =, , 122500 = 350 N, , Also, we know that, cos x =, cos y =, cos z =, , Fx 100, , x = 73.4°, F 350, Fy, F, , , , 150, y = 115.4°, 350, , Fz 300, , z = 31.0°, F 350, , Moment of Forces and its Applications.
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Engineering Mechanics, , www.aktutor.in, , 1–13 C (CE-Sem-3), , Questions-Answers, Long Answer Type and Medium Answer Type Questions, , Que 1.12. Define moment of forces. Also give its applications., , Answer, A., , Moment of Forces : The product of a force and the perpendicular, distance of the line of action of the force from a point is known as, moment of the force about that point., Moment (M) of the force F about O is given by,, M = Fr, Line of action of force, F, Perpendicular distance, , r, O, , Fig. 1.12.1., , B., , Applications : Following are the applications of moment of forces :, , 1., , Used in levers., , 2., , Used in levers safety valve., , 3., , Used in balancing., , Que 1.13. State and prove Varignon’s theorem., , AKTU 2011-12, Marks 05, , Answer, A., , Statement : Varignon’s theorem states that the moment of a force, about any point is equal to the algebraic sum of the moments of its, components about that point., , B., , Proof :, , 1., , Let R be the resultant of forces F1 and F2 and B the moment centre., , 2., , Let d, d1 and d2 be the moment arms of the forces, R, F1 and F2, respectively, from the moment centre B. Then in this case, we have to prove that :, Rd = F1 d1 + F2 d2
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1–14 C (CE-Sem-3), 3., , Introduction to Engineering Mechanics, , www.aktutor.in, , Join AB and consider it as Y-axis and draw X-axis at right angle to it at, A [Fig. 1.13.1(b)]. Denoting by the angle that R makes with X-axis, noting that the same angle is formed by perpendicular to R at B with, AB1, we can write :, Rd = R × AB cos, = AB × (R cos ), = AB × Rx, ...(1.13.1), where Rx denotes the component of R in X direction., B, , d2, d1, , Y, R, , B, , d, , F2, , , , A, A, (a), , d, P2, , F1, , (b ), , R, , B1, F1, X, , Fig. 1.13.1., , 4., , Similarly, if F1x and F2x are the components of F1 and F2, in X direction,, respectively, then, F1 d1 = AB × F1x, , 5., 6., , ...(1.13.2), , and, F2 d2 = AB × F2x, ...(1.13.3), From eq. (1.13.2) and eq. (1.13.3), we have, F1 d1 + F2 d2 = AB (F1x + F2x) = AB × Rx, ...(1.13.4), Since, the sum of x components of individual forces is equal to the x, component of the resultant R. From eq. (1.13.1) and eq. (1.13.4), we can, conclude :, Rd = F1 d1 + F2 d2, , Que 1.14., , Calculate the moment of 90 N force about point O for, , the condition = 15°. Also, determine the value of for which the, moment about O is zero., O, 500 mm, F = 90 N, , A, 600 mm, Fig. 1.14.1., , AKTU 2013-14, (II) Marks 05
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Engineering Mechanics, , www.aktutor.in, , 1–15 C (CE-Sem-3), , Answer, Given : = 15°, F = 90 N, To Find : i. Moment., ii. Value of ., 1., , Taking moment about O by 90 N force,, M = 90 cos 15° × 600 – 90 sin 15° × 500, = 52159.99 – 11646.85, = 40513.14 N, O, , 500 mm, , F = 90 N, , , 600 mm, , F cos , A, F sin , , Fig. 1.14.2., , 2., , According to the question, moment about O due to 90 N is zero., , MO = 0, 90 cos × 600 – 90 sin × 500 = 0, 54 cos = 45 sin , , 54 6, , 45 5, tan = 1.2, = 50.19°, tan =, , Que 1.15. What do you understand by like parallel forces and unlike, parallel forces ?, , Answer, i., , Like Parallel Forces : The parallel forces which are acting in the same, direction are known as like parallel forces. These forces may be equal or, unequal in magnitude., , ii., , Unlike Parallel Forces : The parallel forces which are acting in the, opposite direction are known as unlike parallel forces.
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1–16 C (CE-Sem-3), , Introduction to Engineering Mechanics, , www.aktutor.in, , Couples and Resultant of Force System., , CONCEPT OUTLINE, Couple : Two parallel forces equal in magnitude and opposite in, direction and separated by a definite distance are said to form a couple., Resultant of Several Forces : When a number of coplanar forces, are acting on a rigid body, then these forces can be replaced by a single, force which has the same effect on the rigid body as that of all the, forces acting together, then this single force is known as the resultant, of several forces., , Questions-Answers, Long Answer Type and Medium Answer Type Questions, , Que 1.16., , Derive an expression for the resultant of collinear, , coplanar forces., , Answer, 1., , 2., , The resultant is obtained by adding all the forces if they are acting in the, same direction. If any one of the forces is acting in the opposite direction,, then resultant is obtained by subtracting that force., Fig. 1.16.1, shows three collinear coplanar forces F1, F2 and F3, acting on a rigid body in the same direction, their resultant R will be, the sum of these forces., R = F1 + F2 + F3, , , , F1, , F2, , F3, , Fig. 1.16.1., , 3., , If any one of these forces (say force F2) is acting in the opposite direction,, as shown in Fig. 1.16.2, then their resultant will be given by,, R = F1 – F2 + F3
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Engineering Mechanics, , 1–17 C (CE-Sem-3), , www.aktutor.in, , F1, , F2, , F3, , Fig. 1.16.2., , Que 1.17., , Three collinear horizontal forces of magnitude 200 N,, , 100 N and 300 N are acting on a rigid body. Determine the resultant, of the forces analytically when, i., All the forces are acting in the same direction., ii. The force 100 N acts in the opposite direction., , Answer, Given : F1 = 200 N, F2 = 100 N and F3 = 300 N, To Find :, Resultant, when, i. All the forces are acting in the same direction., ii. The force 100 N acts in the opposite direction., 1., , When all the forces are acting in the same direction, then resultant is, given as,, , 2., , When the force 100 N acts in the opposite direction, then resultant is, given as,, , R = F1 + F2 + F3 = 200 + 100 + 300 = 600 N, , R = F1 + F2 + F3 = 200 – 100 + 300 = 400 N, , Que 1.18., , Derive an expression for the resultant of concurrent, , coplanar forces when two or more than two forces act on a point., , Answer, A., 1., , When Two Forces Act at a Point :, Suppose two forces P and Q act at point O as shown in Fig. 1.18.1 and , is the angle between them. Let is the angle made by the resultant R, with direction of force P., C, , B, Q, , O, , R, , , P, , A, Fig. 1.18.1.
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1–18 C (CE-Sem-3), 2., , Introduction to Engineering Mechanics, , www.aktutor.in, , Forces P and Q form two sides of a parallelogram and according to the, law, the diagonal through the point O gives the resultant R as shown., Thus, the magnitude of resultant is given by,, R=, , 3., , P 2 Q2 2PQ cos , , The direction of the resultant with the force P is given by,, , Q sin , = tan– 1 , P Q cos , B., , When More than Two Forces Act at a Point :, , 1., , According to this method, all the forces acting at a point are resolved, into horizontal and vertical components and then algebraic summation, of horizontal and vertical components is done separately., , 2., , The summation of horizontal component is written as FH and that of, vertical FV. Then resultant R is given by,, R=, , 3., , The angle made by the resultant with horizontal is given by,, tan =, , 4., , (FH )2 (FV )2, FV, FH, , Let four forces F1, F2, F3 and F4 act at a point O as shown in Fig. 1.18.2., Y, F1, , F2, 2, X, , 3, , 1, O, , F3, Y, Fig. 1.18.2., , 5., , X, , 4, , F4, , The inclination of the forces is indicated with respect to horizontal, direction. Let,, 1 = Inclination of force F1 with OX., 2 = Inclination of force F2 with OX., 3 = Inclination of force F3 with OX., 4 = Inclination of force F4 with OX., , 6., , Summation or algebraic sum of horizontal components,
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Engineering Mechanics, , www.aktutor.in, , 1–19 C (CE-Sem-3), , FH = F1 cos 1 – F2 cos 2 – F3 cos 3 + F4 cos 4, 7., , Summation or algebraic sum of vertical components,, FV = F1 sin 1 + F2 sin 2 – F3 sin 3 – F4 sin 4, , 8., , Then the resultant will be given by,, R=, , (FH )2 (FV )2, , And the angle () made by resultant with X-axis is given by,, tan =, , (FV ), (FH ), , Que 1.19. The force system applied to an angle bracket is shown, in Fig. 1.19.1. Determine the magnitude, direction and line of action, of the resultant force., 80 kN, , 2m, 150 kN, , 20 kN, , 30°, , 40 kN, , 4m, Fig. 1.19.1., , AKTU 2013-14, (I) Marks 10, , Answer, Given : Fig. 1.19.1., To Find : Magnitude, direction and line of action of the resultant force., 1., , Considering the equilibrium of force system, we have, FH = 0 – 80 + 150 cos 30° + R cos = 0, R cos = – 49.9 kN (towards negative X-axis), FV = 0 150 sin 30° + R sin + 20 – 40 = 0, R sin = – 55 kN (towards negative Y-axis)
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1–20 C (CE-Sem-3), , Introduction to Engineering Mechanics, , www.aktutor.in, A, , 80 kN, , 150 kN, 2m, , R, 20 kN, R sin, 30°, , , C, 4m, , O, , B, R cos, , x, 40 kN, Fig. 1.19.2., , 2., , ( 49.9)2 (55)2 = 74.26 kN, , R=, 3., , ( R cos )2 ( R sin )2, , Resultant magnitude, R =, , Direction of the resultant,, tan =, , 55, R sin , =, 49.9, R cos , , tan = 1.1022, = 47.78°, 4., , Now for line of action of the resultant taking moment about O, we have, MO = 0, 80 × 2 – R sin × OC + 20 × 4 = 0, 160 – 74.26 × sin 47.78° × x + 80 = 0, x = 4.36 m, Resultant will act at a distance 4.36 m from point O towards B and it will, lie outside the frame., , Que 1.20. The resultant of four forces which are acting at a point, O as shown in Fig. 1.20.1 is along Y-axis. The magnitude of forces F1,, F3 and F4 are 10 kN, 20 kN and 40 kN respectively. The angles made by, 10 kN, 20 kN and 40 kN with X-axis are 30°, 90° and 120° respectively., Find the magnitude and direction of force F2 if resultant is 72 kN.
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1–22 C (CE-Sem-3), , 6., , Introduction to Engineering Mechanics, , www.aktutor.in, , , = tan– 1 1.0899 = 47.46°, Substituting the value of in eq. (1.20.2), we get, F2 sin (47.46°) = 12.36, F2 =, , 12.36, 12.36, , = 16.77 kN, sin(47.46) 0.7368, , Equilibrium of System of Forces, Free Body Diagrams., , CONCEPT OUTLINE, Equilibrium of System of Forces : When some external forces act, on a body but it does not start moving and also does not start rotating, about any point, then the body is said to be in equilibrium., Free Body Diagram : A diagram in which the body under, consideration is freed from all the contact surfaces and all the forces, acting on it are shown on it, is known as free body diagram (FBD)., , Questions-Answers, Long Answer Type and Medium Answer Type Questions, , Que 1.21. State and prove Lami’s Theorem., , AKTU 2011-12, Marks 05, , Answer, A., , Statement : Lami’s theorem states that if three forces acting at a point, are in equilibrium, then each force will be proportional to the sine of the, angle between the other two forces., , B., , Proof of Lami’s Theorem :, , 1., , The three forces acting on a point are in equilibrium and hence they can, be represented by the three sides of the triangle taken in the same, order., , 2., , Now draw the force triangle as shown in Fig. 1.21.1(b)., , 3., , Now applying sine rule, we get, , P, Q, R, , =, sin(180 ), sin(180 ) sin(180 ), 4., , This can also be written as,
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Engineering Mechanics, , 1–23 C (CE-Sem-3), , www.aktutor.in, , P, Q, R, , =, sin , sin sin , Q, , , , , O, , P, , (a), , R, , 80, °–, , ), , , , (1, , R, , Q, , (180° – ), (180° – ), , O, , , , P, , , (b), Fig. 1.21.1., , Que 1.22. Write in short about principle of equilibrium., , Answer, 1., , The principle of equilibrium states that, a stationary body which is, subjected to coplanar forces (concurrent or parallel) will be in equilibrium, if the algebraic sum of all the external forces is zero and also the algebraic, sum of moments of all the external forces about any point in their plane, is zero., , 2., , Mathematically, it is expressed by the following equations, F = 0, , ...(1.22.1), , M = 0, , ...(1.22.2), , 3., , The eq. (1.22.1) is also known as force law of equilibrium whereas the, eq. (1.22.2) is known as moment law of equilibrium., , 4., , The forces are generally resolved into horizontal and vertical components., Hence eq. (1.22.1) is written as
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1–24 C (CE-Sem-3), , Introduction to Engineering Mechanics, , www.aktutor.in, Fx = 0 and Fy = 0, , where,, , Fx = Algebraic sum of all horizontal components., Fy = Algebraic sum of all vertical components., , Que 1.23. Two slender rods of negligible weight are pin connected, at C and attached to two blocks. A and B each of weight 100 N is, shown in Fig. 1.23.1. If coefficient of friction is 0.3 at all surfaces of, contact, find largest value of P for which equilibrium is maintained., P, , 10°, C, , A, , 30°, , 60°, , B, , Fig. 1.23.1., , AKTU 2013-14, (I) Marks 05, , Answer, Given : W = 100 N, = 0.3, To Find : Value of P., 1., , Considering FBD of pin C (Fig. 1.23.2), we have, FCB = P cos 10°, FCA = P sin 10°, FCA, FCB, P, , 10°, 10°, FCB, , P, , FCA, Fig. 1.23.2., , 2., 3., , Now, considering the FBD of block A (Fig. 1.23.3)., For vertical force equilibrium,, FV = 0, RA – 100 – FAC cos = 0, RA = 100 + P sin 10° cos , , 4., , Now,, , FH = 0,, , 0.3 RA – FAC sin = 0, 0.3 [100 + P sin 10° cos ] – P sin 10° sin = 0, , {FAC = FCA}
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Engineering Mechanics, , 1–25 C (CE-Sem-3), , www.aktutor.in, , 30 + 0.3 P sin 10° cos – P sin 10° sin = 0, P sin 10° (sin – 0.3 cos ) = 30, , 30, sin 10(sin 0.3 cos ), , P=, 100 N, , RA, FA, , C, , 90° – , , FAC sin , 0.3 NA, , FA = N A, , FAC cos , , RA, , 100 N, Fig. 1.23.3., , 5., , Let,, , = 60°, so, P=, , 30, sin 10(sin 60 0.3 cos 60), , P = 241.28 N, 6., , Considering FBD of block B (Fig. 1.23.4)., 100 N, , RB, , FBC, 30°, 60°, RB, , FBC cos 60°, , RB, , RB, , FBC sin 60°, 100 N, Fig. 1.23.4., , 7., , For vertical equilibrium,, 100 + FBC sin 60° = RB, RB = 100 + P cos 10° sin 60°, , 8., , For horizontal equilibrium,, 0.3 RB – FBC cos 60°= 0, 0.3 (100 + P cos 10° sin 60°) – P cos 10° cos 60° = 0, 30 + P cos 10° (0.3 sin 60° – cos 60°) = 0, P=, , 30, cos 10(cos 60 0.3 sin 60), , { FBC = FCB}
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1–26 C (CE-Sem-3), , Introduction to Engineering Mechanics, , www.aktutor.in, P = 126.83 N, , 9., , So the largest value of P for which equilibrium is maintained will be,, P = 126.83 N, , Que 1.24. Two smooth spheres each of radius 100 mm and weight, 100 N, rest in a horizontal channel having vertical walls, the distance, between which is 360 mm. Find the reactions at the points of contacts, A, B, C, and D shown in Fig. 1.24.1 below., 360 mm, O1, , A, , B, , , D, , O2, , P, , C, Fig. 1.24.1., , AKTU 2016-17, (II) Marks 07, , Answer, Given : r = 100 mm = 0.1 m, W = 100 N, l = 360 mm = 0.36 m, To Find : Reaction at A, B, C and D., 1., , From Fig. 1.24.1, we have, cos =, , O2 P, 360 – O1 A – O2 D, =, O1O2, O1 B O2 B, , 360 – 100 – 100 160, , 100 100, 200, cos = 0.8, =, , sin =, =, , 1 cos2 1 (0.8)2, , 0.36 = 0.6, RB, , RD, , 100 N, 100 N, RB, RC, (a), Fig. 1.24.2., , (b ), , RA
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Engineering Mechanics, 2., , 3., , www.aktutor.in, , 1–27 C (CE-Sem-3), , Considering FBD of sphere 1 [Fig. 1.24.2(b)]., FV = 0, RB × sin = W, RB × 0.6 = 100, RB = 166.67 N, FH = 0, RA = RB × cos , RA = 166.667 × 0.8 = 133.33 N, Considering FBD of sphere 2 [Fig. 1.24.2(a)]., FV = 0, RC = RB sin + W, RC = 166.67 × 0.6 + 100, RC = 200 N, FH = 0, RD = RB cos , RD = 166.67 × 0.8 = 133.33 N, , Que 1.25., , Two identical rollers, each of weights 1000 N are, , supported by an inclined plane as shown in Fig. 1.25.1. Assuming, smooth surfaces, find the reactions induced at the points of supports., , AKTU 2015-16, (I) Marks 10, , 70º, , 40º, , Fig. 1.25.1., , Answer, Given : Fig. 1.25.1, w = 1000 N, To Find : Reactions at the point of supports, 1., , Considering FBD of sphere 1 (Fig. 1.25.2)., Along axis OO :, R1 cos 20° – 1000 cos 20° – R4 = 0, R1 cos 20° – R4 = 939.69, , ...(1.25.1), , Along axis perpendicular to OO :, R2 – 1000 sin 20° – R1 sin 20° = 0, – R1 sin 20° + R2 = 342.02, , ...(1.25.2)
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Engineering Mechanics, , 1–29 C (CE-Sem-3), , www.aktutor.in, Questions-Answers, , Long Answer Type and Medium Answer Type Questions, , Que 1.26., , Write down the equations of equilibrium for coplanar, , non-concurrent force system and coplanar concurrent force system., , Answer, i., , Equations of Equilibrium for Coplanar Non-concurrent Force, System :, , 1., , A non-concurrent force system will be in equilibrium if the resultant of, all forces and moment is zero., , 2., , Hence the equations of equilibrium are :, Fx = 0,Fy = 0 and M = 0, , ii., , Equations of Equilibrium for Coplanar Concurrent Force, System :, , 1., , For the concurrent forces, the lines of action of all forces meet at a point,, and hence the moment of those forces about that point will be zero or, M = 0 automatically., , 2., , Thus for concurrent force system, the condition M = 0 becomes, redundant and only two conditions, i.e., Fx = 0 and Fy = 0 are required., , Que 1.27. Three parallel forces F1, F2 and F3 are acting on a body, as shown in Fig. 1.27.1 and the body is in equilibrium. If force, F1 = 250 N and F3 = 1000 N and the distance between F1 and F2 = 1.0 m,, then determine the magnitude of force F2 and the distance of F2, from force F3., F1 = 250 N, , F3 = 1000 N, B, C, , A, x, , 1m, F2, Fig. 1.27.1., , Answer, Given : F1 = 250 N, F3 = 1000 N, AB = 1.0 m, To Find : F2 and BC.
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1–30 C (CE-Sem-3), 1., , Introduction to Engineering Mechanics, , www.aktutor.in, , For the equilibrium of the body, the resultant force in the vertical, direction should be zero., , , FV = 0, F1 + F3 – F2 = 0, 250 + 1000 – F2 = 0, F2 = 250 + 1000 = 1250 N, , 2., , For the equilibrium of the body, the moment of all forces about any, point must be zero. Taking moments of all forces about A and considering, distance BC = x, we have, F2 × AB – AC × F3 = 0, 1250 × 1 – (1 + x) × 1000 = 0, , ( AC = AB + BC = 1 + x), , 250 = 1000 x, x=, , 250, = 0.25 m, 1000, , Friction, Types of Friction, Limiting Friction, Laws of Friction, Static and Dynamic Friction, Motion of Bodies., , CONCEPT OUTLINE, Force of Friction : When a solid body slides over a stationary solid, body, a force is exerted at the surface of contact by the stationary body, on the moving body, this force is called force of friction., Static Friction : The force of friction up to which body does not move, is called static friction., Limiting Friction : The force of friction at which body just tends to, start moving is called limiting friction., Kinetic Friction : The force of friction acting on the body when the, body is moving is called kinetic friction., , Questions-Answers, Long Answer Type and Medium Answer Type Questions, , Que 1.28. Define friction. Also explain its types.
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Engineering Mechanics, , www.aktutor.in, , 1–31 C (CE-Sem-3), , Answer, A., , Friction : The property of the bodies by virtue of which a force is, exerted by a stationary body on the moving body to resist the motion of, the moving body is called friction. Friction acts parallel to the surface of, contact and depends upon the nature of surface of contact., , B., , Types of Friction :, , a., , Static and Dynamic Friction : If the two surfaces which are in contact,, are at rest, the force experienced by one surface is called static friction., But if one suface starts moving and the other is at rest, the force, experienced by the moving surface is called dynamic friction., , b., , Wet and Dry Friction : If between two surfaces, which are in contact,, lubrication is used, the fricion, that exists between two surfaces is known, as wet friction. But if no lubricantion is used, then the friction between, two surfaces is called dry friction or solid friction., , Que 1.29. Write down the laws of friction., , Answer, Following are the laws of friction :, 1., , The force of friction acts in the opposite direction in which surface is, having tendency to move., , 2., , The force of friction is equal to the force applied to the surface, so long, as the surface is at rest., , 3., , The limiting frictional force bears a constant ratio to the normal reaction, between two surfaces., , 4., , The limiting frictional force does not depend upon the shape and areas, of the surfaces in contact., , 5., , The ratio between limiting friction and normal reaction is slightly less, when the two surfaces are in motion., , 6., , The force of friction is independent of the velocity of sliding., , Que 1.30. Define the following terms :, i., Coefficient of friction., ii. Angle of friction, and, iii. Angle of repose., , Answer, i., , Coefficient of Friction : It is defined as the ratio of the limiting force, of friction (F) to the normal reaction (R) between two bodies. It is denoted, by ., Mathematically,, , =, , Limiting force of friction F, , Normal reaction, R
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1–32 C (CE-Sem-3), ii., , Introduction to Engineering Mechanics, , www.aktutor.in, , Angle of Friction : It is defined as the angle made by the resultant of, the normal reaction (R) and the limiting force of friction (F) with the, normal reaction (R). It is denoted by ., , F R, , =, R, R, iii. Angle of Repose : It is defined as the maximum inclination of a plane at, which a body remains in equilibrium over the inclined plane by the, assistance of friction only., Mathematically, tan =, , Also, Angle of repose = Angle of friction, , Que 1.31., , Two blocks, as shown in Fig. 1.31.1 slide down at 30°, , incline. If coefficient of friction at all contact surfaces is 0.2, determine, the pressure between the blocks., , 0, 30, , 20, , N, , 0N, , 30°, Fig. 1.31.1., , AKTU 2013-14, (I) Marks 10, , Answer, Given : = 0.2, = 30°, Weight of blocks = 200 N and 300 N, To Find : Pressure between two blocks., 1., , Considering FBD of block of 300 N (Fig. 1.33.2)., a, , R1, R1, 300N, , 300 sin 30°, , 30°, , P, 30°, , 300 cos 30°, 300, Fig. 1.31.2., , 2., , For equilibrium, we have, FV = 0, R1 = 300 cos 30°
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Engineering Mechanics, , 1–33 C (CE-Sem-3), , www.aktutor.in, , Also, FH = 0, 300 sin 30° + P – R1 = ma, 300 sin 30° + P – 0.2 × 300 cos 30° =, , 3., , 300, a, 9.81, , 300, a – P = 98.04, 9.81, Now considering the FBD of block of 200 N (Fig. 1.31.3)., , 20, , 0s, , in, , 30, , ...(1.31.1), , P, R2, , R2, , a, , W, , m g , , 200 N, , °, , 30°, 200 cos 30°, 200 N, , 30°, , Fig. 1.31.3., , 4., , For equilibrium, we have, FV = 0, R2 = 200 cos 30°, FH = 0, 200 sin 30° – P – R2 = ma, 100 – P – 0.2 × 200 cos 30° =, , 5., , 200, a, 9.81, , 200, a + P = 65.36, 9.81, After solving eq. (1.31.1) and eq. (1.31.2), we have, , ...(1.31.2), , a = 3.206 m/sec2 and P = 0, So, no pressure will act between the blocks., , Que 1.32. Determine the force P required to impend the motion of, the block B shown in Fig. 1.32.1. Take coefficient of friction as 0.3, for all contact surface., A, T, , 300 N, P, , B, , 500 N, 400 N, C, , Fig. 1.32.1., , AKTU 2014-15, (II) Marks 10
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1–34 C (CE-Sem-3), , Introduction to Engineering Mechanics, , www.aktutor.in, , Answer, Given : WA = 300 N, WB = 500 N, WC = 400 N, = 0.3, To Find : Value of P., 1., , Considering the FBD of block A (Fig. 1.32.2)., FV = 0, R1 = 300 N, Since F1 is limiting friction,, , , F1 = R1 = 0.3 × 300 = 90 N, FH = 0, gives, T = F1 = 90 N, 300 N, T, , A, R1 = F1, , R1, , Fig. 1.32.2., , 2., , Considering the FBD of block B (Fig. 1.32.3)., R1, P, F2, , B, , F 1 = R 1, , R2, 500 N, Fig. 1.32.3., , FV = 0, R2 – 500 – R1 = 0, R2 – 500 – 300 = 0, R2 = 800 N, , , F2 = R2 = 0.3 × 800 = 240 N, FH = 0, P = F1 + F2, P = 240 + 90, P = 330 N, , Que 1.33. What are the different types of motion of bodies ?, , Answer, Following are the different types of motion of bodies :
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Engineering Mechanics, , www.aktutor.in, , 1–35 C (CE-Sem-3), , i., , Linear Motion : When a body moves in a straight line only, the motion, is called linear motion., ii. Curvilinear Motion : When a body moves along a curved path, the, motion is called curvilinear motion., iii. Rectilinear Motion : When a body posses both linear and circular, motion, it is said to be in rectilinear motion., iv. Periodic Motion : When the motion of a body repeats over a period of, time, it is called periodic motion., v. Oscillatory Motion : To and fro motion of a body about a point is called, oscillatory motion., , Wedge Friction., Questions-Answers, Long Answer Type and Medium Answer Type Questions, , Que 1.34. Define wedge and discuss about the equilibrium of body, placed on wedge., , Answer, A., , Wedge : A wedge is a piece of metal or wood which is usually of a, triangular or trapezoidal in cross-section. It is used for either lifting, loads or used for slight adjustments in the position of a body i.e., for, tightening fits or keys for shafts., , B., , Equilibrium of Body Placed on Wedge :, , 1., , Considering the equilibrium of the wedge. The forces acting on the, wedge are shown in Fig. 1.34.1. They are :, , 2., , i., , The force P applied horizontally on face BC., , ii., , Reaction R1 on the face AC (The reaction R1 is the resultant of, normal reaction on the rubbing face AC and force of friction on, surface AC). The reaction R1 will be inclined at an angle 1 with the, normal., , iii., , Reaction R2 on the face AB (The reaction R2 is the resultant of, normal reaction on the rubbing face AB and force of friction on, surface AB). The reaction R2 will be inclined at an angle 2 with the, normal., , When the force P is applied on the wedge, the surface CA will be moving, towards left and hence force of friction on this surface will be acting, towards right.
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1–36 C (CE-Sem-3), , Introduction to Engineering Mechanics, , www.aktutor.in, , 3., , Similarly, the force of friction on face AB will be acting from A to B., These forces are shown in Fig. 1.34.1., , 4., , Resolving the forces horizontally, we get, R1 sin 1 + R2 sin (2 + ) = P, Resolving the forces vertically, we get, R1 cos 1 = R2 cos (2 + ), No, a, rm, , R2, , l, , 2, , (, , 90, , –, , ), , B, , of, rce, Fo ction, fri, , , P, , C, , A, 1, , Normal, reaction, , Force of, friction R, 1, , Fig. 1.34.1., , Que 1.35., , A uniform ladder 5 m long weighs 180 N. It is placed, , against a wall making an angle of 60º with floor. The coefficient of, friction between the wall and ladder is 0.25 and between the floor, and the ladder is 0.35. The ladder has to support a mass 900 N at its, top. Calculate the horizontal force P to be applied to the ladder at, the floor level to prevent slipping., , AKTU 2014-15, (II) Marks 10, , Answer, Given : W1 = 180 N, W2 = 900 N, a = 0.35, b = 0.25, l = 5 m, = 60°, To Find : Horizontal force P to prevent slipping., 1., , According to Fig. 1.35.1 for the ladder AB placed against a wall and, various force acting on it. P is the horizontal force which has been, applied on the ground level to prevent slipping., , 2., , Resolving all the forces along horizontal and vertical directions, we, have, P + aRa = Rb, Ra + b Rb = W1 + W2 = 180 + 900 = 1080 N, , 3., , Taking moments about the end A,, W2 × OA + W1 × DA = Rb × OB + b Rb × OA, , ...(1.35.1), ...(1.35.2)
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Engineering Mechanics, , 1–37 C (CE-Sem-3), , www.aktutor.in, , W2 = 900 N, Fb = bRb, Rb, , 2.5, , m5m, , B, , P, , W1 = 180 N, , 60º, , F a = aR a, , A, , O, D, , Ra, Fig. 1.35.1., , 4., , From the geometrical configuration,, OA = 5 cos 60º = 2.5 m, DA = 2.5 cos 60º = 1.25 m, OB = 5 sin 60º = 4.33 m, 900 × 2.5 + 180 × 1.25 = Rb × 4.33 + 0.25 Rb × 2.5, Rb (4.955) = 2475, , 2475, = 499.495 N, 4.955, From eq. (1.35.2) and eq. (1.35.1), we have, Rb =, , 5., , Ra =, Ra =, P=, P=, , 1080 – b Rb = 1080 – 0.25 × 499.495, 955.13 N, Rb – a Ra = 499.495 – 0.35 × 955.13, 165.2 N, , Screw Jack and Differential Screw Jack., , CONCEPT OUTLINE, Screw Jack : It is a device used for lifting heavy weights or loads with, the help of a small effort applied at its handle.
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1–38 C (CE-Sem-3), , Introduction to Engineering Mechanics, , www.aktutor.in, Questions-Answers, , Long Answer Type and Medium Answer Type Questions, , Que 1.36. Derive an expression for the effort applied to lift or lower, the load., , Answer, I., , Effort Applied at the End of Handle to Lift the Load :, , 1., , Let,, , W = Weight placed on the screw head,, P = Effort applied at the end of the handle,, L = Length of handle,, p = Pitch of the screw,, d = Mean diameter of the screw,, = Angle of the screw or helix angle,, = Angle of friction, and, = Coefficient of friction between screw and nut = tan , Load = w, L, , Screw head, , P(Power), Handle, Screw with, square threads, Nut, (Also body of jack), d, , Fig. 1.36.1. Simple screw-jack., , 2., , When the handle is rotated through one complete turn, the screw is also, rotated through one turn. Then the load is lifted by a height p (pitch of, screw).
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Engineering Mechanics, , 1–39 C (CE-Sem-3), , www.aktutor.in, , 3., , The development of one complete turn of a screw thread is shown in, Fig. 1.36.2(a). This is similar to the inclined plane. The distance AB will, be equal to the circumference (d) and distance BC will be equal to the, pitch (p) of the screw., , 4., , From the Fig. 1.36.2(a), we have, tan =, , 5., , Let,, , BC, p, , AC d, , ...(1.36.1), , P = Effort applied horizontally at the mean radius, of the screw jack to lift the load W,, r = Mean radius of the screw jack = d/2,, R = Normal reaction, and, F = Force of friction = R., C, , p, A, , , d, B, (a) Development of a screw thread, , R, , of l oa d, i on, f, e ct en t o, r, i, D, m, e, v, mo, , , F = R, , P, , , , , , W, (b) Force acting on body placed on screw jack, Fig. 1.36.2., , 6., , As the load W is lifted upwards, the force of friction will be acting, downwards. All the forces acting on the body are shown in Fig. 1.36.2(b)., , 7., , Resolving forces along the inclined plane, we have, F + W sin = P cos , R + W sin = P cos , , (, , F = R), ... (1.36.2)
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1–40 C (CE-Sem-3), , Introduction to Engineering Mechanics, , www.aktutor.in, , 8., , Resolving forces normal to the inclined plane, we have, , 9., , Substituting the value of R in eq. (1.36.2), we get, , R = W cos + P sin , (W cos + P sin) + W sin = P cos , , sin , (W cos + P sin) + W sin = P cos , cos , W, , sin , , tan cos , , sin cos , sin sin , P, + W sin = P cos , cos , cos , , 10. Multiplying by cos , we get, W sin cos + P sin sin+ W sin cos = P cos cos , W (sin cos + sin cos ) = P(cos cos – sin sin ), W sin( + ) = P cos( + ), P = W, , sin ( ), = W tan ( + ), cos( ), , ...(1.36.3), , 11. Now P is the effort applied at the mean radius of the screw-jack. But in, case of screw-jack, effort is actually applied at the end of the handle as, shown in Fig. 1.36.1. The effort applied at the end of the handle is P., 12. Moment of P about the axis of the screw, = P × Distance of P from the axis of the screw, = P × Mean radius of the screw jack, = P × d/2, 13. Moment of P about the axis of the screw, = P × Distance of P from axis, = P× L, 14. Equating the two moments, we get, , P , , , P = P , , d, PL, 2, d, P, , P, 2L 2L, , ...(1.36.4), , 15. Substituting the value of P from eq. (1.36.3) into eq. (1.36.4), we get, P=, , d, W tan ( ), 2L, , ...(1.36.5), , Eq. (1.36.5) gives the relation between the effort required at the end of, the handle and the load lifted.
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Engineering Mechanics, , 16. Torque required to work the jack, T PL , 17. Now,, , 1–41 C (CE-Sem-3), , www.aktutor.in, , P=, , =, , d, W tan ( ), 2L, Wd tan tan , 2 L 1 tan tan , , p, , , Wd d, , =, 2 L 1 p , , d , =, , d, W tan( ), 2, , Wd, 2L, , p, , , , tan , tan , , d, , p d , d p , , , , ...(1.36.6), , Eq. (1.36.6) gives the value of P in terms of coefficient of friction and, pitch of the screw., II. Effort Required at the End of Screw Jack to Lower the Load :, 1., , The screw jack is also used for lowering the heavy load. When the load, is lowered by the screw jack, the force of friction (F = R) will act, upwards. Fig. 1.36.3 shows all the forces acting on the body., R, F=, , , , , , W, , R, , P, , , , Fig. 1.36.3. Body moving down., , 2., , Resolving forces along the inclined plane,, F + P cos = W sin , R + P cos = W sin , , 3., , Resolving forces normal to the plane, R = W cos + P sin , , 4., , Substituting the value of R in eq. (1.36.7), we get, (W cos + P sin ) + P cos = W sin , W cos + P sin + P cos = W sin , P sin + P cos = W sin – W cos , , ...(1.36.7)
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1–42 C (CE-Sem-3), , Introduction to Engineering Mechanics, , www.aktutor.in, , P sin + cos ) = W (sin – cos ), , sin , , P , sin cos = W sin sin cos , , , cos , cos , , , , , sin , , tan cos , 5., , Multiplying by cos , we get, P(sin sin + cos cos ) = W (sin cos – sin cos ), P[cos ( – )] = W[sin ( – )], , 6., , sin( ), = W tan (– ), cos( ), , , , P = W, , If > , then, , P = W tan (– ), , ...(1.36.8), , But P is the effort applied at the mean radius of the screw jack. But in, actual case, effort is applied at the handle of the jack. Let the effort, applied at the handle is P. Equating the moment of P and P about the, axis of the jack, we get, P × L = P , , , , P=, , d, 2, , d, d, P =, W tan ( ), 2L, 2L, , ...(1.36.9), , Eq. (1.36.9) gives the relation between the efforts required at the end of, the handle to lower the load (W)., 7., , Expression for P in terms of coefficient of friction and pitch of the screw,, P=, , Wd, Wd tan tan , tan( ) , 2L, 2 L 1 tan tan , , p , , Wd u d , d, =, tan , tan , , 2 L 1 p , d , , d , =, , Wd d p , 2 L d p
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Engineering Mechanics, , www.aktutor.in, , 1–43 C (CE-Sem-3), , Que 1.40., a., , Find the effort required to apply at the end of a handle, fitted to, the screw head of screw jack to lift a load of 1500 N. The length of, the handle is 70 cm. The mean diameter and the pitch of the, screw jack are 6 cm and 0.9 cm respectively. The coefficient of, friction is given as 0.095., , b., , If instead of raising the load of 1500 N, the same load is lowered,, determine the effort required so apply at the end of the handle., , Answer, Given : W = 1500 N, L = 70 cm = 0.7 m, d = 6 cm = 0.06 m, p = 0.9 cm = 0.009 m, = 0.095, To Find : i. Effort required to raise the load., ii. Effort required to lower the load., 1., , Effort required to raise the load is given by,, P=, , Wd, 2L, , p d , d p , , , , 1500 0.06 0.009 0.095 0.06 , = 9.22 N, 2 0.70 0.06 0.009 0.095 , Effort required for lowering the load is given by,, =, , 2., , d p , d p , , , 1500 0.06 0.095 0.06 0.009 , =, 2 0.70 0.06 0.009 0.095 , = 3.024 N, , P=, , Wd, 2L, , Que 1.38. Write a short note on differential screw jack with neat, diagram., , Answer, 1., , Differential screw jack consists of two spindles A and B. B externally, threaded and A both internally and externally threaded., , 2., , The internal threads of spindle A meshes with internal threads of spindle, B. Spindle A is screwed to fixed base., , 3., , When the lever is rotated such that spindle A rises, spindle B also rotates, and it will come down.
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1–44 C (CE-Sem-3), , 4., , Velocity ratio,, , Introduction to Engineering Mechanics, , www.aktutor.in, VR =, , Distance moved by the effort, Distance moved by the load, W, , A, P, , L, , B, , Fig. 1.38.1., ,
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Engineering Mechanics, , 2, , www.aktutor.in, , 2–1 C (CE-Sem-3), , Centroid and Centre, of Gravity, , CONTENTS, Part-1, , :, , Centroid, Centre of Gravity, ................. 2–2C to 2–13C, Centroid of Simple Figures, from First Principle, , Part-2, , :, , Centroid of Composite .......................... 2–14C to 2–20C, Sections, Centre of Gravity, and its Implications, , Part-3, , :, , Area Moment of ..................................... 2–20C to 2–21C, Inertia-Definition, Moment, of Inertia of Plane Sections, from First Principle, , Part-4, , :, , Theorems of Moment ............................ 2–22C to 2–23C, of Inertia, , Part-5, , :, , Moment of Inertia of ............................. 2–23C to 2–37C, Standard Sections and, Composite Sections, , Part-6, , :, , Mass Moment of Inertia ....................... 2–37C to 2–45C, of Circular Plate, Cylinder,, Cone, Sphere, Hook
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2–2 C (CE-Sem-3), , Centroid and Centre of Gravity, , www.aktutor.in, , Centroid, Centre of Gravity, Centroid of Simple Figures, from First Principle., , CONCEPT OUTLINE, Centre of Gravity : It is the point at which the whole weight of the, body acts. A body is having only one centre of gravity for all positions, of the body., Centroid : The point at which the total area of a plane figure (like, triangle, rectangle, circle, etc.) is assumed to be concentrated is known, as the centroid of that area., , Questions-Answers, Long Answer Type and Medium Answer Type Questions, , Que 2.1., , Derive the coordinates for the centroid of :, , i., a line,, ii. a straight line, and, iii. a composite line., , Answer, i., 1., , Centroid of a Line :, Consider a homogenous wire of uniform cross-sectional area A, total, length L and density . If we divide it into infinitesimally small elements, then the weight of an element of length dL is given as,, dW = A(dL)g, Y, x, , dL, y, X, , Fig. 2.1.1. Centroid of a line., , 2., , Hence, the weight of the entire wire is obtained by integrating the, above expression over the length,, , W = AgL
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Engineering Mechanics, 3., , 4., , www.aktutor.in, , The first moment of weight of the infinitesimally small element about, the X-axis is given as the weight multiplied by the perpendicular distance,, i.e., Ag(dL)y., Using the principle of moments, the y-coordinate of location of centre of, gravity of the entire wire is determined as, yW =, , 5., , ii., 1., , 2., , Ag(dL) y, Ag(dL) y, , (W = AgL), yAgL =, Since the density and cross-sectional area A are constant throughout, the length of the wire, they can be taken outside the integral sign., , , 6., , 2–3 C (CE-Sem-3), , y =, , ydL, , x =, , xdL, , My =, , xdx , , L, Similarly, the x-coordinate of location of centre of gravity of the wire, can be determined as,, , L, Centroid of a Straight Line :, Consider a straight line of length L along the X-axis. If we take an, infinitesimally small length dx at a distance x from the origin then its, first moment about the Y-axis is,, dMy = x dx, Therefore, the first moment of the entire length about the Y-axis is,, L, , 0, , L2, 2, , Y, , O, , X, dx, Fig. 2.1.2. Centroid of a straight line., , 3., , x, , The x-coordinate of the centroid is given as,, , L2 / 2, = L/2, L, L, 4. From figure 2.1.2, we can readily see that as the line is along the X-axis,, y = 0. Therefore, we can conclude that the centroid of a straight line, lies at the midpoint of the line., iii. Centroid of a Composite Line :, 1. In general, a given curve may not be of regular shape then in that case,, it is divided into finite segments of regular shapes for which positions of, centroids are readily known., , x =, , My, , =
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2–4 C (CE-Sem-3), , Centroid and Centre of Gravity, , www.aktutor.in, , 2., , Let Li be the length of a segment for which the centroid is known and, , 3., , Then the centroid of the composite line is given by,, , (xi , yi ) be the location of its centroid., , x =, and, , Que 2.2., , y =, , L x, i, , i, , L, Li yi, L, , Derive an expression for the centroid of an arc of a, , circle., , Answer, 1., , Consider an arc of a circle symmetric about the X-axis as shown in, Fig. 2.2.1. Let R be the radius of the arc and 2 be the subtended angle., , 2., , Consider an infinitesimally small length dL such that the radius to the, length makes an angle with the X-axis. Then its length dL is given as,, dL = R d, Y, , R, , d, , dL, , , O, , X, , , x, , Fig. 2.2.1. Centroid of an arc of a circle., , 3., , Therefore, the total length of the arc is, , , L=, , R d 2R, , –, , 4., 5., , The first moment of the infinitesimally small length about the Y-axis is,, dMy = x dL = (R cos ) (R d) = R2 cos d, Hence, the first moment of the entire arc about the Y-axis is given as,, , , My =, , R, , 2, , cos d, , –, , , , = R2 sin – = 2R2 sin , 6., , Therefore, the x-coordinate of centroid of the arc is given as,, , x =, , My, L, , , , 2 R2 sin R sin , , 2R, , , ...(2.2.1)
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Engineering Mechanics, 7., 8., , 2–5 C (CE-Sem-3), , www.aktutor.in, , From the Fig. 2.2.1, we can see that due to symmetry of the arc about, X-axis, y = 0., For a semicircular arc, varies from –/ 2 to / 2 hence the location of, its centroid is obtained by substituting = / 2 in eq. (2.2.1), we get, , x = 2R / and, Que 2.3., , y=0, , A wire is bent into a closed loop A–B–C–D–E–A as shown, , in Fig. 2.3.1 in which portion AB is circular arc. Determine the, , AKTU 2011-12, Marks 05, , centroid of the wire., 5m, , C, 5m, , B, , 10 m, , A, , E, 10 m, Fig. 2.3.1., , D, , Answer, Given : Fig. 2.3.1., To Find : Centroid of the wire., 1., 2., , Consider ED as X-axis and AE as Y-axis or AE and ED as reference axes, to determine the centroid., r, 5, Length of arc AB =, = = 7.85 m, 2, 2, Y, , B, , C, 2r, , , A, 2r, , , E, D, Fig. 2.3.2., , X
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2–6 C (CE-Sem-3), 3., , Centroid and Centre of Gravity, , www.aktutor.in, , Position of centroid for arc,, , 2r, 2 5, =5–, = 1.82 m, , , 2r, 2 5, yi = 10 +, = 10 +, = 13.18 m, , , The coordinates for the centroid of various lines and curves are shown, in table given below :, xi = 5 –, , 4., , S. No. Curve/Line, , Length (Li), (in mm), , Centroid Co-ordinate, (in mm), xi, , 1., , AB, , 7.85, , 2., , BC, , 5, , 1.82, , 3., , CD, , 5 + 10 = 15, , 4., , DE, , 5., , EA, , yi, , Li xi, , Li yi, , 13.18, , 14.287, , 103.463, , 10 + 5 = 15, , 37.5, , 75, , 10, , 15, = 7.5, 2, , 150, , 112.5, , 10, , 10, =5, 2, , 0, , 50, , 0, , 10, , 0, , 10, =5, 2, , 0, , 50, , 251.787, , 340.963, , 5+, , 5, = 7.5, 2, , Li = 47.85, , 5. Centroid of the given figure is,, , x, y, , L x L y , = i i , i i, Li Li , 251.787 340.963 , = , ,, = (5.26,7.13), 47.85, 47.85 , , Que 2.4., , Prove that centroid of a rectangle lies at the intersection, , of its diagonals., , Answer, 1., , Consider a rectangle of base length b and height h. If we take a thin strip, parallel to the X-axis at a distance y from the X-axis and of infinitesimally, small thickness dy then its area is given as,, dA = b dy
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Engineering Mechanics, , 2–7 C (CE-Sem-3), , www.aktutor.in, Y, , dy, h, y, X, b, Fig. 2.4.1., , 2., , Hence, the area of the rectangle is,, h, , A=, , h, , dA b dy bh, 0, , 0, , 3., , As each point on this strip is at the same distance y from the X-axis, we, can take moment of area of the strip about the X-axis as,, , 4., , Therefore, the first moment of the entire area about the X-axis is,, , 5., , bh2, 2, 0, Hence, the y-coordinate of the centroid of the rectangle is given as,, , dMx = ydA = yb dy, h, , Mx =, , y(b dy) , , M x bh2 / 2 h, , , A, bh, 2, In a similar manner, we can consider a vertical strip at a distance x from, the Y-axis and of infinitesimally small thickness dx, and obtain the, x-coordinate of the centroid as,, b, x=, 2, Thus, we can see that the centroid of a rectangle lies at the midpoint or, in other words, at the intersection of its two diagonals., , y=, , 6., , 7., , Que 2.5., , Show that centroid of a right angled triangle lies at, , (b/3, h/3) where b and h are the base and height of the triangle, respectively., , Answer, 1., , Consider a right angled triangle of base b and height h. If we take a thin, strip parallel to the base at a distance y from the X-axis and of, infinitesimally small thickness dy then its area is dA = b′ dy, where, b′ is the width of the strip.
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2–8 C (CE-Sem-3), , Centroid and Centre of Gravity, , www.aktutor.in, Y, A, dy, , h–y, E, , D, , h, , b, y, , B, , C, , X, , b, Fig. 2.5.1. Centroid of a right angled triangle., , 2., , 3., , From similar triangles ABC and ADE, we have, b, b, b, =, b (h – y), h– y, h, h, b, , dA = bdy = (h – y) dy, h, Then area of the entire triangle is obtained as,, h, , A=, , b, (h – y) dy, h 0, h, , b, y2 , bh, hy –, , h, 2 0, 2, The first moment of the strip with respect to the X-axis is,, =, , 4., , b, , dMx = ydA = y ( h – y) dy, h, , 5., , Therefore, the first moment of the entire area about the X-axis is given, as,, h, , Mx =, , h, , b, , y dA y h (h – y)dy, 0, , 0, , h, , =, , b, (hy – y2 )dy, h 0, h, , b y 2 y3 , bh2, –, h, =, h 2, 3 0, 6, Therefore, the y-coordinate of the centroid is given as,, =, , 6., , M x bh2 / 6 h, , , A, bh / 2 3, In a similar manner, we can consider a vertical strip of area dA parallel, to the Y-axis and obtain the x-coordinate of the centroid as,, y=, , 7.
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Engineering Mechanics, , 2–9 C (CE-Sem-3), , www.aktutor.in, x=, , My, A, , , , b, 3, , b h, Thus the coordinate of centroid of a right angled triangle is , , 3 3, , Que 2.6., , Find out the centroid of area of a circular sector and, , also find the centroid of a semicircle., , Answer, 1., , Consider an area of a circular sector of radius R with subtended angle, 2, and symmetric about the X-axis. If we take an element of area OCD, at an angle from the X-axis then its area can be determined by considering, OCD as a triangle and is given as,, dA = (1/2) R × Rd=, , R2, d, 2, , Y, , R, , d, , , O, , C, D, X, , , , Fig. 2.6.1. A circular sector., , 2., , The centroid of this triangle lies at a distance of (2/3) R from O., Hence, the x and y-coordinates of the centroid are,, , 2, 2, R cos and y =, R sin , 3, 3, Area of the entire circular sector is obtained by integrating the expression, for dA between limits, i.e.,, x=, , 3., , , , 4., , R2, d R2, – 2, Taking the first moment of the triangle OCD about the Y-axis,, , 5., , 2, R2, R cos , d, 3, 2, Therefore, the first moment of the entire area about the Y-axis is,, , A=, , , , dMy = x dA =, My =, , x dA
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2–10 C (CE-Sem-3), , Centroid and Centre of Gravity, , www.aktutor.in, , , =, , 2, R2, R cos , d, 3, 2, , , , –, , R3, 2 R3 sin , , sin – , , 3, 3, Therefore, the x-coordinate of the centroid is, 2 R sin , x = My / A , 3, , As the sector is symmetric about X-axis,, =, , 6., , 7., , y =0, , , 8., , 9., , ...(2.6.1), , For a semicircular area, we know that varies from – /2 to /2. Hence,, its centroid is obtained by substituting = /2 in eq. (2.6.1) for x ., Therefore, we get, 4R, x=, and y = 0, 3, Similarly, if the area is symmetric about Y-axis then the centroidal, coordinates are, 4R, x = 0 and y , 3, , Que 2.7., , Derive the expression for the centroid of a parabola., , Answer, 1., , Consider a shaded area bounded by a parabola of equation y = kx2,, X-axis and line x = b as shown in Fig. 2.7.1. Then we see that at x = 0,, y = 0 and at x = b, y = h. Therefore,, h, k= 2, b, Y, , h, , y = kx2, y, X, dx, , x, b, Fig. 2.7.1., , 2., , Hence, we can write the equation of the curve as,, h, y = 2 x2, b
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Engineering Mechanics, 3., , 2–11 C (CE-Sem-3), , www.aktutor.in, , Consider a vertical strip parallel to the Y-axis at a distance x from the, origin and of infinitesimally small thickness dx as shown in the, Fig. 2.7.1. Then its elemental area is given as dA = y dx = (h/b2)x2 dx., Therefore, the area under the entire curve is,, b, , h 2, x dx, 2 , , 0, , b, , A=, , h b3 bh, , , b2, 3, 3, We see that the area of the curve is 1/3rd of the area of the enclosed, rectangle., The first moment of the area about the Y-axis is given as,, =, , 4., , My =, , x dA, b, , h, , x b, , =, , 2, , x 2 dx, , 0, , h b4 b2 h, , , b2, 4, 4, Therefore, the x-coordinate of the centroid is given as,, =, , 5., , b2h / 4 3, b, A, bh / 3 4, In a similar manner, we can consider a thin strip parallel to the X-axis, and of infinitesimally small thickness dy as shown in Fig. 2.8.2., My, , x=, , 6., , , , Y, , h, , dy, , y = kx2, , y, X, x, , b–x, , Fig. 2.7.2., , 7., , The elemental area is given as dA = (b – x)dy. Therefore, the first, moment of the area about the X-axis is given as,, h, , Mx =, , y dA y(b – x) dy, 0, , h, , 8., , h, , y2, b y5/ 2 , bh 2, b 1/ 2 , , 0 y b – h1/ 2 y dy = b 2 – h1/ 2 5 / 2 10, 0, Therefore, the y-coordinate of the centroid is given as,, =, , y =, , M x bh2 / 10 3, , , h, A, bh / 3, 10
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2–12 C (CE-Sem-3), , Que 2.8., , Centroid and Centre of Gravity, , www.aktutor.in, , Determine the centroid of a semi circular segment given, , that a = 100 mm and = 45°., , a, , Fig. 2.8.1., , AKTU 2013-14 (I), Marks 05, , Answer, Given : a = 100 mm = 0.1 m, = 45°, To Find : Centroid of semi circular segment., 1., , 2., , Let us consider an element at a distance r from the centre O of the, semi circle, radial width being dr and bound by radii at and + d., Area of element = rddr, Its moment about X-axis is given by,, Y, , dr, , a, r, , Q, P, , , , , O, , + d, , X, , a, Fig. 2.8.2., , , , rd dr × r sin = r2 sin dr d, Total moment of area about X-axis is,, a, , , , r 2 sin dr d =, , 0 0, , , , , 0, , a, , r3 , sin d, 3 0, , =, , a3, a3, – cos 0 = 3 [– cos cos 0], 3, , =, , (100)3, [– cos 45 1] = 97631.073 mm3, 3
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Engineering Mechanics, 4., , www.aktutor.in, , 2–13 C (CE-Sem-3), , Area of the sector = a2 , 360 , , 45 , 2, mm = 3927 mm2, = (100)2 , 360 , 5., , The position of centroid y =, , Moment of area about X - axis, Total area, , 97631.073, 39267, y = 24.86 mm, Now consider an elementary strip OPQ that subtends an angle d at O., PQ = a d, As angle d is very small, consider it as a triangle., =, , 6., 7., , Area of the elementary strip =, , 1, ( ad) a, 2, , a2, d, 2, Centroid of this triangular strip lies on a line that joins O to the mid, dA =, , 8., , point of PQ and at a distance, 9., , 2, a from O., 3, , Distance x of centroid from Y-axis =, , 2, a cos , 3, , 10. Moment of area of elementary strip about Y-axis =, , a2 d 2, a cos , 2, 3, , 1 3, a cos d, 3, 11. The x-coordinate of the centroid of the lamina from Y-axis will be,, dMy =, , x=, , Moment of area about Y - axis, Total area of section, , , , =, , 0, , 1 3, a cos d, 3, , , , 0, , a2, d, 2, , , , =, , 2 sin 0, a, 3, 0, , 2a sin , =, 3 , , x=, , 2 100 sin 45, = 60.02 mm, , 3, , 45 , , 180
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2–14 C (CE-Sem-3), , Centroid and Centre of Gravity, , www.aktutor.in, , Centroid of Composite Sections, Centre of, Gravity and its Implications., Questions-Answers, Long Answer Type and Medium Answer Type Questions, , Que 2.9., , Discuss in brief about centroid of composite figures., , Answer, 1., , 2., 3., , In engineering work, we frequently need to locate the centroid of a, composite area. Such an area may be composed of regular geometric, shapes such as rectangle, triangle, circle, semicircle, quarter circle, etc., In such cases, we divide the given area into regular geometric shapes, for which the positions of centroids are readily known., Let Ai be the area of an element and (xi , yi ) be the respective centroidal, coordinates. Then for the composite area,, , Ax = A1 x1 A2 x2 ... An xn, , 4., , , , x=, , Similarly,, , y=, , Ax, i, , i, , A, Ai yi, , A, where the total area, A = A i, in which the areas are added up, algebraically., , Que 2.10. Find out the centroid of an L-section of 120 mm × 80 mm, × 20 mm as shown in Fig. 2.10.1., 20, , 120, 20, 80, Fig. 2.10.1., , AKTU 2014-15 (I), Marks 10
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Engineering Mechanics, , 2–15 C (CE-Sem-3), , www.aktutor.in, , Answer, Given : Fig. 2.10.1., To Find : Centroid of L-section., 1., , The given L-section is not symmetrical about any section. Hence, in this, case, there will be two axes of references. The lowest line of the figure, (i.e., line GF) will be taken as axis of reference for calculating y and the, left line of the L-section (i.e., line AG) will be taken as axis of reference, for calculating x ., A, , 20, B, All dimensions in mm, 1, , 120, , 100, , C, , E, , D, 2, G, , 2., 3., , 80, Fig. 2.10.2., , 20, F, , The given L-section is split up into two rectangles ABCD and DEFG, as, shown in Fig. 2.10.2., A1 = Area of rectangle ABCD = 100 × 20 = 2000 mm2, y1 = Distance of centroid of rectangle ABCD from bottom line GF., , 100, = 20 + 50 = 70 mm, 2, A2 = Area of rectangle DEFG = 80 × 20 = 1600 mm2, y2 = Distance of centroid of rectangle DEFG from bottom line GF., y1 = 20 , , 20, = 10 mm, 2, By using the formula, we have, A y A2 y2, y = 1 1, , where A = A1 + A2, A, 2000 70 1600 10, =, = 43.33 mm, 2000 1600, Let, x1 = Distance of the rectangle ABCD from left line AG., =, , 4., , 5., , 20, = 10 mm, 2, x2 = Distance of the rectangle DEFG from left line AG., =
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2–16 C (CE-Sem-3), , Centroid and Centre of Gravity, , www.aktutor.in, , 80, = 40 mm, 2, Using formula, we have, A x A2 x2, x = 1 1, A, 2000 10 1600 40, = 23.33 mm, =, 2000 1600, Hence, the centroid of the L-section is at a distance of 43.33 mm from, the bottom line GF and 23.33 mm from the left line AG., =, , 6., , Que 2.11., , Locate the centroid of the shaded area shown in, , Fig. 2.11.1. All dimensions are in meters., 60, , 60, 60, , R=, 80, , 80, , 40, 80, Fig. 2.11.1., , AKTU 2014-15 (II), Marks 10, , Answer, Given : Fig. 2.11.2, To Find : Centroid of the shaded area., 1., , Shaded area, ABCED = Rectangle AOCD, + Triangle DCE – Quarter circle OBC, Y, , E, , (60, 140), 60, , 60, , 60, , 120, , (0, 80) D, , C (120, 80), R=80, , 80, A, , B, , O, , 40, 80, Fig. 2.11.2., , X
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Engineering Mechanics, 2., , 2–17 C (CE-Sem-3), , www.aktutor.in, , The coordinates of the centroid for various sections are shown in the, table given below :, Centroid coordinate, Shape, , Area, Ai, (mm 2), , xi, (mm), , yi, (mm), , Ai x i, (mm 3), , Ai y i, (mm 3), , Rectangle, AOCD, , 120 × 80, = 9600, , 120/2, = 60, , 80/2, = 40, , 576 ×103, , 384 × 103, , Triangle, , 1, × 120 × 60, 2, , 0 120 60, 3, , 80 140 80, 3, , 216, , 360 × 103, , ×, DEC, , = 3600, , Quarter, circle, , (80)2, 4, , BOC, , = – 5026.55, , = 60, 40 +, , = 100, , 4 80, 3, , = 73.95, , 4 80, 3, , – 371.7, , –170.65, , × 103, , × 103, , Aixi =, , Aiyi =, , = 33.95, , Ai = 8173.4, , 3., , 103, , 420 × 103 573.35×103, , Centroid of shaded portion, ( x , y), , Ai xi Ai yi , ,, = , Ai , Ai, , 420 103 573.35 103 , = , ,, , 8173.45, 8173.45 , = (51.4, 70.15), Que 2.12., , Locate the centroid of the T-section shown in the, , Fig. 2.12.1., 100, O, A1, , G1, , X, 20, , y, G, G2, A2, , 20, Y, Fig. 2.12.1., , 100
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2–18 C (CE-Sem-3), , Centroid and Centre of Gravity, , www.aktutor.in, , Answer, Given : Fig. 2.12.1., To Find : Centroid of T section., 1., , Selecting the axis as shown in Fig. 2.12.1, we can say due to symmetry, centroid lies on Y-axis, i.e., x = 0., , 2., , Now the given T-section may be divided into two rectangles A1 and A2, each of size 100 × 20 mm and 20 × 100 mm. The centroid of A1 and A2 are, G1(0, 10) and G2(0, 70) respectively., , 3., , The distance of centroid from top is given by,, , y =, , 100 20 10 20 100 70, = 40 mm, 100 20 20 100, , Hence, centroid of T-section is on the symmetric axis at a distance, 40 mm from the top., , Que 2.13., , For the semi-annular area shown in Fig. 2.13.1,, , determine the ratio of a to b so that y =, , 3, b., 4, , Y, , a, , b, X, , Fig. 2.13.1., , AKTU 2015-16 (I), Marks 10, , Answer, 3, b, 4, To Find : Ratio of a to b., Given : y =
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2–20 C (CE-Sem-3), , 7., , Centroid and Centre of Gravity, , www.aktutor.in, , ab , 4 (a b)2 ab 4 , 3, b=, , , a b a b, 3 , ab, 4, 3 , , 3 , , y b, 4, a , a, 32 , = 1, ...(2.13.1), a b , 42, b, After solving eq. (2.3.1), we get, , a, = 1.34, b, , Area Moment of Inertia-Definition, Moment of Inertia of Plane, Sections from First Principle., Questions-Answers, Long Answer Type and Medium Answer Type Questions, , Que 2.14. Write a short note on area moment of inertia., , Answer, 1., , Consider the area shown in Fig. 2.14.1(a). dA is an elemental area with, coordinates as x and y. The term yi2 dAi is called moment of inertia of, the area about X axis and is denoted as IXX. Similarly, the moment of, inertia about y axis is, , 2., , IYY = yi2 dAi, In general, if r is the distance of elemental area dA from the axis AB, [Fig. 2.14.1(b)], the sum of the terms r2 dA to cover the entire area is, called moment of inertia of the area about the axis AB., Y, x, , dA, , dA, , B, r, , y, X, ( a), , 3., , A, Fig. 2.14.1., , (b ), , Though moment of inertia of plane area is a purely mathematical term,, it is one of the important properties of areas. The strength of members, subject to bending depends on the moment of inertia of its cross-sectional, area.
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Engineering Mechanics, 4., , 2–21 C (CE-Sem-3), , www.aktutor.in, , The moment of inertia is a fourth dimensional term since it is a term, obtained by multiplying area by the square of the distance. Hence, its SI, unit is m4., , Que 2.15. Define the following terms :, i., , Polar moment of inertia, and, , ii., , Radius of gyration., , Answer, i., 1., , Polar Moment of Inertia :, If an elemental area dA is at a distance r from origin of the coordinate, axes then its polar moment of inertia is given by,, JO =, where,, , 2., , r, , 2, , dA, , As, , JO = Polar moment of inertia of the area A with, respect to the pole O., r2 = x2 + y2, , Hence,, , JO =, , where, , JO = IX + IY, IX = Moment of inertia of the area about X-axis., IY = Moment of inertia of the area about Y-axis., , (x, , 2, , y2 ) dA y2 dA x 2 dA, , Y, dA, r, O, , y, x, A, , X, , Fig. 2.15.1., , 3., , In other words we can say that polar moment of inertia of an area is the, moment of inertia of the area about Z-axis., , ii., , Radius of Gyration : Radius of gyration is defined as the distance, which is when squared and multiplied by area gives the moment of, inertia of that area., Mathematically,, where,, , I = k2A k =, , I, A, , k = Radius of gyration,, I = Moment of inertia, and, A = Cross-sectional area.
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2–22 C (CE-Sem-3), , Centroid and Centre of Gravity, , www.aktutor.in, , Theorems of Moment of Inertia., , Questions-Answers, Long Answer Type and Medium Answer Type Questions, , Que 2.16. State and prove perpendicular axis theorem., , Answer, A., , Perpendicular Axis Theorem :, , 1., , The moment of inertia of an area about an axis perpendicular to its, plane (i.e., polar moment of inertia) at any point O is equal to the sum of, moment of inertia about any two mutually perpendicular axis through, the same point O and lying in the plane of the given area., , B., , Proof :, , 1., , Consider an elemental area dA at distance r from point O., , 2., , Let dA have coordinates x and y, then from the definition,, IZZ = r2dA, = (x2 + y2)dA = x2dA + y2dA ( r2 + x2 + y2), IZZ = IXX + IYY, IZZ is also called polar moment of inertia., Y, x, , dA, , r, , O, , y, , X, Fig. 2.16.1., , Que 2.17. State and prove parallel axis theorem., , Answer, A., , Parallel Axis Theorem :, , 1., , According to this theorem, moment of inertia about any axis in the, plane of an area (or lamina) is equal to the sum of moment of inertia
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2–24 C (CE-Sem-3), , Centroid and Centre of Gravity, , www.aktutor.in, Questions-Answers, , Long Answer Type and Medium Answer Type Questions, , Que 2.18., , Find the moment of inertia of following shapes about, , base and centroidal axis :, i., Rectangle,, ii. Triangle, and, iii. Circle., , Answer, i., , Rectangle :, , 1., , Consider a rectangle of width b and depth d (Fig. 2.18.1)., dy, d/2, y, X, , X, , d/2, , A, , b, , A, , Fig. 2.18.1., , 2., , Consider an elemental strip of width dy at a distance y from the, X-X axis. Moment of inertia of the elemental strip about the centroidal, axis X-X is,, dIXX = y2dA = y2b dy, d/2, , , , IXX =, , d/2, , y3 , y2 b dy b , 3 – d/2, – d/2, , , , d 3 d3 , bd 3, = b, , = 12, 24 24 , , db3, 12, Now moment of inertia about base,, Similarly,, , 3., , IYY =, , IAA = ICG + Ah2
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2–26 C (CE-Sem-3), , Centroid and Centre of Gravity, , www.aktutor.in, IXX =, , bh3, 36, , iii. Circle :, 1. Let dA be an elemental ring of radius r and thickness dr., So, elemental area, dA = 2rdr, 2. Now, moment of inertia of thin ring about its central axis or polar, moment of inertia,, Y, dr, , r, , X', , X, R, , Y', D, Fig. 2.18.3., , IZZ = 0R r 2 dA = 0R r 2 (2 r ) dr, IZZ =, =, 3., , R4, 2, , D, , R , 2, , , D4, 32, , By perpendicular axis theorem,, IZZ = IXX + IYY, , 4., , Due to symmetry along X-X and Y-Y axes we have, IXX = IYY, IXX’ = IYY =, , IZZ ' D4, , 2, 64, , Que 2.19. Find the moment of inertia of a semicircle and quarter, circle., , Answer, i., , Moment of Inertia of a Semicircle :, , a., , About Diametral Axis :, , 1., , If the limit of integration is put as 0 to instead of 0 to 2 in the, derivation for the moment of inertia of a circle about diametral axis the, moment of inertia of a semicircle is obtained.
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Engineering Mechanics, 2., , www.aktutor.in, , 2–27 C (CE-Sem-3), , It can be observed that the moment of inertia of a semicircle (Fig. 2.19.1), about the diametral axis AA is,, IAA =, , 1 d 4 d 4, , , 2 64 128, , G, X, , X, yc, A, , A, d, Fig. 2.19.1., , b., , About Centroidal Axis X-X :, , 1., , Now, the distance of centroidal axis yc from the diametral axis is given, by,, 4 R 2d, yc =, , 3 3, Area, A =, , 2., , 1 d 2 d 2, , , 2, 4, 8, , From parallel axis theorem,, IAA = IXX Ayc2, , d 4, d 2 2 d , , = I XX , 3 , 128, 8, IXX =, , 2, , d 4, d4, –, 128 18, , = 0.00686 d4, ii., , Moment of Inertia of a Quarter of a Circle :, , a., , About the Base :, , 1., , If the limit of integration is put as 0 to /2 instead of 0 to 2 in the, derivation for moment of inertia of a circle, the moment of inertia of a, quarter of a circle is obtained.
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2–28 C (CE-Sem-3), , Centroid and Centre of Gravity, , www.aktutor.in, , G, X, , X, A, , A, , 4R, 3, , R, Fig. 2.19.2., , 2., , It can be observed that moment of inertia of the quarter of a circle about, the base AA is,, , 1 d 4 d 4, , , 4 64 256, About Centroidal Axis X-X :, Now, the distance of centroidal axis yc from the base is given by,, 4 R 2d, yc =, , 3 3, 1 d 2 d 2, , Area, A = , 4, 4, 16, From parallel axis theorem,, IAA =, , b., 1., , 2., , IAA = I XX Ayc2, , d 4, d 2, = I XX , 256, 16, IXX =, , 2d , , , 3 , , 2, , d 4, d4, = 0.00343 d4, –, 256 36 , , Que 2.20. Discuss the procedure of finding the moment of inertia, of composite sections., , Answer, , 1., 2., 3., , 4., , Moment of inertia of composite sections about an axis can be found by, the following steps :, Divide the given figure into a number of simple figures., Locate the centroid of each simple figure by inspection or using standard, expressions., Find the moment of inertia of each simple figure about its centroidal, axis. Add the term Ay2, where A is the area of the simple figure and y is, the distance of the centroid of the simple figure from the reference axis., This gives moment of inertia of the simple figure about the reference axis., Sum up moments of inertia of all simple figures to get the moment of, inertia of the composite section.
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Engineering Mechanics, , 2–29 C (CE-Sem-3), , www.aktutor.in, , Que 2.21. Determine the moment of inertia of the L section shown, in Fig. 2.21.1 about its centroidal axis parallel to the legs. Also find, the polar moment of inertia., , 125 mm, , 10 mm, , 10 mm, 85 mm, Fig. 2.21.1., , AKTU 2016-17 (I), Marks 10, , Answer, Given : Fig. 2.21.1., To Find : i. Moment of inertia about centroid axis., ii. Polar moment of inertia., 1., , 2., , The given section is divided into two rectangles A1 and A2., Area, A1 = 125 × 10 = 1250 mm2, Area, A2 = 75 × 10 = 750 mm2, Total Area = 2000 mm2, First, the centroid of the given section is to be located. Two reference, axis (1)-(1) and (2)-(2) are chosen as shown in Fig. 2.21.2., 1, 10, , Y, , 125, , A1, , G1, , X, , G, X, , A2, , G2, 10, , 2, , 75, 1, , Y, Fig. 2.21.2., , 3., , The distance of centroid from the axis (1)-(1),, , 2
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2–30 C (CE-Sem-3), , x=, , 4., , Centroid and Centre of Gravity, , www.aktutor.in, , Sum of moment of areas A1and A2 about (1)-(1), Total area, , 75 , , 1250 5 750 10 , , , 2 = 20.94 mm, x =, 2000, Similarly, the distance of the centroid from the axis (2)-(2),, , 125, 750 5, 2, y =, = 40.94 mm, 2000, With respect to the centroidal axis X-X and Y-Y, the centroid of, A1 is G1 (15.94, 21.56) and that of A2 is G2 (26.56, 35.94)., 1250 , , 5., , , , IXX = Moment of inertia of A1 about X-X axis + Moment of inertia, of A2 about X-X axis, , , , IXX =, , i.e.,, 6., , 10 1253, 75 103, + 1250 × 21.562 +, +750 × 35.942, 12, 12, IXX = 3183658.9 mm4, , Similarly,, , 125 103, 10 753, + 1250 × 15.942 +, + 750 × 26.562, 12, 12, IYY = 1208658.9 mm4, IYY =, 7., , Polar moment of inertia, Izz = Ixx + Iyy, = 3183658.9 + 1208658.9, IZZ = 4392317.8 mm4, , Que 2.22. Determine the area moment of inertia of the composite, area ABOC about given X and Y axes., Y, A, 30, , m, , m, 30, , C, , O, Fig. 2.22.1., , m, , m, , B, , X, , AKTU 2013-14 (II), Marks 10
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Engineering Mechanics, , 2–31 C (CE-Sem-3), , www.aktutor.in, , Answer, Y, 30 – 12.73, D, , A, m, , m, , 30 – 12.73, , 30, , O, , C, , B, , 30 mm, , X, , Fig. 2.22.2., , Given : Fig. 2.22.1., To Find : Area moment of inertia., 1., , For quarter circle OAB :, i. Moment of inertia about X-axis,, , R4 (30)4, = 159043.13 mm4, , 16, 16, ii. Moment of inertia about Y-axis,, IXX =, , IYY =, 2., , R4 (30)4, = 159043.13 mm4, , 16, 16, , For square AOCD :, i. Moment of inertia about, centroidal X-axis,, , b4 (30)4, = 67500 mm4, , 12, 12, Moment of inertia about X-axis,, IXG =, , ii., , 30 , IXX = IXG + A , 2, , 2, , 2, , 30 , = 67500 + (30)2 × = 270000 mm4, 2, iii. Moment of inertia about centroidal Y-axis,, b4 (30)4, = 67500 mm4, , 12, 12, iv. Moment of inertia about Y-axis,, IYG =, , 2, , 30 , 30 , IYY = IYG + A 67500 (30)2 , 2, 2, = 270000 mm4, , 2
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Engineering Mechanics, , 2–33 C (CE-Sem-3), , www.aktutor.in, , Answer, Given : Fig. 2.23.1., To Find : Moment of inertia., 1., , Let’s consider a horizontal strip of small thickness dy at distance y, from X-axis, as shown in Fig. 2.23.2., Y, , B, , A, y=x, dy, , 8m, x, y, , X, , O, 2m, Fig. 2.23.2., , 2., , The area of the strip is,, , 3., , The moment of inertia about the X-axis,, , dA = x dy, dIXX = y2dA, 4., , We know that,, So,, , y = x3 x = y1/ 3, dIXX = y2 x dy, dIXX = y2 y1/ 3 dy, , 5., , ...(2.23.1), , Integrating the eq . (2.23.1) within the limits 0 to 8, we get, 8, , 8, , y(7/ 3) 1 , y7/ 3 dy , , (7 / 3) 1 0, , IXX =, , , , IXX =, , 3, (8)10/ 3, 10, , 0, , IXX = 307.2 m4, , Que 2.24. Determine the area moment of inertia of the composite, area shown in Fig. 2.24.1 about X and Y axis.
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2–34 C (CE-Sem-3), , Centroid and Centre of Gravity, , www.aktutor.in, Y, , 20 mm, 30 mm, 20 mm, X, 10 mm, Fig. 2.24.1., , AKTU 2013-14 (I), Marks 10, , Answer, Given : Fig. 2.24.1., To Find : Area moment of inertia., 1., 2., , The given area can be obtained by subtracting a quarter circle and a, triangle from a rectangle., Area moment of inertia of rectangle OEBF :, i., About X-axis,, (IXX)OEBF =, ii., , 1, 1, bh3 =, × 30 × (40)3 = 6.4 × 105 mm4, 3, 3, , About Y-axis,, , 1 3, 1, b h=, × (30)3 × 40 = 3.6 × 105 mm4, 3, 3, Moment of inertia of triangle ABF :, i., Moment of inertia of right angled triangle ABF about its centroidal, axis along X1,, (IYY)OEBF =, , 3., , IX1X1 =, ii., , hb3, 1, , 30 (10)3 = 833.33 mm4, 36 36, , About X-axis,, (IXX)ABF = IX1X1 + AdX12, A = Area of =, , 1, 1, bh =, × 30 × 10 = 150 mm2, 2, 2, , dX1 = 36.67 mm, (IXX) ABF = 833.33 + 150 × (36.67)2, (IXX)ABF = 2.0254 × 105 mm4, iii., , About Y-axis,, 3, 3, (IYY)ABF = bh 10 (30) = 0.225 × 105 mm4, 12, 12
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2–36 C (CE-Sem-3), , Centroid and Centre of Gravity, , www.aktutor.in, , Que 2.25. Determine the moment of inertia about X-X and Y-Y axis, passing through the centroid of the symmetrical I- section as shown, in Fig. 2.25.1., 20 cm, 5 cm, , 5 cm, , 15 cm, , 5 cm, 30 cm, Fig. 2.25.1., , AKTU 2015-16 (I), Marks 15, , Answer, Given : Fig. 2.25.1., To Find : Moment of inertia about X-X and Y-Y axis., 1., , 2., , Since the section is divided into three rectangles as shown in, Fig. 2.27.2., A1 = 20 × 5 = 100 cm2, A2 = 15 × 5 = 75 cm2, A3 = 30 × 5 = 150 cm2, Total Area, A = A1 + A2 + A3 = 325 cm2, Due to symmetry, centroid lies on axis Y-Y. The bottom fiber 1-1 may be, chosen as reference axis to locate the centroid., Y, 20 cm, A1, , 5 cm, , G1, A2, , 5 cm, G2, , X, , 15 cm, X, 10.96 cm, , A3, 1, , G3, , 30 cm, , 5 cm, 1, , Y, Fig. 2.25.2.
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Engineering Mechanics, 3., , =, , 5., , 2–37 C (CE-Sem-3), , The distance of the centroid from 1-1,, , y =, , 4., , www.aktutor.in, , S um of moments of the areas of the rectangles about axis 1 - 1, Total area of section, , 100 (20 2.5) 75 (7.5 5) 150 2.5, = 10.96 cm, 325, , With reference to the centroidal axis X-X and Y-Y, the centroid of, rectangle A1 is G1 (0.0, 11.54), that of A2 is G2 (0.0, 1.54) and that of A3 is, G3 (0.0, – 8.46)., , 20 53, 5 153, , IXX = , 100 11.54 2 , 75 1.54 2 , 12, 12, , 30 53, , +, 150 ( 8.46)2 , 12, , IXX = 26157.8533 cm4, , 6., , IYY =, , 5 203 15 53 5 303, , , 12, 12, 12, , IYY = 14739.5833 cm4, , Mass Moment Inertia of Circular Plate, Cylinder, Cone,, Sphere, Hook., , CONCEPT OUTLINE, Mass Moment of Inertia : Mass moment of inertia of a body about, an axis is defined as the sum total of product of its element masses and, square of their distance from the axis., , Questions-Answers, Long Answer Type and Medium Answer Type Questions
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2–38 C (CE-Sem-3), , Que 2.26., , Centroid and Centre of Gravity, , www.aktutor.in, , Derive the expression of mass moment of inertia of, , circular disc about its diametral axis., , AKTU 2014-15 (II), Marks 10, , Answer, 1., , Consider an elemental area r dθ dr and thickness dr as shown in, Fig. 2.26.1., Mass of the element, dm = r d dr t = t r d dr, where,, , = Density of the circular plate., t = Thickness of the plate., , Its distance from X axis = r sin , 2., , Now,, , IXX =, , (r sin ), , 2, , dm, , R 2, , =, , r, , 2, , sin 2 t r d dr, , 0 0, , R 2, , 3 1 – cos 2 , = t r , drd, 2, , , 0 0, , R, , = t , 0, , r3, 2, , 2, , R, , R4, r4 , = t = t, 4, 4 0, 3., , Mass of the plate, M = × R2t, , , IXX =, , MR2, 4, , Similarly,, , IYY =, , MR2, 4, , R, , 3, , r, sin 2 , , – 2 dr = t 2 2 dr, , 0, 0
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Engineering Mechanics, , www.aktutor.in, , 2–39 C (CE-Sem-3), , Y, , d, X, , r, , , dr, X, , Y, Fig. 2.26.1., 2, , MR, is the moment of inertia of circular plate about any, 4, diametral axis in the plate., Actually I =, , 4., , To find IZZ, consider the same element,, R 2, , IZZ =, , 2, r dm , , r, , 2, , t r dr d, , 0 0, , R, , R, , 3, 3, 2, = t r []0 dr = t 2r dr, 0, , 0, , R, , r4 , R4, R4, t, = t 2 = t2, 4, 2, 4 0, 5., , But total mass,, , , Que 2.27., , M = tR 2, IZZ =, , MR2, 2, , Derive an expression for mass moment of inertia of a, , solid cylinder about its longitudinal axis and its centroidal axes., , Answer, Let us consider a solid cylinder of base radius R, length L and uniform, mass density .
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2–40 C (CE-Sem-3), , Centroid and Centre of Gravity, , www.aktutor.in, Y, X, C, , R, , Z, , z, dz, L, Fig. 2.27.1., , i., , Mass Moment of Inertia about Longitudinal Axis :, , 1., , In the Fig. 2.27.1, Z-axis which passes from the centroid of the cylinder, and is along the length of cylinder is termed as longitudinal axis of, cylinder., , 2., , Now consider a solid circular disc of infinitesimal thickness dz, perpendicular to Z-axis of a distance z from the origin., , 3., , Mass of the infinitesimal disc, dm = R2 dz, , 4., , Mass moment of inertia about the Z-axis, dIZZ’ = dm, , 5., , Now,, , dIZZ’ = dm, , R2, 2, , R2, R2, R4, = R2 dz, = , dz, 2, 2, 2, , L/ 2, , dIZZ' =, , R4, R4 L / 2, dz =, [ z] L / 2, 2, 2, L / 2, , , , 4, IZZ’ = R L, 2, , IZZ’ =, , MR 2, 2, , M R L, 2, , Here, M = Mass of the solid cylinder., ii., 1., , Mass Moment of Inertia about Centroidal Axes :, Mass moment of inertia of the solid circular disc about an axis (i.e., X-X, or Y-Y axis) lying on its plane is,, dIX’X’ = dm, , 2., , R2, 4, , Now using parallel axis theorem, we have, dIXX = dIX’X’ + z2 dm = dm, L/2, , dI XX, , =, , , L / 2, , R2, + z2 dm, 4, , L/2, , R2 dz, , R2, , R2 z2 dz, 4 L/ 2
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Engineering Mechanics, , www.aktutor.in, , 2–41 C (CE-Sem-3), L/2, , z3 , R4 L / 2, z L / 2 R2 , IXX =, 4, 3 L / 2, =, IXX =, 3., , R 4, R2 3, R 2 L, L, L =, [3 R2 L2 ], 4, 12, 12, , M, [3 R2 L2 ], 12, , { M R2 L}, , As the cylinder is symmetrical about X-Z and Y-Z plane,, IXX = IYY =, , , , Que 2.28., , M, [3R2 + L2], 12, , Find the mass moment of inertia of a hollow cylinder, , about its axis. The mass of cylinder is 5 kg, inner radius 10 cm, outer, , AKTU 2012-13, Marks 05, , radius 15 cm and height 20 cm., , Answer, Given : M = 5 kg, R2 = 10 cm = 0.1 m, R1 = 15 cm = 0.15 m, L = 20 cm, = 0.2 cm, To Find : Mass moment of inertia of hollow cylinder., 1., , Mass moment of inertia of hollow cylinder about longitudinal axis is, given by,, IZZ =, , 5, M, [R12 + R22] = [(0.15)2 + (0.1)2], 2, 2, , IZZ = 0.08125 kg-m2, 2., , Mass moment of inertia of hollow cylinder about its centroidal axis is, given by,, IXX = IYY =, =, , M, [3(R12 + R22) + L2], 12, , 5, [3(0.15)2 + 3(0.1)2 + (0.2)2], 12, , = 0.05729 kg-m2 0.0573 kg-m2, , Que 2.29. Calculate the mass moment of inertia of the cylinder of, radius 0.5 m, height 1 m and density 2400 kg/m3 about the centroidal, axis Fig. 2.29.1., , AKTU 2013-14 (I), Marks 10
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2–42 C (CE-Sem-3), , Centroid and Centre of Gravity, , www.aktutor.in, Y, , Z, 1.0 m, , X, , X, Z, 0.5 m, , Y, Fig. 2.29.1., , Answer, Given : R = 0.5 m, L =1 m, = 2400 kg/m3, To Find : Mass moment of inertia of the cylinder about centroidal axis., , 1., , We know that,, , Izz =, =, , 1, M (3R2 + L2), 6, , 1, R2L (3R2 + L2), 6, , ( M = R2L), , 1, × 2400 × × 0.52 × 1 × (3 × 0.52 + 12), 6, = 549.78 kg-m2, =, , Que 2.30. Determine the mass moment of inertia of a right circular, solid cone of base radius R and height h about the axis of rotation., , AKTU 2013-14 (I), Marks 10, , Answer, 1., , Consider a solid cone of height h and radius R. If is the density of the, material of the cone, then, Mass of the cone, M = Density × Volume, M= ×, , 1, R2 h, 3, , 2., , Consider an element of thickness dy and radius r at distance y from the, apex A., , 3., , Mass of the elemental strip, dm = r2 dy, , 4., , Mass moment of inertia of the elemental strip about axis YY, = (1 / 2) × Mass moment of inertia about polar axis
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Engineering Mechanics, , www.aktutor.in, , =, , 1 2, 1 2, (r dm) =, r (r2 dy), 2, 2, , =, , 1, (r4 dy), 2, Y, , 2–43 C (CE-Sem-3), , A, dy, , y, , E, , D, , h, , r, , B, , C, , R, , Y, , X, , Fig. 2.30.1., , 5., , Since the integration is to be done with respect to y within the limits 0 to, h., In triangles ADE and ABC, , r, y, y, =, , r=R×, R, h, h, h, , , , IYY =, , 1, , 4, , Ry , dy, h , , 2 , 0, , h, , Que 2.31., , =, , R4 h, R4 y5 , =, , , 10, 2h4 5 0, , =, , R2 h, 3, ×, R2, 3, 10, , =, , 3, MR2, 10, , 1, , 2 , M R h, 3, , Derive the expression for mass moment of inertia of a, , sphere about centroidal axis., , AKTU 2015-16 (I), Marks 10, , Answer, 1., , Consider a solid sphere of radius R with O as centre. If is the density, of the material of the sphere, then
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Engineering Mechanics, , Que 2.32., , www.aktutor.in, , 2–45 C (CE-Sem-3), , Determine the mass moment of inertia of uniform, , density sphere of radius 5 cm about its centroidal axes., , AKTU 2013-14 (II), Marks 10, , Answer, Given : R = 5 cm = 0.05 m, To Find : Mass moment of inertia., 1., , Assume uniform density of solid sphere is . So, mass of sphere,, M= ×V = , , 2., , 4, 4, R3 53, 3, 3, , = 523.6 kg, Mass moment of inertia about centroidal axis in terms of mass M,, IXX = IYY =, =, , I ZZ 2, MR2, 2, 5, , 2, × 523.6 × (5)2 = 5236 cm4, 5, ,
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Engineering Mechanics, , 3, , www.aktutor.in, , 3–1 C (CE-Sem-3), , Basic Structural, Analysis, , CONTENTS, Part-1, , :, , Basic Structural Analysis .......................... 3–2C to 3–3C, , Part-2, , :, , Equilibrium in Three Dimensions ........... 3–3C to 3–3C, , Part-3, , :, , Analysis of Simple Trusses ....................... 3–3C to 3–7C, by Method of Sections, , Part-4, , :, , Analysis of Simple Trusses ..................... 3–7C to 3–19C, by Method of Joints, , Part-5, , :, , Zero Force Member ................................ 3–19C to 3–20C, , Part-6, , :, , Simple Beams and Support ................... 3–20C to 3–28C, Reactions
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3–2 C (CE-Sem-3), , www.aktutor.in, , Basic Structural Analysis, , Basic Structural Analysis., , CONCEPT OUTLINE, Truss : A structure made up of several members riveted or welded, together is known as truss., Frame : If the members of the structure are hinged or pin-joined,, then the structure is known as frame., , Questions-Answers, Long Answer Type and Medium Answer Type Questions, , Que 3.1., , What are the different types of frames ?, , Answer, Following are the different types of frames :, i., , Perfect Frame :, , 1., , The frame which is composed of such members, which are just sufficient, to keep the frame in equilibrium, when the frame is supporting an, external load is known as perfect frame., , 2., , For a perfect frame, the number of joints and number of members are, given by,, n = 2j – 3, where,, n = Number of members., j = Number of joints., , ii., , Imperfect Frame :, , 1., , A frame in which number of members and number of joints are not, given by n = 2j – 3 is known as imperfect frame. This means that, number of members in an imperfect frame will be either more or less, than (2j – 3)., , 2., , If the number of members in a frame are less than (2j – 3), then the, frame is known as deficient frame., , 3., , If the number of members in a frame are more than (2j – 3), then the, frame is known as redundant frame., , Que 3.2., , What do you understand by the analysis of frame ? Also, , write down the assumptions made in the analysis of frame.
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Engineering Mechanics, , www.aktutor.in, , 3–3 C (CE-Sem-3), , Answer, a., , Analysis of a Frame :, , 1., , Analysis of a frame consists of :, i., , Determinations of the reactions at the supports., , ii., , Determinations of the axial forces in the members of the frame., , 2., , The reactions are determined by the condition that the applied load, system and the induced reactions at the supports form a system in, equilibrium., , 3., , The forces in the members of the frame are determined by the condition, that every joint should be in equilibrium and so, the forces acting at, every joint should form a system in equilibrium., , b., , Assumptions made in the Analysis of Frame :, , 1., , The frame should be perfect., , 2., , The frame carries load at the joints., , 3., , All the members are pin-joined and joints are smooth., , Equilibrium in Three Dimensions., Questions-Answers, Long Answer Type and Medium Answer Type Questions, , Que 3.3., , Write down the equations for the equilibrium of a body, , in three dimension., , Answer, 1., , 2., , There are six equations expressing the equilibrium of a body in three, dimensions. These are :, i., , Sum of forces : Fx = 0, Fy = 0 and Fz = 0, , ii., , Sum of moments : Mx = 0, My = 0 and Mz = 0, , The above six equations can be resolved into components to solve the, given problems., , Analysis of Simple Trusses by Method of Sections.
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3–4 C (CE-Sem-3), , www.aktutor.in, , Basic Structural Analysis, , Questions-Answers, Long Answer Type and Medium Answer Type Questions, , Que 3.4., , Write the procedure of method of section in truss, , analysis., , Answer, Procedure of method of sections is as follows :, Step 1 : The truss is split into two parts by passing an imaginary section., Step 2 : The imaginary section has to be such that it does not cut more, than three members in which the forces are to be determined., Step 3 : The conditions of equilibrium Fx = 0, Fy = 0, and M = 0 are, applied for one part of the truss and the unknown forces in the member, is determined., Step 4 : While considering equilibrium, the nature of force in any, member is chosen arbitrarily to be tensile or compressive., i., , If the magnitude of a particular force comes out positive, the, assumption in respect of its direction is correct., , ii., , However, if the magnitude of the force comes out to be negative,, the actual direction of the force is opposite to that what has been, assumed., , Que 3.5., , A truss of 12 m span is loaded as shown in Fig. 3.5.1., , Determine the forces in the members DG, DF and EF, using method, of sections., , Answer, Given : Length of truss = 12 m, Fig. 3.5.1., To Find : Forces in members DG, DF and EF., 1., , In triangle AEC, AC = AE cos 30°, = 4 × 0.866 = 3.464 m, , 2., , Now length,, , 3., , Now taking the moments about A, we get, , AD = 2 × AC = 2 × 3.464 = 6.928 m, RB × 12 = 2 × AC + 1 × AD + 1 × AE, = 2 × 3.464 + 1 × 6.928 + 1 × 4 = 17.856, , , , RB =, , 17.856, = 1.49 kN, 12
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Engineering Mechanics, , 3–5 C (CE-Sem-3), , www.aktutor.in, 1 kN, D, , 1, , 2 kN, C, , 1 kN, A, , G, , 30°, , 60°, , 60°, , E, , HA, 4m, , 30°, , B, , F, 1, , 1 kN, , 4m, , 4m, , RA, , RB, Fig. 3.5.1., , 4., , Now draw the section line (1-1), passing through members DG, DF and, EF in which the forces are to be determined. Consider the equilibrium, of the right part of the truss. This part is shown in Fig. 3.5.2., , 5., , Taking moments of all forces acting on right part about point F, we get, 1, D, FDG, F FD, E, FFE, , G, , F, , 1, , 30°, B, 1.49 kN, 4m, RB, , Fig. 3.5.2., , RB × 4 + FDG × FG = 0, 1.49 × 4 + FDG × (4 × sin 30°) = 0, FDG =, , 6., , ( FG = 4 × sin 30°), , 1.49 4, = – 2.98 kN, 4 sin 30, , FDG = 2.98 kN (Compressive), , Now taking the moments about point D, we get, RB × BD cos 30° = FFE × BD sin 30°, RB × cos 30° = FFE × sin 30°, , , FFE =, , 1.49 cos 30 1.49 0.866, , sin 30, 0.5, , = 2.58 kN (Tensile)
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3–6 C (CE-Sem-3), 7., , www.aktutor.in, , Basic Structural Analysis, , Now taking the moments of all forces acting on the right part about, point B, we get, FFD × Perpendicular distance between FFD and B = 0, , , FFD = 0 ( Perpendicular distance between FFD and B cannot be zero), , Que 3.6., , Find forces in the members EC, FC and FD of the truss, , shown in Fig. 3.6.1., G, , H, , 1000 N, , 5m, , , 3m, F, , E, FEC, , FFC, , , , 1, C, , 3m, , D, 5m, , A, , HA, , FFD, 1, , , 3m, B, , RA, , 4m, , RB = 2250 N, , Fig. 3.6.1., , Answer, Given : Fig. 3.6.1., To Find : Forces in the members EC, FC and FD., 1., , 2., , From geometry of Fig. 3.6.1., 3, = 0.6, cos =, 5, 4, and, sin =, = 0.8, 5, Draw the FBD of the portion above section 1 – 1 (Fig. 3.6.2)., , 3., , Consider the equilibrium of the FBD of the drawn portion,, MF = 0, – FEC × 4 + 1000 × 3 = 0, FEC = 750 N
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Engineering Mechanics, , 3–7 C (CE-Sem-3), , www.aktutor.in, G, , H, , 1000N, , , FFC sin , , F, , E, FFC, , , , FEC, , FFD, , FFC cos , Fig. 3.6.2., , 4., , Consider the condition of equilibrium at point F,, Fx = 0, FFC sin = 1000, FFC × 0.8 = 1000, , , FFC = 1250 N, , and, , Fy = 0, , FEC + FFD + FFC cos = 0, 750 + FFD + 1250 × 0.6 = 0, FFD = – 1500 N, 5., , So, direction of FFD is opposite to our assumed direction hence it is, compressive in nature., , Analysis of Simple Trusses by Method of Joints., Questions-Answers, Long Answer Type and Medium Answer Type Questions, , Que 3.7., , Write the procedure of method of joints in truss analysis., , Answer, Procedure of method of joints is as follows :, Step 1 : Determine the inclinations of all inclined members., Step 2 :, 1., , Look for a joint at which there are only two unknowns., , 2., , If such a joint is not available, determine the reactions at the, supports, and then at the supports these unknowns may reduce to, only two.
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3–8 C (CE-Sem-3), , Basic Structural Analysis, , www.aktutor.in, , Step 3 :, 1., , Now there are two equations of equilibrium for the forces meeting, at the joint and two unknown forces. Hence, the unknown forces, can be determined., , 2., , If the assumed direction of unknown force is opposite, the value, will be negative. Then reverse the direction and proceed., , Step 4 : On the diagram of the truss, mark arrows on the members near, the joint analysed to indicate the forces on the joint. At the other end,, mark the arrows in the reverse direction., Step 5 : Look for the next joint where there are only two unknown, forces and analyse that joint., Step 6 : Repeat steps 4 and 5 till forces in all the members are found., Step 7 :, 1., , Determine the nature of forces in each member and tabulate the, results., , 2., , Note that if the arrow marks on a member are towards each other,, then the member is in tension and if the arrow marks are away from, each other, the member is in compression as shown in Fig. 3.7.1., Tension, Compression, Fig. 3.7.1., , Que 3.8., , Using method of joint determine the forces in each, , member of the truss shown in Fig. 3.8.1., , AKTU 2013-14, (II) Marks 10, E, , 4m, A, , D, B, , C, 4m, , 3m, 10000 N, , 3m, 6000 N, , Fig. 3.8.1., , Answer, Given : Fig. 3.8.1., To Find : Forces in each member of truss.
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Engineering Mechanics, , 1., , From AFE,, , 2., , In BEF,, , 3–9 C (CE-Sem-3), , www.aktutor.in, , EF, EF, 4, 4, , , , AF AB BF 3 2 5, = 38.66°, , tan =, , EF 4, =2, BF 2, = 63.43°, Fy= 0, RA + RD = 10000 + 6000 = 16000 N, tan =, , ...(3.8.1), , E, , 4m, A, , , , , , , B, , F, 4m, , 3m, RA, , 10000 N, , , C, , D, 3m, , 6000 N, , RD, , Fig. 3.8.2., , 3., , Taking moment about A, MA = 0, 10000 × 3 + 6000 × 7 – RD × 10 = 0, , 72000, = 7200 N, 10, From eq. (3.8.1), we have, RA = 8800 N, Considering equilibrium of joint A,, RD =, , 4., 5., , F AE, , A, , 38.66°, FAB, 8800 N, Fig. 3.8.3., , Fx = 0, FAE cos 38.66° + FAB = 0, ...(3.8.2), Fy = 0, FAE sin 38.66° + 8800 = 0, FAE = – 14086.81 N(Compressive), From eq. (3.8.2), we get, FAB = – FAE cos 38.66° = 10999.92 11000 N (Tensile)
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Engineering Mechanics, Que 3.9., , www.aktutor.in, , 3–11 C (CE-Sem-3), , Determine the force in each member of the simple, , equilateral truss Fig. 3.9.1., A, 2m, B, , 2m, 2m, , 75 kg, C, Fig. 3.9.1., , AKTU 2014-15, (I) Marks 10, Answer, Given : Fig. 3.9.1., To Find : Force in each member of truss., 1., , Consider the equilibrium of the entire frame,, MA = 0, RC × 2 – 735 × 2 cos 30º = 0, RC = 636.53 N, Fx = 0, RC + HA = 0, RA, HA, , A, 2m, , F3, , 60º, , 2m, 60º, RC, , F1, 60º, , B, , F2, 2m, 75 kg, , C, 735 N, Fig. 3.9.2., , HA = – RC = – 636.53 N, HA = – 636.53 N, Fy = 0
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3–12 C (CE-Sem-3), , www.aktutor.in, , Basic Structural Analysis, , RA – 735 = 0, RA = 735 N, 2., , Considering equilibrium of joint A,, RA = 735 N, A, F1, , HA = – 636.53 N 60°, , Fig. 3.9.3., , Fy = 0, F3 + F1 cos 60º = 735, , ...(3.9.1), , Fx = 0, – 636.53 + F1 sin 60° = 0, F1 =, 3., , 636.53, = 735 N (Tensile), sin 60º, , From eq. (3.9.1), we get, F3 = 735 – F1 cos 60°, = 735 – 735 cos 60°, F3 = 367.5 N (Tensile), , 4., , Considering equilibrium of joint C,, , F3, 60º, 30º, RC, , F2, , C, Fig. 3.9.4., , Fx = 0, RC – F2 cos 30º = 0, 636.53 = F2 cos 30º, F2 = 735 N (Tensile), Que 3.10., , Compute the forces in all the members for the given, , truss as shown in Fig. 3.10.1. Distance between A and C is 12 m.
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Engineering Mechanics, , www.aktutor.in, , 3–13 C (CE-Sem-3), , B, 30º, 60º D, , A, 5m, , C, , 500 N, Fig. 3.10.1., , AKTU 2015-16 , Marks 10, Answer, Given : Fig. 3.10.1, AC = 12 m, To Find : Forces in all members., 1., , Consider the equilibrium of entire truss,, Fy = 0, RA + RC = 500 N, , ...(3.10.1), , B, 90º 30º, , A, RA, , 120º, 60º, D 500 N, 5m, , 30º, , 12 m, Fig. 3.10.2., , 2., , Taking moment about A, MA = 0, 500 × 5 = RC × 12, RC = 208.33 N, , 3., , From eq. (3.10.1), we get, , 4., , Considering equilibrium of joint A,, , RA = 500 – 208.33 = 291.67 N, F AB, 30°, A, , FAD, RA, Fig. 3.10.3., , C, RC
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3–14 C (CE-Sem-3), , www.aktutor.in, , Basic Structural Analysis, , Fy = 0, RA + FAB sin 30° = 0, 291.67 + FAB × sin 30° = 0, FAB = 583.34 N (Compressive), Fx = 0, FAD + FAB cos 60° = 0, FAD + (– 583.34) cos 30° = 0, FAD = 505.19 N (Tensile), 5., , Considering the equilibrium of joint C,, FCB, 30°, , C, , FCD, , RC, , Fig. 3.10.4., , Fy = 0, FCB sin 30° + RC = 0, FCB × sin 30 + 208.33 = 0, FCB = 416.66 N (Compressive), Fx = 0, FCD + FCB cos 30° = 0, FCD + (– 416.66) cos 30° = 0, FCD = 360.84 N (Tensile), 6., , Considering the equilibrium of joint D,, FDB, 60°, , D, FDC, , FDA, , 500 N, Fig. 3.10.5., , Fy = 0, FDB sin 60° = 500, FDB = 577.35 N (Tensile), Que 3.11., , Determine the forces in all members of the truss as, , shown in Fig. 3.11.1.
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Engineering Mechanics, , 3–17 C (CE-Sem-3), , www.aktutor.in, , FDC = 10 cos 30° = 8.66 kN (Tensile), 9., , Consider the equilibrium at joint B,, B, , F AB, , –, , 30º, 45º, , 15º, , =, , 60º, 30º, 2, .3, 17, , FBC, FBD, , 20 kN, Fig. 3.11.5., , Fy = 0, 20 + FAB cos 30º + FBC cos 15º + FBD cos 60º = 0, 20 – 17.32 cos 30º + FBC cos 15º – 10 cos 60º = 0, FBC 0, , Que 3.12. Determine the forces in all member of the truss shown, in Fig. 3.12.1 and indicate the magnitude and nature of forces on the, diagram of the truss. All inclined members are at 60° to horizontal, , AKTU 2016-17, (I) Marks 10, , and length of each member is 2 m., 40 kN, , 50 kN, , B, , C, , A, , E, 2m, , D, 2m, , 60 kN, Fig. 3.12.1., , Answer, Given : Fig. 3.12.1, length of each member = 2 m, To Find : Magnitude and nature of all the forces in the members of, truss., 1., , Consider the equilibrium of the entire frame,, MA = 0, RD × 4 – 40 × 1 – 60 × 2 – 50 × 3 = 0, , RD = 77.5 kN, Fy = 0, RA + 77.5 = 40 + 60 + 50
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Engineering Mechanics, , www.aktutor.in, , 3–19 C (CE-Sem-3), , FBC = 60.622 kN (Compressive), 40 kN, F BC, , B, 60°, , 60°, , 60°, FBE, , F AB, , Fig. 3.12.4., , 6., , Considering equilibrium at joint C,, Fy = 0, FCE sin 60° + 50 + FDC sin 60° = 0, FCE =, , ( 77.5) – 50, = 31.754 kN (Tensile), sin 60, 50 kN, , FBC, , C, , 60°, , 60°, , 60°, , FCD, , FCE, Fig. 3.12.5., , Now the forces in all the members are known. The results are shown on, the diagram of the truss in Fig. 3.12.6., , 489, 89., , 528, 37., , 31., 754, , 40 kN, 50 kN, 60.622, B, C, 83., 716, , 7., , A, , D, 41.858, , E, , 44.745, , 60 kN, RA = 72.5 kN, , RD = 77.5 kN, Fig. 3.12.6., , Zero Force Member.
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3–20 C (CE-Sem-3), , www.aktutor.in, , Basic Structural Analysis, , Questions-Answers, Long Answer Type and Medium Answer Type Questions, , Que 3.13. Write a short note on zero force members., Answer, 1., , Zero force members are the members in which there is no force., , 2., , After knowing the members of zero forces, they can be eliminated while, calculating the forces in the members., 0, , 0, 0, , F, , F, , 0, , F, , F, , Fig. 3.13.1., , 3., , A member that joins two other collinear members, at right angles and if, no load is acting at that joint, then it will be a zero force member (member, with ‘L’, ‘T’ and ‘Y’ shapes)., , Simple Beams and Support Reactions., Questions-Answers, Long Answer Type and Medium Answer Type Questions, , Que 3.14. What are the different types of beams ?, Answer, i., 1., 2., , Following are the different types of beams :, Cantilever Beam :, A beam which is fixed at one end and free at the other end is known as, cantilever beam (Fig. 3.14.1)., At the fixed end, there will be fixing moment. Also at the fixed end,, there can be horizontal and vertical reactions, depending upon the type, of load acting on the beam.
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Engineering Mechanics, , www.aktutor.in, , 3–21 C (CE-Sem-3), , Fig. 3.14.1. Cantilever beam., , ii., , Simply Supported Beam : A beam supported or resting freely on the, supports at its both ends is known as simply supported beam (Fig. 3.14.2)., , Fig. 3.14.2. Simply supported beam., , iii. Overhanging Beam : If the end portion of a beam is extended beyond, the support, such beam is known as overhanging beam (Fig. 3.14.3)., Simply supported, portion, , Overhanging, portion, , Support, Fig. 3.14.3. Overhanging beam., , iv. Fixed Beam : A beam whose both ends are fixed or built-in-walls, is, known as fixed beam (Fig. 3.14.4). A fixed beam is also known as a builtin or encastre beam. At the fixed ends, there will be fixing moments and, reactions., , Fig. 3.14.4. Fixed beam., , v., , Continuous Beam : A beam which is provided more than two supports, as shown in Fig. 3.14.5, is known as continuous beam., , Fig. 3.14.5. Continuous beam., , Que 3.15., , Discuss in short about the various types of supports., , Answer, Following are the various types of supports :
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3–22 C (CE-Sem-3), i., , www.aktutor.in, , Basic Structural Analysis, , Simple Support or Knife Edge Support : A beam supported on the, knife edges A and B is shown in Fig. 3.15.1. The reactions at A and B in, case of knife edge support will be normal to the surface of the beam., Beam, A, , B, (a), , RA, , RB, , (b ), Fig. 3.15.1., , ii., , Roller Support : A beam supported on the rollers at points A and B is, shown in Fig. 3.15.2(a). The reaction in case of roller supports will be, normal to the surface on which roller is placed as shown in Fig. 3.15.2(b)., A, , B, (a), , RA, , RB, , (b ), Fig. 3.15.2., , iii. Pin Joint (or Hinged) Support : A beam, which is hinged (or pinjoint) at point A, is shown in Fig. 3.15.3. The reaction at the hinged end, may be either vertical or inclined depending upon the type of loading., A, Fig. 3.15.3., , iv. Smooth Surface Support : Fig. 3.15.4 shows a body in contact with a, smooth surface. The reaction will always act normal to the support as, shown in Fig. 3.15.4., Body, , A, Smooth, surface, , RA, , Fig. 3.15.4.
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Engineering Mechanics, v., 1., 2., , 3–23 C (CE-Sem-3), , www.aktutor.in, , Fixed or Built-in Support :, Fig. 3.15.5, shows the end A of a beam, which is fixed. Hence the support, at A is known as a fixed support., The fixed support prevents the vertical movement and rotation of the, beam. Hence at the fixed support there can be horizontal reaction and, vertical reaction. Also there will be fixing moment at the fixed end., , A, Fig. 3.15.5., , Que 3.16. What are the different types of loading ? Explain., Answer, Following are the different types of loading :, i., 1., , 2., , Concentrated or Point Load :, Fig. 3.16.1 shows a beam AB, which is simply supported at the ends A, and B. A load W is acting at the point C. This load is known as point load, (or concentrated load)., Hence any load acting at a point on a beam, is known as point load., W, , B, , A, , C, RA, , RB, Fig. 3.16.1., , ii., , Uniformly Distributed Load :, , 1., , If a beam is loaded in such a way that each unit length of the beam, carries same intensity of the load, then that type of load is known as, uniformly distributed load which is written as UDL., , 2., , Fig. 3.16.2 shows a beam AB, which carries a uniformly distributed load., B, , A, RA, , RB, Fig. 3.16.2., , iii. Uniformly Varying Load :, 1., , Fig. 3.16.3 shows a beam AB, which carries load in such a way that the, rate of loading on each unit length of the beam varies uniformly. This, type of load is known as uniformly varying load.
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3–24 C (CE-Sem-3), 2., , www.aktutor.in, , Basic Structural Analysis, , The total load on the beam is equal to the area of the load diagram. The, total load acts at the center of gravity of the load diagram., , A, RA, , RB, Fig. 3.16.3., , Que 3.17., , Determine the reactions at A for the cantilever beam, , subjected to the distributed and concentrated loads Fig. 3.17.1., Y, 2 kN, X, , 4 kN/m, A, 3m, , 1.5 m, , 1.5 m, , Fig. 3.17.1., , AKTU 2014-15, (I) Marks 10, Answer, Given : Fig. 3.17.1., To Find : Reaction at A., 1., , Consider the equilibrium of the beam,, Fy = 0, RA –, , 1, ×3×4–2= 0, 2, 2 kN, 4 kN/m, M, A, , B, 3m, , C, 1.5 m, , RA, Fig. 3.17.2., , D, 1.5 m
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Engineering Mechanics, , www.aktutor.in, , 3–25 C (CE-Sem-3), , RA = 2 + 6 = 8 kN, 2., , Taking moments about A,, MA = 0, , 2, 1, , M – 3 4 3 – 2 × 4.5 = 0, 3, 2, , M = 21 kN-m, 3., , So, the reaction at A, RA = 8 kN, Moment, M = 21 kN-m, , Que 3.18., , Determine the reaction at support A and D in the, , structure shown in Fig. 3.18.1., 80 kN, A, , B, C, D, 3m, , 1m, , 0.5 m, , 4m, , Fig. 3.18.1., , AKTU 2014-15, (I) Marks 10, Answer, Given : Fig. 3.18.1., To Find : Reaction at support A and D., 1., , Consider the FBD of the given beam for the section AB and consider its, equilibrium,, 80 kN, A, , B, 3m, , 1m, , RA, , RB, Fig. 3.18.2., , Fy = 0, RA + RB = 80 kN, 2., , Taking moment about A, MA = 0, 80 × 3 = 4 × RB, RB = 60 kN
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3–26 C (CE-Sem-3), , www.aktutor.in, , RA = 80 – RB = 80 – 60 = 20 kN, , , 3., , Basic Structural Analysis, , Consider the FBD of the given beam for section BD and consider its, equilibrium,, 60 kN, C, , B, , D, , 0.5, , 4m, RC, , RD, , Fig. 3.18.3., , Fy = 0, RC + RD = 60 kN, 4., , Taking moment about D, we have, MD = 0, RC × 4 = 60 × 4.5, RC = 67.5 kN, RD = 60 – RC = 60 – 67.5 = – 7.5 kN, , , , Here negative sign means that reaction will at C in downward direction., , Que 3.19. Determine the reactions at B and E of the beam, loaded, as shown in Fig. 3.19.1 below., 40 kN, , 20 kN, , 10 kN/m, , A, , B, 2m, , C, 2m, , RB, , 10 kN/m, , D, 2m, , E, , F, , 2m, , Fig. 3.19.1., , 2m, RE, , AKTU 2016-17 , (II) Marks 10, Answer, Given : Fig. 3.19.1., To Find : Reactions at B and E.
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Engineering Mechanics, 1., , 3–27 C (CE-Sem-3), , www.aktutor.in, , Considering the equilibrium of the beam,, Fy = 0, RB + RE = 10 × 2 + 40 + 10 × 2 + 20, RB + RE = 100 kN, , 2., , ...(3.19.1), , Now taking moment about B, we have, MB = 0, – 10 × 2 × 1 + 40 × 2 + 10 × 2 × 5 – RE × 6 + 20 × 8 = 0, RE = 53.33 kN, , 3., , From eq. (3.19.1), we get, RB = 100 – RE = 100 – 53.33, RB = 46.67 kN, , Que 3.20. Calculate the support reactions in the given cantilever, beam as shown in Fig. 3.20.1., 20 kN, , 40 kN/m, 50 kNm, , B, , A, 3m, , 1m, Fig. 3.20.1., , 1m, , AKTU 2015-16, (I) Marks 10, Answer, Given : Fig. 3.20.1., To Find : Support reactions., 1., , Considering the equilibrium of the beam,, Fy = 0, RA –, , 1, × 3 × 40 – 20 = 0, 2, RA = 80 kN
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3–28 C (CE-Sem-3), 2., , www.aktutor.in, , Basic Structural Analysis, , Taking moment about point A, MA = 0, MA –, , 1, 2, × 3 × 40 ×, × 3 – 50 – 20 × 5 = 0, 2, 3, MA = 120 + 50 + 100, MA = 270 kN-m, 60 kN, , 20 kN, , 40 kN/m, , MA, , 50 kNm, A, , RA, , B, , 2m, 3m, , 1m, , Fig. 3.20.2., , , , 1m
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Engineering Mechanics, , 4, , www.aktutor.in, , 4–1 C (CE-Sem-3), , Review of Particle, Dynamics, , CONTENTS, Part-1, , :, , Review of Particle Dynamics – ............. 4–2C to 4–11C, Rectilinear Motion, , Part-2, , :, , Plane Curvilinear Motion ..................... 4–11C to 4–23C, (Rectangular, Path and Polar, Coordinates), , Part-3, , :, , Work, Kinetic Energy, Power ............. 4–23C to 4–29C, Potential Energy, , Part-4, , :, , Impulse, Momentum (Linear, ............. 4–29C to 4–34C, Angular), , Part-5, , :, , Impact (Direct and Oblique) ................ 4–34C to 4–38C
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4–2 C (CE-Sem-3), , Review of Particle Dynamics, , www.aktutor.in, , Review of Particle Dynamics – Rectilinear Motion., , CONCEPT OUTLINE, Dynamics of Particle : The study of motion of a particle is, known as dynamics of particle. It is further divided into kinematics, and kinetics., Rectilinear Motion : The motion of the body along a straight, line is called rectilinear motion. It is also known as onedimensional motion., , Questions-Answers, Long Answer Type and Medium Answer Type Questions, , Que 4.1., , What are the parameters used for defining the rectilinear, , motion of a particle ?, OR, Define the following terms :, i., Displacement., ii. Velocity., iii. Acceleration., , Answer, Following are the parameters used for defining the rectilinear motion :, i., , Displacement : It is defined as the change in position of the particle, during given interval of time. The displacement is a vector quantity and, it is dependent only on the initial and final position of the particle., , ii., , Velocity : Velocity of a particle can be defined as the rate of change of, displacement with time., , ds, dt, iii. Acceleration : Acceleration of a particle can be defined as the rate of, change of the velocity with time., Mathematically,, , v=, , Mathematically,, , a=, , dv, dt
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www.aktutor.in, , Engineering Mechanics, Que 4.2., , 4–3 C (CE-Sem-3), , Derive the equation of motion for a body moving in a, , straight line by the method of integration., , Answer, The equation of motion of a body moving in a straight line may be, derived by integration as given below :, , 1., , 1, at2 :, 2, Let a body is moving with a uniform acceleration a., , 2., , We know that,, , 3., , d ds , d2 s, = a or, = a, 2, dt, dt, dt, ds , d = a dt, or, dt , Integrating the above equation,, , i., , Derivation of s = ut +, , ds, = at + C1, dt, where, C1 = Constant of integration., ds , , d dt , , 4., , 6., , a dt, , or, , ...(4.2.1), , ds, = Velocity at any instant., dt, When t = 0, the velocity is known as initial velocity which is represented, by u., But, , ds, = Initial velocity = u, dt, Substituting these values in eq. (4.2.1), we get, u = a × 0 + C1, C1 = u, Substituting the value of C1 in eq. (4.2.1), we get, , , 5., , =, , At, t = 0,, , ds, = at + u, dt, Now, integrating eq. (4.2.2), we get, at 2, + ut + C2, 2, where, C2 = Another constant of integration., s=, , ...(4.2.2), , ...(4.2.3), , 7., , When t = 0, then s = 0. Substituting these values in eq. (4.2.3), we get, , 8., , a, × 0 + u × 0 + C2, 2, , C2 = 0, Substituting this value of C2 in eq. (4.2.3), we get, 0=
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4–4 C (CE-Sem-3), , Review of Particle Dynamics, , www.aktutor.in, s = ut +, , ii., 1., , 1 2, at, 2, , Derivation of v = u + at :, From eq. (4.2.2), we have, , ds, = at + u, dt, 2., , ...(4.2.4), , ds, represents the velocity at any time. After the time ‘t’ the velocity, dt, is known as final velocity, which is represented by v., But, , ds, after time ‘t’ = Final velocity = v, dt, ds, 3. Substituting the value of, = v in eq. (4.2.4), we get, dt, v = u + at, iii. Derivation of v2 = u2 + 2as :, 1. We know that, acceleration a is given by, , , v dv, ds, v dv = ads, Integrating eq. (4.2.5), we get, a=, , 2., , 3., 4., , ...(4.2.5), , v2, = as + C3, 2, where,, C3 = Constant of integration., When s = 0, the velocity is known as initial velocity., At s = 0,, v=u, Substituting these values in eq. (4.2.6), we get, , ...(4.2.6), , u2, = a × 0 + C3, 2, u2, 2, Substituting the value of C3 in eq. (4.2.6), we get, , , 5., , C3 =, , v2, u2, = as +, or v2 = u2 + 2as, 2, 2, Que 4.3., , Acceleration of a ship moving along a straight curve, , varies directly as the square of its speed. If the speed drops from, 3 m/sec to 1.5 m/sec in one minute, find the distance moved in this, period., , AKTU 2013-14 , (II) Marks 10
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www.aktutor.in, , Engineering Mechanics, , 4–5 C (CE-Sem-3), , Answer, Given : a v2, v1 = 3 m/sec , v2 = 1.5 m/sec, t = 1 min = 60 s, To Find : Distance moved., 1., , 2., , Acceleration is given by,, a = Kv2, , ...(4.3.1), , dv, = Kv2, dt, dv, = Kdt, v2, On integrating both sides,, 1.5, 60, dv, 3 v2 = K 0 dt, 1.5, , v 21 , 60, , = K[ t ]0, 2 1 3, , 1 , v , , 3., , 4., , 1.5, , = K × 60, 3, , 1, 1, 1.5 3 = 60 K, 1, = 60 K, 3, K = – 5.55 × 10–3, From eq. (4.3.1.), we get, a = – 5.55 × 10–3 v2, vdv, = – 5.55 × 10–3 v2, ds, , dv, = – 5.55 × 10–3 ds, v, On integrating both sides,, –3 s] s, [ln v]1.5, 3 = [– 5.55 × 10, 0, , ln 1.5 – ln 3 = – 5.55 × 10–3 s, s = 124.89 m, Que 4.4., , Derive the formula for the distance travelled in nth, , second., , Answer, 1., , Let,, , u = Initial velocity of a body., a = Acceleration of the body., sn = Distance covered in n second.
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4–6 C (CE-Sem-3), , Review of Particle Dynamics, , www.aktutor.in, , sn – 1 = Distance covered in (n – 1) seconds., 2., , Then distance travelled in the nth seconds, = Distance travelled in n seconds –, Distance travelled in (n – 1) seconds, = sn – sn – 1, , 3., , Distance travelled in n seconds is obtained by substituting t = n in the, following equation,, , 1 2, at, 2, 1, , sn = un + an2, 2, 1, Similarly,, sn – 1 = u(n – 1) + a(n – 1)2, 2, Distance travelled in the nth seconds, s = ut +, , 4., , = sn – sn – 1, , 1, 1, , , , = un an2 u(n 1) a( n 1)2 , , , 2, 2, , = un , , 1 2 , 1, , an un u a(n2 1 2n) , 2, , 2, , , 1 2, 1, 1, 1, an un u an 2 a a 2n, 2, 2, 2, 2, 1, a, = an u a u (2n 1), 2, 2, = un , , Que 4.5., , Write down the equation of motion due to gravity., , Answer, Following are the equation of motion due to gravity :, i., , For Downward Motion :, a= +g, v = u + gt, s = h = ut +, , ii., , 1 2, gt, 2, , v2 – u2 = 2gh, For Upward Motion :, a= –g, v = u – gt, s = h = ut –, v2 – u2 = – 2gh, , 1 2, gt, 2
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www.aktutor.in, , Engineering Mechanics, Que 4.6., , 4–7 C (CE-Sem-3), , A stone is dropped into a well and is heard to strike the, , water after 4 seconds. Find the depth of the well if the velocity of, , AKTU 2014-15, (II) Marks 05, , sound is 350 m/sec., , Answer, Given : t = 4 sec, velocity of sound = 350 m/sec, To Find : Depth of the well., 1., , Let,, , h = Depth of well., t1 = Time taken by stone to strike water., t2 = Time taken by sound to reach from surface of, water to top of well., , 2., 3., , 4., , So, total time,, t = t1 + t2 = 4, ...(4.6.1), Considering downward motion of stone and using the relation, 1, s = ut +, gt2, we have, 2, 1, h = 0 × t1 +, × 9.81 × t12, 2, h = 4.905 t12, Considering the motion of sound, the time taken by the sound to reach, from surface of water to top of well is given by,, t2 =, , 4.905 t12, Depth of well, h, , , Speed of sound 350, 350, ( h = 4.905 t12 ), , 5., , From eq. (4.6.1), we have, t1 +, , 4.905 t12, =4, 350, , 350 t1 + 4.905 t12 = 1400, , 6., , 4.905 t12 + 350 t1 – 1400 = 0, , Solution of the quadratic equation given as,, t1 =, , 350 350 2 4 4.905 1400 350 387.26, , 2 4.905, 9.81, , Taking the +ve root ; t1 = 3.798 sec, 7., , Depth of well, h = 4.905 t12 = 4.905 × (3.798)2 = 70.75 m
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4–8 C (CE-Sem-3), , Que 4.7., , Review of Particle Dynamics, , www.aktutor.in, , Acceleration of particle is defined by a = 21 – 21s2, where, , a is acceleration in m/sec2 and s is in metres. The particle starts, with rest at s = 0, Determine (a) velocity when s = 1.5 m, (b) the, position where velocity is again zero, (c) the position where the, , AKTU 2013-14, (I) Marks 10, , velocity is maximum., , Answer, Given : a = 21 – 21 s2, u = 0, To Find :, , i. Velocity when s = 1.5 m., ii. The position where velocity is again zero., iii. The position where the velocity is maximum., , 1., , Velocity when s = 1.5 m,, , dv, = 21 – 21 s2, ds, , a= v, , , , v, , 0, , vdv =, , , , 1.5, , 0, , (21 21 s2 )ds, 1.5, , 2., , 3., , , v2, 21 s3 , = 21s , , 2, , 3 0, , (1.5)3 , v2 = 2 21 1.5 21 , , 3 , , v2 = 15.75, v = 3.97 m/sec, Position where velocity is again zero,, dv, = 21 – 21s2, v, ds, 2, v d v = (21 21s ) ds, , v2, 21s 3, = 21s –, +C, 2, 3, Here C is integration constant., At s = 0, v = 0, from eq. (4.7.1), we have, C=0, Put the value of C in eq. (4.7.1), we get, , , 21s3 , v2 = 2 21s , 3 , , 4., , For v = 0, we have, , 21s 3, =0, 3, 3, 3s – s = 0, s(3 – s2) = 0, , 21s –, , ...(4.7.1), , ...(4.7.2)
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Engineering Mechanics, , 5., , www.aktutor.in, , s = 0,, s2 = 3 s = , The velocity will be again zero at s = 1.732 m, On differentiating the eq. (4.7.2) w.r.t s,, 2v, , 4–9 C (CE-Sem-3), , 3, , , s2 , dv, = 2 21 21 3 , 3, ds, , , dv, 21 21s2, =, ds, v, , ...(4.7.3), , dv, =0, ds, , 6., , For maximum or minimum velocity,, , 7., , 0 = (21 – 21s2), s2 = 1, s=±1, At s = 1 m, from eq. (4.7.2), , 8., , , 21 , 2 (21 1) = 5.29 m/sec, 3 , , Now, again differentiating the eq. (4.7.3) w.r.t s, we get, , 9., , d2 v d v 2 , v 2 , = (– 21 × 2 s), ds , ds, , At s = 1 m, from eq. (4.7.3), , v=, , ...(4.7.4), , dv, 21 21s 2 21 21, =, =0, , ds, v, v, dv, 10. Now substituting, = 0 and s = 1 in eq. (4.7.4), we get, ds, 42, 42, d2 v, , = , = – 7.94, 2, v, 5.29, ds, d2 v, As, at s = 1 m is negative, therefore velocity is maximum at 1 m., ds2, , Que 4.8., , A car starts from rest on a curved road of 200 m radius, , and accelerates at a constant tangential acceleration of 0.5 m/sec2., Determine the distance and time which the car will travel before, the total acceleration attained by it becomes 0.75 m/sec2., , AKTU 2013-14, (II) Marks 05, Answer, Given : R = 200 m, at = 0.5 m/sec2, a = 0.75 m/sec2, u = 0, To Find : Distance and time for which the car travel., 1., , We know that,, Final acceleration2 = Normal acceleration2 + Tangential acceleration2
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4–10 C (CE-Sem-3), , Review of Particle Dynamics, , www.aktutor.in, a2 = an2 + at2, , (0.75)2 = an2 + (0.5)2, an2 = 0.3125 m/sec2, 2., , 3., , an = 0.559 m/sec2, Normal acceleration is given by,, 2, an = v, R, v2 = an × R = 0.559 × 200, v2 = 111.8, v = 10.57 m/sec, We know that,, v = u + at t, ( Initial velocity, u = 0), 10.57 = 0.5 t, , t=, 4., , 10.57, = 21.14 sec, 0.5, , Also we known that,, v2 = u2 + 2at s, 2, , v 2 10.57, , 2at, 2 0.5, s = 111.725 m, s=, , Que 4.9., , ( u = 0), , An automobile is accelerated at the rate of 0.8 m/sec2 as, , it travels from station A to station B. If the speed of the automobile, is 36 km/h as it passes station A, determine the time required for, automobile to reach B and its speed as it passes station B. The, distance between A and B is 250 m., , AKTU 2013-14, (II) Marks 05, , Answer, 5, = 10 m/sec, 18, To Find : Time taken to reach B and speed as it passes station B., Given : a = 0.8 m/sec2, s = 250 m , u = 36 km/h = 36 , , 1., , We know that,, , 1 2, at, 2, 1, 250 = 10t +, × 0.8 × t2, 2, 250 = 10t + 0.4t2, 0.4t2 + 10t – 250 = 0, s = ut +, , t=, , – 10 , , (10)2 – 4 0.4 (– 250), 2 0.4
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Engineering Mechanics, , www.aktutor.in, , 4–11 C (CE-Sem-3), , 100 400, 0.8, t = 15.45 sec, =, , 2., , – 10 , , Also, we know that, v2 = u2 + 2as, v2 = (10)2 + 2 × 0.8 × 250 = 500, v = 22.36 m/sec, , Plane Curvilinear Motion (Rectangular, Path and, Polar Coordinates)., , CONCEPT OUTLINE, Curvilinear Motion : The motion of a body in a plane along a, circular path is known as plane curvilinear motion., Equation of Motion for Curvilinear Motion :, = 0 + t, 2 = 02 + 2, , 1 2, t, 2, 0 = Initial angular velocity (rad/sec), = 0t +, , where,, , = Final angular velocity (rad/sec), = Angular acceleration (rad/sec2), = Angular displacement (rad), t = Time (sec), Projectile Motion : Curvilinear motion with constant acceleration, can be considered as the combination of two rectilinear motions, occurring simultaneously along two mutually perpendicular x and y, directions. This motion is known as projectile motion., Example : Motion of a missile or a ball hit in air., , Questions-Answers, Long Answer Type and Medium Answer Type Questions, , Que 4.10., , What are the parameters required for defining the, , curvilinear motion of a body ?, OR
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4–12 C (CE-Sem-3), , Review of Particle Dynamics, , www.aktutor.in, , Define the following terms :, i., Angular displacement., ii. Angular velocity., iii. Angular acceleration., , Answer, Following are the parameters required for defining the curvilinear, motion of the body :, i., , Angular Displacement : The displacement of a body in rotation is, called angular displacement, and it is measured in terms of the angle, through which the body moves from the initial state., , ii., , Angular Velocity : The rate of change of angular displacement of a, body with respect to time is called angular velocity. If the body traverses, angular distance d over a time interval dt, then the average angular, velocity is given by,, , d, dt, iii. Angular Acceleration : The rate of change of angular velocity of a, body with respect to time is called angular acceleration., =, , Mathematically,, , Que 4.11., , =, , d d d d 2, , dt dt dt dt2, , Write down the relationship between angular motion, , and linear motion., , Answer, 1., 2., , If r is the distance of the particle from the centre of rotation, then, s = r, The tangential velocity of the particle is called as linear velocity and is, denoted by v. Then, , ds, d, r, dt, dt, The linear acceleration of the particle in tangential direction at is given, by, v=, , 3., , at =, , Que 4.12., , dv, d 2, r 2, dt, dt, , If crank OA rotates with an angular velocity of, , = 12 rad/sec, determine the velocity of piston B and the angular, velocity of rod AB at the instant shown in the Fig. 4.12.1., , AKTU 2014-15, (I) Marks 10
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Engineering Mechanics, , www.aktutor.in, , 4–13 C (CE-Sem-3), , A, 0.6 m, 0.3 m, , 12 rad/sec, 30º, , O, , B, Fig. 4.12.1., , Answer, Given : OA = 12 rad/sec, OA = 0.3 m, AB = 0.6 m, = 30º,, To Find : i. Velocity of piston B., ii. Angular velocity of rod AB., 1., , Applying the sine rule in the OAB (Fig. 4.12.2),, , OA, AB, =, sin 30, sin , 0.3, 0.6, =, (1 / 2), sin , sin = 1, = 90°, So this is the right angle triangle., A, , 0., 6, , 0.3 m, , m, , 12 rad/sec, , , 30º, B, , O, Fig. 4.12.2., , 2., , The length of the link,, OB =, , AB2 OA2
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4–14 C (CE-Sem-3), , Review of Particle Dynamics, , www.aktutor.in, , 0.62 0.32 = 0.5196 m = 0.52 m, , =, 3., , Velocity of point A,, , 4., , vA = rOA = OA × OA = 0.3 × 12 = 3.6 m/sec, Angular velocity of rod AB,, , 5., , vA, 3.6, =, = 6 rad/sec, AB, 0.6, And the velocity at point B, vB = OB × OA, AB =, , = 0.52 × 12 = 6.24 m/sec, , Que 4.13., , A wheel that is rotating at 300 rpm attains a speed of, , 180 rpm after 20 seconds. Determine the acceleration of the flywheel, assuming it to be uniform. Also determine the time taken to come to, rest from a speed of 300 rpm if the acceleration remains the same, and number of revolutions made during this time., , AKTU 2015-16, (I) Marks 10, Answer, Given : N0 = 300 rpm, 0 =, , 2 (300), = 31.4159 rad/sec, 60, , 2 180, = 18.8495 rad/sec, t = 20 sec, 60, i. Acceleration of the flywheel., , N = 180 rpm, =, To Find :, , ii. Time taken to come to rest from a speed of 300 rpm., iii. Number of revolutions., 1., , We know that,, , = 0 + t, , ...(4.13.1), , 18.8495 = 31.4159 + × (20), = – 0.62832 rad/sec2, 2., , If, , = 0, (for rest) then from eq. (4.13.1),, 0 = 31.4159 + (– 0.62832) × t, t = 49.99984 = 50 sec, , Hence time taken to come to rest = 50 sec., 3., , Also we know that,, , 2 = 02 + 2, 0 = (31.4159)2 + 2 × (– 0.62832) × , = 785.395 rad, , 4., , Total revolutions made by flywheel, =, , 785.395, = 124.999 125, 2
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www.aktutor.in, , Engineering Mechanics, , 4–15 C (CE-Sem-3), , Que 4.14. Discuss the curvilinear motion of a body in rectangular, coordinates., , Answer, 1., , Consider a particle moving in the XY-plane. Let its position at an instant, , , of time be A, whose position vector is r as shown in Fig. 4.14.1., Y, A, , x, , , , r, , y, , X, Fig. 4.14.1., , 2., , If x and y be the rectangular coordinates of the point A, then its position, , , vector r can be expressed as, , , 3., , ˆ, ˆ, ...(4.14.1), r = xi yj, Then velocity vector can be obtained by differentiating eq. (4.14.1) with, respect to time, i.e.,, , , d r dx ˆ dy ˆ, v = dt dt i dt j, = v x iˆ v y ˆj, , , , ...(4.14.2), , , where vx and vy are x and y components of velocity v (Fig. 4.14.2)., , vy, , , , v, , Y, A, v, , , r, , X, Fig. 4.14.2., , , vx
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4–16 C (CE-Sem-3), 4., , Review of Particle Dynamics, , www.aktutor.in, , The magnitude and direction of instantaneous velocity can be expressed, in terms of its components as,, v=, , vy , v2x v 2y and v = tan– 1 , vx , , The direction of this instantaneous velocity is tangential to the path of, the particle at that instant., 5., , If the equation of path of the particle is known in the form, y = f(x), then, it can be proved that the direction of velocity vector coincides with the, slope of the curve or tangent to the curve at that point., , 6., , Similarly, the acceleration vector can be obtained by differentiating, eq. (4.14.2) with respect to time, i.e.,, , , , , , , , , a = d v d v x iˆ d v y ˆj, dt, dt, dt, =, , d2 x ˆ d2 y ˆ, i 2 j, dt 2, dt, , = ax iˆ ay ˆj, where axand ay are x and y components of acceleration., 7., , The magnitude and direction of instantaneous acceleration in terms of, its components are,, a=, , Que 4.15., , ay , a2x a2y and a = tan– 1 , ax , , The x and y coordinates of the position of a particle, , moving in curvilinear motion are defined by x = 2 + 3t2 and y = 3 + t3., Determine the particle’s position, velocity and acceleration at t = 3 sec., , Answer, Given : x = 2 + 3t2, y = 3 + t3, To Find : Particle’s position, velocity and acceleration at t = 3 sec., 1., , It is given that,, x = 2 + 3t2 and y = 3 + t3, Therefore, the x and y components of velocity and acceleration can be, obtained by differentiating successively the above expressions with, respects to time., , , vx =, , dx, dy, = 6t, vy =, = 3t2, dt, dt
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Engineering Mechanics, , 4–17 C (CE-Sem-3), , d2 x, d2 y, = 6t, 2 = 6, ay =, dt, dt 2, Particle’s position at t = 3 sec,, and, , 2., , www.aktutor.in, ax =, , x(3) = 2 + 3(3)2 = 29 m, y(3) = 3 + (3)3 = 30 m, 3., , Magnitude and direction of position vector at t = 3 sec are,, r=, and, , 4., , x 2 y2 =, , 292 302 = 41.73 m, , 1 y, 1 30 , r = tan = tan = 45.97°, x, 29, , Particle’s velocity at t = 3 sec,, vx(3) = 6(3) = 18 m/sec, vy(3) = 3(3)2 = 27 m/sec, , 5., , Magnitude of velocity at time t = 3 sec is given by,, v=, , 6., , v2x v 2y =, , 182 272 = 32.45 m/sec, , Its inclination with respect to the X-axis is given by,, , vy , 27 , 1, v = tan = tan 1 = 56.31°, 18 , vx , 7., , Particle’s acceleration at t = 3 sec,, ax(3) = 6 m/sec2, ay(3) = 6(3) = 18 m/sec2, , 8., , Magnitude of acceleration at t = 3 sec is given by,, a=, , 9., , a2x a2y =, , 62 182 = 18.97 m/sec2, , Its inclination with respect to the X-axis is given by,, a , 18 , a = tan 1 y = tan 1 = 71.57°, 6, ax , , Que 4.16., , Write down the equations of projectile motion and, , derive expression for the various terms associated with projectile, motion., , Answer, A., , Equation of Motion for Projectile Motion :, , i., , Motion along the X-direction (Uniform Motion) :, ax = 0, , ...(4.16.1), , vx = v0 cos , , ...(4.16.2)
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4–18 C (CE-Sem-3), , Review of Particle Dynamics, , www.aktutor.in, x = (v 0 cos ) t, , ii., , ...(4.16.3), , Motion along the Y-direction (Uniform Accelerated Motion) :, ay = – g, , ...(4.16.4), , vy = v0 sin – gt, , ...(4.16.5), , vy2 = (v 0 sin )2 – 2gy, y = (v 0 sin ) t –, , ...(4.16.6), , 1 2, gt, 2, , ...(4.16.7), , Y, vy, , v 0 sin , , v, , vx, , vx, , v0, , v x, v y, , , , v, X, , O v cos , 0, Fig. 4.16.1. Projectile motion., , B., , Derivation of Various Terms :, , i., , Time Taken to Reach Maximum Height and Time of Flight :, , 1., , When the particle reaches the maximum height, we know that the, vertical component of velocity i.e., vy is zero. Therefore, from the, eq. (4.16.5), we have, , 2., , Hence, the time taken to reach the maximum height is,, , 0 = v0 sin – gt, , t=, 3., , v0 sin , g, , ...(4.16.8), , Since the time of ascent is equal to the time of descent, the total time, taken for the projectile to return to the same level of projection is,, T=, , 2 v0 sin , g, , ii., , Maximum Height Reached :, , 1., , Substituting the value of time of ascent in the eq. (4.16.7), we get, , v sin 1 v 0 sin , y = v 0 sin 0, 2 g , , , g, g, , 2
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Engineering Mechanics, , www.aktutor.in, =, , 2., , 4–19 C (CE-Sem-3), , v02 sin 2 1 v02 sin 2 , v2, g, = 0 sin 2 , 2, , g, 2 , g, , 2g, , Hence, the maximum height reached is,, hmax =, , v02 sin 2 , 2g, , iii. Range :, 1., , The horizontal distance between the point of projection and point of, return of projectile to the same level of projection is termed as range., , 2., , Hence, range is obtained by substituting the value of total time of flight, in the eq. (4.16.3),, R = (v 0 cos )T, , 2 v sin , = (v0 cos ) 0, , g, , , 3., , Since, sin 2 = 2 sin cos , we can write,, R=, , Que 4.17., , v20 sin 2, g, , A ball is thrown from the ground with a velocity of, , 20 m/sec at an angle of 30° to the horizontal. Determine :, i, The velocity of the ball at t = 0.5 sec and t = 1.5 sec., ii. Total time of flight of the ball., iii. Maximum height reached., iv. Range of the ball., v. Maximum range., , Answer, Given : v0 = 20 m/sec, = 30°, To Find :, , i. The velocity of the ball at t = 0.5 s and t = 1.5 sec., ii., iii., iv., v., , 1., , Total time of flight of the ball., Maximum height reached., Range of the ball., Maximum range., , The initial velocity of the ball can be resolved into horizontal and vertical, components as,, v0x = v0 cos = 20 cos 30° = 17.32 m/sec, and, , v0y = v0 sin = 20 sin 30° = 10 m/sec
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4–20 C (CE-Sem-3), , Review of Particle Dynamics, , www.aktutor.in, , Y, , 5.1 m/sec, , 20 sin 30°, , 20, , 18.06 m/sec, , , 20 cos 30°, , 4.72 m/sec, , 20 cos 30°, 17.95 m/sec, , = 30°, O, , X, , 20 cos 30°, 35.31 m, Fig. 4.17.1., , 2., , We know that the horizontal component of velocity always remains, constant and only the vertical component of velocity varies with time., Thus,, vy(0.5) = v0 sin – gt, = 10 – 9.81 (0.5) = 5.1 m/sec, , 3., , The total velocity at that instant is obtained by,, v(0.5 sec) =, =, , v2x v 2y, , (17.32)2 (5.12 ) = 18.06 m/sec, , And its inclination with respect to the X-axis is obtained by,, , v , = tan 1 y , vx , 1 5.1 , = 16.41°, = tan , 17.32 , , 4., , Similarly,, , , vy(1.5 sec) = 10 – 9.81(1.5) = – 4.72 m/sec, v=, =, , v2x v 2y, , (17.32) 2 ( 4.72)2 = 17.95 m/sec, , v , = tan 1 y = tan 1, vx , , 5., , 4.72 , , = 15.24°, 17.32 , , We know that total time of flight of the ball is given by,, T=, , 2v0 sin 2(10), , = 2.04 sec (v0 sin = 10), g, 9.81
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Engineering Mechanics, 6., , www.aktutor.in, , Maximum height reached by the ball is given by,, h=, , 7., , v20 sin 2 , (10)2, , = 5.1 m, 2g, 2 9.81, , Range of the projectile is given by,, R=, , 8., , 4–21 C (CE-Sem-3), , Maximum range, Rmax =, , Que 4.18., , (20)2 sin 60, v20 sin 2, =, = 35.31 m, 9.81, g, , v20 (20)2, , = 40.77 m, g, 9.81, , (2 = 90°), , Discuss the curvilinear motion of a body in polar, , coordinates., , Answer, 1., , Consider a particle moving in a curvilinear path as shown in Fig. 4.18.1(a)., , 2., , Let it be at a point A at a particular instant of time. Its position is then, , , , , specified by the radial vector r and inclination or r with respect X-axis,, , , i.e., . The instantaneous velocity v of the particle is tangential to the, path at that instant., 3., , This tangential velocity can be resolved into orthogonal components, along the radial and transverse directions., , 4., , For this, let us consider unit vector eˆr and ê along the radial and, transverse directions respectively as shown in Fig. 4.18.1 (a)., , , v, , , v, , ^, e, , , , e^r, , ^, e, , , A, , r, , ^e, r, O, , , vr, , ^, e, , , , ^, e, , ^, er, , , , , , ^, er, , X, (a), , (b ), Fig. 4.18.1., , 5., , As the particle moves from point A to another point in a small interval of, time, we can see that the directions of unit vectors also change. To, determine this change in unit vectors, we proceed as follows.
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www.aktutor.in, , Engineering Mechanics, , Que 4.19., , 4–23 C (CE-Sem-3), , The motion of a particle is defined as r = 2t2 and = t,, , where r is in metres, is in radians and t is in seconds. Determine, the velocity and acceleration of the particle at t = 2 sec., , Answer, Given : r = 2t2, = t, To Find : Velocity and acceleration of the particle at t = 2 sec., 1., , Differentiating the radial and angular displacement functions, we have, , , , , r = 4t, = 1, , , 2., , , , r = 4, = 0, We know that velocity vector is given as,, , , 3., , , , , , v = r eˆr r eˆ, Substituting the values, we have, , , 4., , 2, v = (4 t) eˆr (2t )(1) eˆ, Hence, the velocity at t = 2 sec is obtained by,, , , , 2, v = (4 2) eˆr 2 (2) (1) eˆ, = 8 eˆr 8 eˆ, , , , 5., , |v| = 82 82 = 11.31 m/sec, The acceleration vector is given by,, , , , , , , , , , , a = [ r r()2 ] eˆr [r 2 r ] eˆ, , 6., , = [(4) 2t2 (1)2 ] eˆr [2t 2 (0) 2(4 t) (1)] eˆ, Hence, the acceleration at t = 2 s is obtained by,, , , a = [(4) 2(2)2 ] eˆr [(8)(2)] eˆ = 4 eˆr 16eˆ, , , , | a| =, , 2, (4)2 (16)2 = 16.49 m/sec, , Work, Kinetic Energy, Power, Potential Energy., , Questions-Answers, Long Answer Type and Medium Answer Type Questions
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4–24 C (CE-Sem-3), , Review of Particle Dynamics, , www.aktutor.in, , Que 4.20. Define work done and discuss its special cases., Answer, A., , Work Done : Work done in general is defined as a product of the, component of the force in the direction of motion and the displacement., Mathematically,, W = (F cos )s, where,, F = Force in the direction of motion., s = Displacement., s, F, , Force acting at an angle , Fig. 4.20.1., , B., , Special Cases :, , i., , When the Displacement (s) is Zero :, , 1., , Even though forces may act on a particle, if there is no displacement of, the particle then no work is done on the particle., , 2., , Consider a block resting on a table. In its free-body diagram, we see that, even though its weight W and normal reaction R are acting on it, they do, no work on the block as there is no displacement of the block., W, , R, Fig. 4.20.2., , ii., , When the Motion is at Right Angle to the Direction of the Forces :, , 1., , When the motion is at right angle to the direction of the forces, we see, that = 90° and hence, cos = 0. Thus, work done is zero., , 2., , Consider a block moving along a horizontal plane as shown in, Fig. 4.20.3. Since the displacement is at right angles to the direction of, the forces, namely, its weight and normal reaction, the two forces do not, work on the block., P, , W, P, R, , Fig. 4.20.3., , iii. When the Motion is in the Direction of Force :, 1., , In this case,, , , = 0°, W = Fs, , ( cos 0° = 1)
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Engineering Mechanics, , Que 4.21., , www.aktutor.in, , 4–25 C (CE-Sem-3), , A block of 10 kg mass resting on a rough horizontal, , plane is pulled by an inclined force P as shown Fig. 4.21.1, at a, constant velocity over a distance of 5 m. The coefficient of kinetic, friction between the contact planes is 0.2. Sketch the free body, diagram of the block showing all the forces acting on it. Also,, determine (i) the work done by each force acting on the free body,, and (ii) the total work done on the block., P, 30°, , Fig. 4.21.1., , Answer, Given : m = 10 kg, s = 5 m, = 30°, = 0.2, To Find :, i. Work done by each force acting on the free body., ii. Total work done on the block., 1., , The free-body diagram of the block is shown in Fig. 4.21.2. As there is no, motion along the Y-direction,, , , Fy = 0, R + P sin 30° – 10 g = 0, , , 2., , R = 10 g – P sin 30°, , ...(4.21.1), , Since the block is moving with constant velocity along the horizontal, direction, its acceleration in that direction is zero. Hence, we can write, Fx = 0, P cos 30° – F = 0, P cos 30° – R = 0, , ( F = R), , P sin 30°, 10 g, , ...(4.21.2), , P, , P cos 30°, F, R, Fig. 4.21.2., , 3, , Substituting the value of R from eq. (4.21.1) in eq. (4.21.2), we have, P cos 30° – (10g – P sin 30°) = 0
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4–26 C (CE-Sem-3), P=, , 4., 5., 6., , Review of Particle Dynamics, , www.aktutor.in, , 10 g, 10(0.2)(9.81), =, [cos 30 sin 30] [cos 30 (0.2) sin 30], , = 20.31 N, Force of friction is given by,, F = P cos 30° = 20.31 cos 30° = 17.59 N, Work done by the horizontal component of P, i.e., P cos is,, (W)Pcos = 20.31 cos 30° × 5 = 87.95 J, Work done by the frictional force is given by,, WF = – 17.59 × 5 = – 87.95 J, , 7., , Since the other forces acting on the block, P sin, mg and R are all, perpendicular to the direction of displacement of the block, the work, done by each of them is zero., , 8., , The total work done on the block is the algebraic sum of works done by, each of the forces acting on the block., W = 87.95 – 87.95 = 0, , , , Alternatively, we could say that as the block moving with constant, velocity, the resultant force acting on it is zero, hence, the work done on, the block is zero., , Que 4.22. Define kinetic energy and also derive an expression for, it., , Answer, A., , Kinetic Energy : The energy that a body possesses by the virtue of its, motion is known as kinetic energy., Mathematically,, , KE =, , 1, m v2, 2, , B., , Mathematical Expression for Kinetic Energy :, , 1., , Consider a body of mass m starting from rest. Let it be subjected to an, accelerating force F and after covering a distance s, its velocity, becomes v., Initial velocity,, , u=0, , 2., , Now, work done on the body = Force × Distance, , 3., , But,, , = Fs, , 4., , F = ma, , Substituting the value of F in eq. (4.22.1), we get, Work done = m × (as), , 5., , ...(4.22.1), , Force = Mass × Acceleration, , But from equation of motion, we have, , ...(4.22.2)
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Engineering Mechanics, , www.aktutor.in, , v2 – u2 = 2as, , or, , 4–27 C (CE-Sem-3), , v2 – 02 = 2as, , ( u = 0), , 2, , v, 2, Substituting the value of as in eq. (4.22.2), we have, as =, , 6., , v2, 2, But work done on the body is equal to KE possessed by the body., Work done, , 7., , , , =m, , KE =, , 1, mv2, 2, , Que 4.23. Write a short note on power., Answer, 1., , Power is defined as the rate at which work is done. The capacity of an, engine or a machine used to do work is normally expressed as its rated, power., , 2., , If W is the total work done in a time interval t, then average power is, given by,, Pavg =, , 3., , Total work done W, , T ime taken, t, , ...(4.23.1), , The instantaneous power, i.e., power at a particular instant of time is, given by,, , dW d( Fs), ...(4.23.2), , dt, dt, The force can be assumed to be constant over this infinitesimally small, time interval dt. Hence, we can write the above expression as :, P=, , 4., , Fds, ...(4.23.3), Fv, dt, In SI system of units, the unit of power is Joule per second (J/sec), also, called watt (W)., P=, , 5., , Que 4.24. A car of 2 ton mass starts from rest and accelerates at a, uniform rate to reach a speed of 60 kmph in 20 seconds. If the, frictional resistance is 600 N/ton, determine the driving power of, the engine when it reaches a speed of 60 kmph., , Answer, Given : u = 0, v = 60 kmph = 16.67 m/sec, m = 2 ton = 2000 kg,, f = 600 N/ton, To Find : Power.
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4–28 C (CE-Sem-3), 1., , Review of Particle Dynamics, , www.aktutor.in, , We know that,, v = u + at, a=, , , 2., , v u 16.67 0, , = 0.8335 m/sec2, t, 20, , The kinetic equation of motion of the car is given by,, F – f = ma, where,, , F = Driving force., , , , F = f + ma, , f = Force of friction., = (600)(2) + (2 × 103)(0.8335) = 2867 N, 3., , Driving power of the engine when the car is moving at 60 kmph is given, by,, P = Fv, = (2867)(16.67) = 47792.89 W = 47.8 kW, , Que 4.25., , Define potential energy and also give principle of, , conservation of mechanical energy., , Answer, A., , Potential Energy : It is defined as the capacity to do work by virtue of, its position. There are many types of potential energies such as, gravitational, electrical, elastic, etc., Mathematically,, , PE = mgh, , B., , Principle of Conservation of Mechanical Energy :, , 1., , If a body is subjected to a conservative system of forces, (say gravitational, force) then its mechanical energy remains constant for any position in, the force field., , 2., , Consider a body either sliding down a smooth incline or freely falling., Since it is initially at rest, all of its energy is potential energy., , 3., , As it accelerates downwards, some of its potential energy is converted, into kinetic energy., , 4., , At the bottom of the incline or at the ground level, the energy will be, purely kinetic, assuming the bottom of the slope or the ground level as, the datum for potential energy., , 5., , By the principle of conservation of energy, we see that the loss in potential, energy is equal to the gain in kinetic energy., Mathematically,, (PE)i – (PE)f = (KE)f – (KE)i, , 6., , On rearranging, we have
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Engineering Mechanics, , www.aktutor.in, , 4–29 C (CE-Sem-3), , (PE)i + (KE)i = (PE)f + (KE)f, (PE) + (KE) = Constant, 7., , Thus, we see that the total mechanical energy, i.e., sum of potential and, kinetic energies remain constant. This is known as principle of, conservation of mechanical energy., , Que 4.26. A ball is dropped from the top of a tower. If it reaches the, ground with a velocity of 30 m/sec, determine the height of the tower, by the conservation of energy method., , Answer, Given : v = 30 m/sec, To Find : Height of the tower., 1., , By the principle of conservation of energy, we know that the total, mechanical energy remains constant. Hence, the total energy at the top, of the tower must be equal to that at the base of the tower i.e.,, (KE + PE)top = (KE + PE)base, , 2., , Since the ground surface is taken as the datum, the potential energy at, the top is mgh [where h is height of the tower] and that at the bottom is, zero. If v is the velocity of the ball at the base, we can write, 0 + mgh =, h=, , 1, m v2 0, 2, , v2, (30)2, =, = 45.87 m, 2g, 2(9.81), , Impulse, Momentum (Linear and Angular)., , CONCEPT OUTLINE, Momentum : The product of mass and velocity of a body is known as, momentum. Mathematically, p = mv, Impulse : The product of the force and time is known as impulse., Mathematically, I = Ft, Conservation of Linear Momentum : When no external forces act, on bodies forming a system, the momentum of the system is conserved, i.e., the initial momentum of the system is equal to final momentum of, the system.
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4–30 C (CE-Sem-3), , Review of Particle Dynamics, , www.aktutor.in, Questions-Answers, , Long Answer Type and Medium Answer Type Questions, , Que 4.27. Derive impulse-momentum equation., Answer, 1., , Let,, , F = Net force acting on a rigid body in the direction of motion, through CG of the body., m = Mass of the rigid body., , 2., , We know that,, , a = Acceleration of the body., , F = ma = m, , dv, dt, , d v, , a , , , dt , , Fdt = mdv, 3., , Integrating the above equation, we get, , , , t2, , t1, , Fdt =, , , , v2, , v1, , mdv, , = m(v2 – v1), Impulse = mv2 – mv1, Impulse = Final momentum – Initial momentum, , Que 4.28., , A football of mass 200 gm is at rest. A player kicks the, , balls which move with a velocity of 20 m/sec at an angle of 30° with, respect to ground level. Find the force exerted by the player on the, ball of duration of strikes is 0.02 seconds., , Answer, Given : m = 200 gm = 0.2 kg , t = 0.02 sec, = 30°, To Find : Force exerted on the ball., 1., , Initially the ball is at rest. Hence, ux = 0 and uy = 0., , 2., , The ball leaves with a velocity of 20 m/sec at an angle of 30°, (Fig. 4.28.1(b))., , 3., , Writing impulse-momentum equation along X- and Y-directions, we get, i., , For X-direction,, Fxt = m (vx – ux),, Fx × 0.02 = 0.2 (20 cos 30° – 0)
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www.aktutor.in, , Engineering Mechanics, , Fx =, , 4–31 C (CE-Sem-3), , 0.2 20 cos 30, = 173.2 N, 0.02, vy, , uy, , 20 m/sec, vx, ux, , 30°, (b) Final position of ball., Ball moves with a velocity of 20 m/sec, (vx = 20 cos 30°, vy= 20 sin 30°), , ( a) Initial position of ball., Ball at rest, (ux = 0, uy = 0), , Fig. 4.28.1., , ii., , For Y-direction,, Fyt = m (vy – uy), Fy × 0.02 = 0.2 (20 sin 30° – 0), , 0.2 20 sin 30, = 100 N, 0.02, Hence, the resultant impulse force exerted by the player on the ball,, Fy =, , 4., , F=, , Que 4.29., , Fx2 Fy2 (173.2) 2 1002 = 199.99 200 N, , A bullet of mass 50 gm is fired into a freely suspended, , target to mass 5 kg. On impact, the target moves with a velocity of, 7 m/sec along with the bullet in the direction of firing. Find the, velocity of bullet., , Answer, Given : m1 = 50 gm = 0.05 kg, m2 = 5 kg, u2 = 0, m = 5 + 0.05 = 5.05 kg,, v = 7 m/sec, To Find : Velocity of bullet., 1., , Total initial momentum (i.e., momentum before impact),, = m1u1 + m2u2 = 0.05 × u1 + 5 × 0, = 0.05 u1, , 2., , Total final momentum (i.e., momentum after impact),, = Total mass × Common velocity = mv, = (5.05) × 7, , 3., , According to conservation of momentum,
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4–32 C (CE-Sem-3), , Review of Particle Dynamics, , www.aktutor.in, , Initial momentum = Final momentum, 0.05 u1 = 0.05 × 7, u1 =, , , , Que 4.30., , 5.05 7, = 707 m/sec, 0.05, , Derive an expression for angular momentum., , Answer, 1., 2., , The product of mass moment of inertia and angular velocity of a rotating, body is known as moment of momentum or angular momentum., If, = Angular velocity of a body rotating about an axis., I = Moment of inertia of the body about the axis., Then, angular momentum = I, , 3., , Consider a body of mass ‘m’ rotating in a circle about its centre O., , 4., , Let,, , dm = Mass of the elementary strip., r = Radius of the mass dm., = Angular velocity of the body or angular velocity of the, mass dm., v = Linear velocity of mass dm., Y, , R, , dr, r, X, , O, , X, , Y, Fig. 4.30.1., , 5., , Now momentum of elementary mass, = Elementary mass × Velocity = dm × v, = dm × r, , 6., , ( v = r), , Moment of momentum of elementary mass dm about O, = Elementary mass × Radius, = (dm × r) × r, = dm × r2, , ...(4.30.1)
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Engineering Mechanics, 7., , www.aktutor.in, , The moment of momentum of the entire mass about O is obtained by, integrating eq. (4.30.1)., Moment of momentum of the entire mass, =, But, , 8., , 4–33 C (CE-Sem-3), , r, , 2, , dm × r, , 2, , = r2dm, , ...(4.30.2), , dm = Moment of inertia of the whole body about O = I., , Substituting the value in eq. (4.30.2), we get, Moment of momentum of the entire mass = I, , Que 4.31. At a given instant the 5 kg slender bar has the motion, shown in Fig. 4.31.1. Determine the angular momentum about point, , AKTU 2014-15, (I) Marks 10, , G (vA = 2 m/sec)., , A, 4m, G, 30º, , B, , O, , Fig. 4.31.1., , Answer, Given : m = 5 kg, vA = 2 m/sec, L = 4 m, To Find : Angular momentum about point G., 1., , In OBA,, , AB = 4 m, OA = 4 sin 30° = 2 m, OB = 4 cos 30° = 3.464 m, A, , O, , 2m, 4m, , 60º, , vA = 2 m/sec, , C, 2m, , B, vB, , 2., , 30º, Fig. 4.31.2., , Velocity of point A, vA = 2 m/sec, , 90º, , O
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4–34 C (CE-Sem-3), , Review of Particle Dynamics, , www.aktutor.in, , vA, 2, =, = 1 rad/sec, OA 2, v, Velocity at point B, AB = B, OB, vB, 1=, 3.464, vB = 3.464 rad/sec, AB =, , 3., , 4., , Angular momentum about G, = I =, , ML2, ×1, 12, , , ML2 , I , , 12 , , 2, = 5 4 × 1 = 6.67 rad/sec2, 12, , Impact (Direct and Oblique)., , CONCEPT OUTLINE, Direct Impact : During collision, when the direction of motion of, each body is along the line joining their centres, the impact is called, direct impact, Oblique Impact : During collision, when the direction of motion of, either one or both bodies is inclined to the line joining their centres,, the impact is called oblique impact., , Questions-Answers, Long Answer Type and Medium Answer Type Questions, , Que 4.32. Derive an expression for the final velocities of the body, during direct impact., , Answer, 1., , Consider two smooth spheres of masses m1 and m2 moving with initial, velocities u1 and u2 respectively., , 2., , Let them collide with each other along the line joining their centres and, let v1 and v2 be their respective velocities after collision.
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4–36 C (CE-Sem-3), , Review of Particle Dynamics, , www.aktutor.in, , Que 4.33. If a ball overtakes a ball of twice its mass moving 1/7th of, its velocity and if the coefficient of restitution between them is 3/4,, show that the first ball after striking the second ball will remain at, rest., , Answer, Given : m1 = m, m2 = 2 m, u1 = u, u2 = u/7, e = 3/4, To Prove : First ball after striking the second ball will remain at rest i.e.,, v1 = 0, 1., , It is given that the velocity of the second ball is 1/7th of the velocity of the, first ball. Hence, applying the conservation of momentum equation,, u/7, u, m, , 2m, , Before impact, Fig. 4.33.1., , m1 u1 + m2 u2 = m1 v1 + m2 v2, mu + 2m, , u, = mv1 + 2mv2, 7, , 9u, 7, Coefficient of restitution is given as,, v1 + 2v2 =, , 2., , –e=, , v1 v2, v v2, = 1, u1 u2, uu/7, , v1 – v2 = –, =–, 3., , ...(4.33.1), , 6, eu, 7, , 6 3, 9, u, u, 7 4 , 14, , From eq. (4.33.1) and eq. (4.33.2) solving for v1, we get, v1 = 0, , Que 4.34. Discuss in brief about oblique impact., , ...(4.33.2)
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Engineering Mechanics, , www.aktutor.in, , 4–37 C (CE-Sem-3), , Answer, 1., , Consider two smooth spheres of masses m1 and m2 approaching each, other with velocities u1 and u2 such that their directions are inclined to, the line joining their centres at the instant of impact at and respectively., , 2., , Let v1 and v2 be the respective velocities immediately after impact and, their directions be inclined to the line joining centres at and , respectively as shown in Fig. 4.34.1., u2, , u1, , m1, , , , m2, , , , , v1, , v2, , Fig. 4.34.1. Oblique impact., , 3., , As the spheres are smooth, there is no impulsive force acting on each, body along their common tangential plane during their time of collision, thus, there is no change in momentum of individual bodies in that, direction., , 4., , Hence, we can write,, v1 sin = u1 sin , and, , v2 sin = u2 sin , , 5., , As the impulsive force exerted by each sphere on the other in the, direction of line joining their centres is equal and opposite, the momentum, of the system is conserved. Thus, we can write, m1(u1 cos ) + m2(u2 cos ) = m1(v1 cos ) + m2(v2 cos ), , 6., , We know that,, –e=, , Que 4.35., , v1 cos v2 cos , u1 cos u2 cos , , A smooth sphere moving at 10 m/sec in the direction, , shown in Fig. 4.35.1 collides with another smooth sphere of double, its mass and moving with 5 m/sec in the direction shown. If the, coefficient of restitution is 2/3, determine their velocities after, collision., 5 m/sec, , 10 m/sec, 30°, , 60°, m, , 2m, , Fig. 4.35.1.
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4–38 C (CE-Sem-3), , Review of Particle Dynamics, , www.aktutor.in, , Answer, Given : m1 = m, m2 = 2 m, u1 = 10 m/sec, u2 = 5 m/sec, e = 2/3, = 30°,, = 60°, To Find : Velocities after collision., 1., , 2., , We know that,, v1 sin = u1 sin = 10 sin 30°, , ...(4.35.1), , v2 sin = u2 sin = 5 sin 60°, , ...(4.35.2), , According to conservation of momentum,, m1 (u1 cos ) + m2 (u2 cos ) = m1 (v1 cos ) + m2 (v2 cos ), m(10 cos 30°) – 2m(5 cos 60°) = m(1 cos ) + 2m(2 cos ), v1 cos + 22 cos = 3.66, , 3., , ...(4.35.3), , Also, we know that,, –e=, , v1 cos v2 cos , 10 cos 30 ( 5 cos 60), , v1 cos – v2 cos = – 7.44, 4., , ( e = 2 / 3), ...(4.35.4), , From eq. (4.35.3) and eq. (4.35.4) solving for v1 cos and v2 cos , we get, and, , v1 cos = – 3.74 m/sec, v2 cos = 3.7 m/sec, , ...(4.35.5), ...(4.35.6), , 5., , From eq. (4.35.1) and eq. (4.35.5), we get v1 = 6.24 m/sec in the direction, opposite to that of the initial velocity at an angle of = 53.2° to the line, joining their centres., , 6., , Similarly, from eq. (4.35.2) and eq. (4.35.6), we get v2 = 5.7 m/sec at an, angle of = 49.49° of the line joining their centres., ,
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Engineering Mechanics, , 5, , www.aktutor.in, , 5–1 C (CE-Sem-3), , Introduction to, Kinetics of Rigid, Bodies, , CONTENTS, Part-1, , :, , Introduction to Kinetics of ........................ 5–2C to 5–6C, Rigid Bodies, Basic Terms,, General Principles in Dynamics,, Types of Motion, , Part-2, , :, , Instantaneous Centre of ......................... 5–6C to 5–10C, Rotation in Plane Motion, and Simple Problems, , Part-3, , :, , D’Alembert’s Principle and its .............. 5–10C to 5–17C, Applications in Plane Motion, and Connected Bodies, , Part-4, , :, , Work-Energy Principle and .................. 5–17C to 5–21C, its Application in Plane Motion, of Connected Bodies, , Part-5, , :, , Kinetics of Rigid Body Rotation ........... 5–21C to 5–26C, , Part-6, , :, , Virtual, Virtual, Virtual, System, , Part-7, , :, , Applications of Energy Method ............ 5–31C to 5–34C, for Equilibrium, , Part-8, , :, , Stability of Equilibrium ......................... 5–34C to 5–34C, , Work and Energy Method, ..... 5–26C to 5–31C, Displacement, Principle of, Work for Particle and Ideal, of Rigid Bodies
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5–2 C (CE-Sem-3), , Introduction to Kinetics of Rigid Bodies, , www.aktutor.in, , Introduction to Kinetics of Rigid Bodies, Basic Terms, General, Principles in Dynamics, Types of Motion., , CONCEPT OUTLINE, Kinetics : It is that branch of engineering mechanics which deals with, the force system which produces acceleration and resulting motion of, bodies., Newton’s Laws of Motion : When a body is at rest or moving in a, straight line or rotating about an axis, the body obeys certain laws of, motion. These laws are called Newton’s law of motion., , Questions-Answers, Long Answer Type and Medium Answer Type Questions, , Que 5.1., , Discuss the various terminologies related with kinetics, , of rigid body., , Answer, Following are the some terminologies related with the kinetics of rigid, body :, i., , Force : It is defined as an agent which tends to change the state of rest or, motion of a body to which it is applied. The SI unit of force is Newton (N)., , ii., , Mass : The quantity of matter combined in a body is known as the mass, of the body. Mass is a scalar quantity. The SI unit of mass is kilogram (kg)., , iii. Acceleration : It is defined as the rate of change of velocity of a body., Its SI unit is m/sec2., , Change of velocity d v, , Time, dt, iv. Weight : Weight of a body is defined as the force by which the body is, attracted towards the centre of the earth. Mathematically weight of a, body is given by,, , , Acceleration =, , Weight = Mass × Acceleration due to gravity = mg, v., , Momentum : The product of the mass of a body and its velocity is, known as momentum of the body. Momentum is a vector quantity., Mathematically, momentum is given by,, Momentum = Mass × Velocity = mv
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Engineering Mechanics, Que 5.2., , www.aktutor.in, , 5–3 C (CE-Sem-3), , State the various Newton’s law of motion., , Answer, Various laws of motion are as follows :, i., , Newton’s First Law of Motion : It states that a body continues in its, state of rest or of uniform motion in a straight line unless it is compelled, by an external force to change that state., , ii., , Newton’s Second Law of Motion : It states that the rate of change of, momentum of a body is proportional to the external force applied on the, body and takes place in the direction of the force., , iii. Newton’s Third Law of Motion : It states that to every action, there, is always an equal and opposite reaction., Que 5.3., , Discuss in detail about Newton’s second law of motion., , Answer, 1., , Newton’s second law of motion enables us to measure a force., , 2., , Let a body of mass m is moving with a velocity u along a straight line. It, is acted upon by a force F and the velocity of the body becomes v in the, time t., , 3., , Initial momentum of the body = Mass × Initial velocity = mu, Final momentum of the body = mv, Change in momentum = Final momentum – Initial momentum, = mv – mu = m(v – u), , 4., , Rate of change of momentum =, , 5., , But we know that,, , Change of momentum m(v u), , Time, t, ...(5.3.1.), , vu, =a, t, 6., , 7., , (i.e., linear acceleration), , v u, Substituting the value of , in eq. (5.3.1), we get, t , Rate of change of momentum = ma, But according to Newton’s second law of motion, the rate of change of, momentum is directly proportional to the external force acting on the, body., , , F ma or F = kma, , where,, , k = Constant of proportionality., , Que 5.4., , Discuss the various types of plane motion., , ...(5.3.2.)
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5–4 C (CE-Sem-3), , Introduction to Kinetics of Rigid Bodies, , www.aktutor.in, , Answer, Generally a body undergoes the following three types of plane motion :, i., , Translation :, , 1., , During translation, the particles have the same velocity and acceleration,, and a straight line drawn on the moving body remains parallel to its, original position at any time., Path, , Path, A, , A, , A, , B, , B, , A, , B, , B, , ( a) Rectilinear translation, , (b) Curvilinear translation, Fig. 5.4.1., , 2., , If path traced by the particles during motion is a straight line, then the, motion is said to be rectilinear translation (Fig. 5.4.1(a))., , 3., , If particle traces a curved path, the motion is called curvilinear translation, (Fig. 5.4.1(b))., , ii., , Rotation :, , 1., , During rotation, the body rotates about a fixed point and all the particles, constituting the body move in a circular path., , 2., , The fixed point about which the body rotates is called the point of rotation, and the axis passing through the point of rotation is called the axis of, rotation., B, , B, , B, , A, , A, B, , A, , A, O, (a), , O, (b ), Fig. 5.4.2.
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Engineering Mechanics, , www.aktutor.in, , 5–5 C (CE-Sem-3), , iii. General Plane Motion (Combined Motion of Translation and, Rotation) :, 1., , Whe n a body posse ss bo th translation and rotation motions, simultaneously at a particular instant, the motion is called general plane, motion., Example : (i) Motion of roller without slipping, motion of wheel of a, locomotive train, truck and car etc., (ii) A rod sliding against a wall at one, end and floor at the other end., , Que 5.5., , A particle of mass 1 kg moves in a straight line under, , the influence of a force which increases linearly with time at the, rate of 60 N per sec. At time t = 0, the initial force may be taken, as 50 N. Determine the acceleration and velocity of the particle, 4 sec after it started from the rest at the origin., , Answer, dF, = 60 N/sec, At t = 0, F = 50 N, t = 4 sec, dt, i. Velocity, , Given : m = 1 kg,, To Find :, , ii. Acceleration, 1., , Force is increasing linearly with time. Hence applied force on the particle, is a function of time., Let,, , F = At + B, , ...(5.5.1), , where, A and B are constant., 2., , When t = 0, F = 50 N. Now eq. (5.5.1) becomes,, 50 = A × 0 + B = B, , , 3., , Differentiating eq. (5.5.1), we get, , But, , 4., , B = 50 N, , dF, =A+0, dt, dF, = 60 N/sec, dt, A = 60 N/s, , Substituting the value A and B in eq. (5.5.1), we get, F = 60t + 50, , 5., , We know that,, , F = ma = m, , ...(5.5.2), , dv, dt, , Substituting this value of F in eq. (5.5.2), we get, , d v, , a , , dt
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5–6 C (CE-Sem-3), , 6., , Introduction to Kinetics of Rigid Bodies, , www.aktutor.in, , m×, , dv, = 60t + 50, dt, , ...(5.5.3), , 1×, , dv, = 60t + 50, dt, , ( m = 1 kg), , dv, = 60t + 50, dt, , ...(5.5.4), , Integrating the eq. (5.5.4) w.r.t time, we get, , dv, , =, , v=, , (60t 50) dt, , , , 4, , 0, , (60 t 50) dt, 4, , 7., , 8., , 60 t2, , 50 t 30 4 2 50 4 480 200, v= , 2, 0, = 680 m/sec, From eq. (5.5.3), we have, dv, = 60t + 50, dt, dv, , a, , a = 60t + 50, , dt, , Acceleration after 4 sec, a = 60 × 4 + 50 = 290 m/sec2, , Instantaneous Centre of Rotation in Plane Motion and Simple, Problems., Questions-Answers, Long Answer Type and Medium Answer Type Questions, , Que 5.6., , Define instantaneous centre of rotation and also write, , the procedure for locating the position of instantaneous centre of, rotation., , Answer, A., , Instantaneous Centre of Rotation :, , 1., , Instantaneous centre is the point about which motion of a body having, both rotatory and translatory motion is assumed to be purely rotational., It is also known as virtual centre.
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Engineering Mechanics, 2., , www.aktutor.in, , 5–7 C (CE-Sem-3), , The angular velocity of any point about instantaneous centre is given, by,, , v, I, = Angular velocity., =, , where,, , v = Linear velocity., I = Instantaneous centre., B., , Locating the Position of Instantaneous Centre of Rotation :, , 1., , If the directions of the velocities of two particles P and Q of the body are, known and if they are different, the instantaneous centre is obtained by, drawing the perpendicular to vP through P and perpendicular to vQ, through Q. The intersection point of these two perpendiculars is known, as instantaneous centre of rotation., , 2., , If the velocities vP and vQ of two particles P and Q are perpendicular to, the line PQ and the magnitudes of vP and vQ are known, the instantaneous, centre of rotation can be found by intersection point of line PQ with the, line joining the extremities of the vectors vP and vQ., , 3., , If the velocities vP and vQ are parallel and have different magnitude or if, the velocities vP and vQ are perpendicular to line PQ and have equal, magnitude, the instantaneous centre O will be at an infinite distance, and will be zero and all the points of the body will have the same, velocity., , Fig. 5.6.1., , Que 5.7., , A compound wheel rolls without slipping between two, , parallel plates A and B as shown in Fig. 5.7.1. At the instant, A moves to the right with a velocity of 1.2 m/sec and B moves to the, left with a velocity of 0.6 m/sec. Calculate the velocity of centre of, wheel and the angular velocity of wheel. Take r1 = 120 mm and, r2 = 360 mm.
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5–8 C (CE-Sem-3), , Introduction to Kinetics of Rigid Bodies, , www.aktutor.in, A, , vA, D, , r1, , r2, , 30º, , B, , vB, , Fig. 5.7.1., , AKTU 2015-16, (I) Marks 10, Answer, Given : vA = 1.2 m/sec, vB = 0.6 m/sec, r1 = 120 mm = 0.12 m,, r2 = 360 mm = 0.36 m, To Find : Velocity of centre of wheel and angular velocity of wheel., 1., , The instantaneous centre I is the point of intersection of the line joining, A and B with line joining the extremities of the velocity vectors vA and, vB as shown in Fig. 5.7.2., vA = 1.2 m/sec, , A, , (r1 + r2 – x), , I, vC, , x, , C, B, vB = 0.6 m/sec, , Fig. 5.7.2., , 2., , Every vector on the wheel will appear to rotate about the instantaneous, centre I with an angular velocity ., , v A vB, , IA IB, v, vA, = B, x, (r1 r2 x), =, ,
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Engineering Mechanics, , www.aktutor.in, , 5–9 C (CE-Sem-3), , 0.6, 1.2, =, x, (480 x), 3., , ...(5.7.1), , Solving eq. (5.7.1), we get, x = 160 mm, IB = 160 mm, , , Also, , IA + IB = 480, IA = 480 – IB = 480 – 160 = 320 mm, , , 4., , Now angular velocity of the disc,, =, , 5., , From Fig. 5.7.2,, , vA, 1.2, , = 3.75 rad/sec, IA 320 , , , 1000, , IC = x – CB, = 160 – 120 = 40 mm, , 6., , Velocity of the centre C,, vC = IC = 3.75 ×, , 40, 1000, , vC = 0.15 m/sec, , Que 5.8., , A slender bar AB slides down a circular surface and on a, , horizontal surface as shown in Fig. 5.8.1. At an instant, when, = 45°, velocity of the end A is 2 m/sec. Determine the angular velocity, of the bar and the velocity of point of contact on the circular surface., B, , 20 cm, , , A, , 2 m/sec, , Fig. 5.8.1., , AKTU 2012-13, Marks 10, Answer, Given : = 45°, vA = 2 m/sec, To Find : Angular velocity of the bar and velocity of point of contact, on the circular surface.
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5–10 C (CE-Sem-3), , Introduction to Kinetics of Rigid Bodies, , www.aktutor.in, , 1., , Instantaneous centre I is obtained by drawing perpendicular on vA and, vC., , 2., , Now,, , vC = vA ×, , IC, = vA × cos 45°, IA, , B, I, 45°, , C, vC, , 20 cm, , 45°, , A, Fig. 5.8.2., , O, , vC = 2 ×, , , 3., , 1, , 2, vA = o × IA, , Also,, , v A = 2 m/sec, , = 1.414 m/sec, , vA, IA, , o =, From ICA,, , IA = 20 2 cm = 28.28 cm = 0.2828 m, ( CA = 20 cm), , Hence, , o =, , 2, = 7.072 rad/sec, 0.2828, , D’Alembert’s Principle and its Applications in Plane, Motion and Connected Bodies., , CONCEPT OUTLINE, D’Alembert’s Principle : It states that the net external forces acting, on the system and the resultant inertia force are in equilibrium., Mathematically, F – ma = 0, where,, , F = External force., ma = Resulting inertia force., , Questions-Answers, Long Answer Type and Medium Answer Type Questions
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Engineering Mechanics, , Que 5.9., , www.aktutor.in, , 5–11 C (CE-Sem-3), , Illustrate D’Alembert’s principle with respect to, , connected bodies., , Answer, Following cases can be considered for illustrating D’ Alembert’s principle :, a., , Motion of a Lift :, , 1., , Let the tension in the string be T, acceleration of the lift be a and W be, the weight of lift plus persons in the lift., , 2., , Considering upward motion of the lift (Fig. 5.9.1 (a))., T W , , W, a =0, g, , a, T = W 1 , g, , T = m (g + a), where,, 3., , ( W = mg), , m = Mass equivalent of weight W., , Considering downward motion of the lift (Fig. 5.9.1 (a)), W T , , W, a =0, g, , a, T = W 1 , g, T = m (g – a), T, , a, , a, , W, ( a), , T, , W, ( b), Fig. 5.9.1., , b., , Motion of Two Connecting Weights over a Smooth Pulley :, , 1., , Let m1 > m2 and the acceleration of the system be a, m1 obviously, moving downwards. According to D’Alembert’s principle,, For block of mass m1,, m1 g – T = m1 a, , ...(5.9.1), , For block of mass m2,, T – m2 g = m2 a, 2., , Adding eq. (5.9.1) and eq. (5.9.2), we get, , ...(5.9.2)
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5–12 C (CE-Sem-3), , Introduction to Kinetics of Rigid Bodies, , www.aktutor.in, a=, , 3., , (m1 m2 ) g, (m1 m2 ), , Subtracting eq. (5.9.1) from eq. (5.9.2), we get, T=, , 2 gm1 m2, m1 m2, , R, T, T, m2, , T, T, m1, , Fig. 5.9.2., , c., , Motion of Two Interconnected Bodies on an Inclined Plane :, This motion can be divided in two cases., Case I :, , 1., , Let two bodies A and B be joined by any inextensible string and the, composite system moved down the inclined with a common acceleration, B, T, T, , A, , sin, W1, , , sin, , W2, , , , W, , , , W, , s, co, , 2, , s, co, 1, , , W1, , W2, Fig. 5.9.3., , For block B,, T W2 sin W2 cos , , W2, a= 0, g, , ...(5.9.3), , W1, a=0, g, , ...(5.9.4), , For block A,, W1 sin T W1 cos , , 2., , From eq. (5.9.3) and eq. (5.9.4), we get, , (W1 W2 ) sin (W1 W2 ) cos , , a, (W1 W2 ) = 0, g
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Engineering Mechanics, , www.aktutor.in, , sin – cos –, , 5–13 C (CE-Sem-3), , a, =0, g, , sin ( ), a, =, cos , g, , 3., , where,, = Angle of friction., If the coefficients of friction are different for A and B, i.e., 1 and 2, then, T W2 sin 2W2 cos , , W2 a, =0, g, , and W1 sin T 1W1 cos , 4., , ( = tan ), , W1 a, =0, g, , Which on simplification gives, a, (W1 W2 ) sin (1W1 2W2 ) cos , =, g, W1 W2, , and, , T=, , W1W2 ( 2 1 cos ), W1 W2, , Case II :, 1., , Motion of two connected masses, one of which moves on the inclined, plane, while the other falls freely being connected to the former by a, string running over a pulley., T, , T, , a, m2, , m1, , m2g, , a, , m1g, , Fig. 5.9.4., , 2., , Let the two masses accelerate with acceleration a in the direction of m1, as shown in Fig. 5.9.4. Considering no friction,, For block of mass m1,, m1 g – T – m1 a = 0, , ...(5.9.5), , For block of mass m2,, m2 g sin – T – m2 a = 0, 3., , ...(5.9.6), , From eq. (5.9.5) and eq. (5.9.6), we have, a=, , g( m1 m2 sin ), m1 m2, , T=, , 2 gm1 m2 sin , m1 m2
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5–14 C (CE-Sem-3), , Introduction to Kinetics of Rigid Bodies, , www.aktutor.in, , d., , Motion of Two Connected Bodies One on each of the Two Smooth, Inclined Planes :, , 1., , Let the motion be on the m2 side of the body as shown in Fig. 5.9.5., , 2., , Then by D’Alembert’s principle,, For block of mass m2,, m2 g sin 2 – T – m2 a = 0, , ...(5.9.7), , For block of mass m1,, T – m1 g sin 1 – m1 a = 0, 3., , ...(5.9.8), , From eq. (5.9.7.) and eq. (5.7.8), we get, a=, , g(m2 sin 2 m1 sin 1 ), m2 m1, , T=, , m1 m2 (sin 1 sin 2 ) g, m2 m1, , T, , T, , m1, co, s, , 1, , m2, , a, , co 2, s, , , 2, , m, , m1g sin 1, , 2g, , 1g, , 1, , m, , 2, , 1, m1g, , m2g sin 2, , m2g, , Fig. 5.9.5., , Que 5.10. Two bodies of masses 80 kg and 20 kg are connected by, a thread along a rough horizontal surface under the action of a, force 400 N applied to the first body of mass 80 kg as shown in, Fig 5.10.1. The coefficient of friction between the sliding surfaces of, the bodies and plane is 0.3. Determine the acceleration of two bodies, and tension in the thread using D’Alembert’s principle., 1, , 2, A, , 20 kg, , B, , 80 kg, 400 N, , Fig. 5.10.1., , AKTU 2014-15, (II) Marks 10
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Engineering Mechanics, , www.aktutor.in, , 5–15 C (CE-Sem-3), , Answer, Given : m1 = 80 kg, W1 = 80 × 9.81 = 784.8 N, m2 = 20 kg, W2 = 20 × 9.81, = 196.2 N, F = 400 N, = 0.3, To Find : i. Acceleration of two bodies, ii. Tensions in the thread., 1., 2., , Let us consider, both the blocks are moving with acceleration a and, tension developed in thread is T., Considering FBD of Block 1 (Fig. 5.10.2 (a)), Using D’Alembert’s principle,, 400 – T – R = 80 a, 400 – T – 0.3 × 784.8 = 80 a, 164.56 – T = 80 a, , 3., , ...(5.10.1), , Considering FBD of Block 2 (Fig. 5.10.2(b)), Using D’Alembert’s principle,, T – R = 20 a, T – 0.3 × 196.2 = 20 a, T – 58.86 = 20 a, , 4., , ...(5.10.2), , On solving the eq. (5.10.1) and eq. (5.10.2), we get, a = 1.057 m/sec2, and, , T = 80 N, 1, , R, 2, , 80 a, , 20 a, T, , 400 N, , T, , R, 196.2 N, , 784.8 N, , R, (a), , (b ), Fig. 5.10.2., , Que 5.11., , A system of weight connected by string passing over, , pulleys A and B shown in Fig. 5.11.1. Find the acceleration of three, weights. Assuming string is weightless and ideal condition for, pulleys.
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5–16 C (CE-Sem-3), , Introduction to Kinetics of Rigid Bodies, , www.aktutor.in, , Pulley, A, T, 15 kN, , Pulley, B, T2, , T2, , 4 kN, , 6 kN, Fig. 5.11.1., , AKTU 2014-15, (II) Marks 10, Answer, Given : Fig. 5.11.1., To Find : Acceleration of three weight., 1., , Considering FBD for block 4 kN (Fig. 5.11.2), T2, 4000, 9.81, , a1, , 4 kN, Fig. 5.11.2., , T2 – 4000 =, 2., , 4000, a1, 9.81, , ...(5.11.1), , Considering FBD for block 6 kN (Fig. 5.11.3), 6000 – T2 =, , 6000, a1, 9.81, , ...(5.11.2)
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Engineering Mechanics, , www.aktutor.in, , 5–17 C (CE-Sem-3), , T2, 6000, 9.81, , a1, , 6 kN, Fig. 5.11.3., , 3., , From eq. (5.11.1) and eq. (5.11.2), we get, T2 = 4800 N, , 4., , ,, , a1 = 1.962 m/sec2, , Considering FBD for pulley A,, T = 2 T2, T = 2T2 =, 2 × 4800 =, , 15000, a2 + 15000, 9.81, , 15000, a2 + 15000, 9.81, , 15000, a2, 9.81, a2 = – 3.5316 m/sec2, , – 5400 =, , T = 2 T2, , 15000, 9.81, , a2, , 15 kN, Fig. 5.11.4., , Negative sign of accelerations indicates that the direction is opposite to, the direction as shown in Fig. 5.11.4., , Work-Energy Principle and its Application in Plane Motion, of Connected Bodies., , Questions-Answers, Long Answer Type and Medium Answer Type Questions, , Que 5.12. State and prove work-energy principle.
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5–18 C (CE-Sem-3), , Introduction to Kinetics of Rigid Bodies, , www.aktutor.in, , Answer, A., , Statement : Work-energy principle states that the change in kinetic, energy of a body during any displacement is equal to the work done by, the net force acting on the body or we can say that work done is equal to, change in kinetic energy of the body., , B., , Proof :, , 1., , We know that,, , F = ma, , where,, , F = Resultant of all forces acting on a body., , ...(5.12.1), , m = Mass of the body., a = Acceleration in the direction of resultant force., , dv, ds, Substituting the value of a in eq. (5.12.1), we get, a= v, , 2., , d v, F = m v, or F ds = mv dv, ds , , ...(5.12.2), , 3., , But F ds is the work done by the resultant force F in displacing the body, by a small distance ds. The total work done by the resultant force F in, displacing the body by a distance s is obtained by integrating the, eq. (5.12.2)., , 4., , Hence, integrating eq. (5.12.2) on both sides, we get, , , , s, , 0, , F ds =, , , , v, , u, , m v dv, v, , , , v2 , m, m v 2 mu2, F s = m [v 2 u2 ] , , 2 u 2, 2, 2, , Work done by resultant force = Change in kinetic energy, , Que 5.13. A body of mass 30 kg is projected up an incline of 30°, with an initial velocity of 10 m/sec. The friction coefficient between, the contacting surfaces is 0.2. Determine distance travelled by the, body before coming to rest., , AKTU 2013-14, (II) Marks 05, , Answer, Given : m = 30 kg, u = 10 m/sec, v = 0 (rest), = 0.2, To Find : Distance travelled by the body before coming to rest.
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Engineering Mechanics, , www.aktutor.in, R, , mg sin , , 30°, , 5–19 C (CE-Sem-3), R, , mg cos , , mg, , Fig. 5.13.1., , 1., , Resultant force acting on the block,, F = mg sin – R, = mg sin – mg cos , = 30 × 10 × sin 30° – 0.2 × 30 × 10 cos 30°, F = 98.04 N, , 2., , Using the work-energy balance equation,, Work done by the block = Kinetic energy of the block, Fx =, 98.04 × x =, , 1, m(u2 – v2 ), 2, 1, × 30 [102 – 02], 2, , x = 15.30 m, , Que 5.14. The speed of a flywheel rotating at 200 rpm is uniformly, increased to 300 rpm in 5 seconds. Determine the work done by the, driving torque and the increase in kinetic energy during this time., Take mass of the flywheel as 25 kg and its radius of gyration as, 20 cm., , Answer, 2 200, = 6.67 rad/sec,, 60, t = 5 sec, m = 25 kg, k = 20 cm = 0.2 m, N = 300 rpm,, , Given : N0 = 200 rpm, 0 =, , 2 300, = 10 rad/sec, 60, To Find : i. Work done by the driving torque., =, , ii. Increase in kinetic energy., 1., , Mass moment of inertia of the flywheel about its centroidal axis is,, I = mk2 = (25)(0.2)2 = 1 kg m2
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5–20 C (CE-Sem-3), 2., , Introduction to Kinetics of Rigid Bodies, , www.aktutor.in, , Since the angular acceleration is uniform, we can use the kinematic, equation,, = 0 + t, , 0, t, 10 6.67 , =, 2.09 rad/ sec 2, 5, , =, , 3., , Also we know that,, 2 = 02 + 2, =, , 2 02, 2, , 4., , (10)2 (6.67 ) 2, 131.07 rad, 2(2.09), Since the angular acceleration is constant, the driving torque is constant, and hence applying the kinetic equation of motion about fixed axis, we, have, , 5., , Work done by the driving torque is given by,, , =, , M = I= (1)(2.09) = 2.09 N-m, W = M(2 – 1), = (2.09)(131.07) = 273.94 J, 6., , The increase in kinetic energy is given by,, (KE) = (KE)f – (KE)i, =, , 1 1 2, I I 0, 2, 2, , =, , 1, I ( 2 20 ), 2, , =, , 1, (1)[(10 )2 (6.67 )2 ] 273.94 J, 2, , Que 5.15. A constant force of 100 N is applied as shown tangentially, on a cylinder at rest, whose mass is 50 kg and radius is 10 cm, for a, distance of 5 m. Determine the angular velocity of the cylinder and, the velocity of its centre of mass. Assume that there is no slip., , Answer, Given : F = 100 N, m = 50 kg, r = 10 cm = 0.1 m, s = 5 m, To Find :, , i. Angular velocity of the cylinder., ii. Velocity of centre of mass.
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Engineering Mechanics, 1., , www.aktutor.in, , 5–21 C (CE-Sem-3), , Since the applied force is horizontal and the displacement is in the, direction of the force, the work done by the force in causing a, displacement s is given by,, W = Fs, 100 N, 10 cm, , Fig. 5.15.1., , 2., , Applying the work-energy principle, we have, Work done = Change in kinetic energy, , , 3., , Fs =, , 1, 1, m v 2 I 2, 2, 2, , Fs =, , 1, 1 mr 2 2, mr 2 2 , , 2, 2 2, , Fs =, , 3, mr 2 2, 4, , 2 =, , 4 Fs, 4(100)(5), , 1333.33, 3mr 2 3(50)(0.1)2, , , mr 2 , v r, I , , 2 , , = 36.51 rad/sec, , Velocity of the centre of mass is given as,, vcm = r, = (0.1)(36.51) = 3.651 m/sec, , Kinetics of Rigid Body Rotation., , Questions-Answers, Long Answer Type and Medium Answer Type Questions, , Que 5.16. Discuss and describe the laws of motion applied to planar, translation and rotation., , AKTU 2014-15, (II) Marks 05, , Answer, A., , Laws of Translation : Refer Q. 5.2, Page 5–3C, Unit-5.
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5–22 C (CE-Sem-3), , Introduction to Kinetics of Rigid Bodies, , www.aktutor.in, , B., , Laws of Rotation : Following are the laws as applied to rotary motion :, , i., , First Law : It states that a body continues in its state of rest or of, rotation about an axis with constant angular velocity unless it is, compelled by an external torque to change the state., , ii., , Second Law : It states that the rate of change of angular momentum, of a rotating body is proportional to the external torque applied on the, body and takes place in the direction of the torque., , iii. Third Law : It states that to every torque there is always an equal and, opposite torque., , Que 5.17. Derive an expression for kinetic energy due to rotation., Answer, 1., , Consider a rigid body rotating about O as shown in Fig. 5.17.1., , 2., , Let,, , = Angular velocity of the body., dm = Elementary mass of the body., r = Radius of elementary mass from O., v = Tangential velocity of elementary mass., , 3., , KE of the elementary mass is,, , 1, 1, × Mass × Velocity2 =, dmv2, ...(5.17.1), 2, 2, KE of the whole body is obtained by integrating the eq. (5.17.1). Hence, KE of the body,, =, , 4., , 1, ( v = r), dm (r) 2, 2, 1, 1, = 2 r 2 dm 2 r 2 dm ( is a constant), 2, 2, =, , 1, , 2 dm v, , 2, , , , , v, dm, , Rigid, body, Fig. 5.17.1., , 5., , But, , , r, , 2, , dm = I = Moment of inertia of the body about O., , KE of the body =, , 1 2, I, 2
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Engineering Mechanics, , www.aktutor.in, , 5–23 C (CE-Sem-3), , Que 5.18. A uniform homogeneous cylinder rolls without slip along, a horizontal level surface with a translational velocity of 20 cm/sec., If its weight is 0.1 N and its radius is 10 cm, what is its total kinetic, energy ?, , Answer, Given : v = 20 cm/sec = 0.20 m/sec, W = 0.1 N, m =, , W 0.1, , kg, g 9.81, , r = 10 cm = 0.1 m, To Find : Total kinetic energy., 1., , We know that,, , I=, =, =, , 2., , Total kinetic energy =, , mr2, 2, 0.1 0.12, = 0.000051, , 9.81, 2, , v 0.20, = 2 rad/sec, , r 0.10, , 1, 1, I 2 m v 2, 2, 2, , 1, 1 0.1, 0.000051 22 , 22, 2, 2 9.81, = 0.000102 + 0.0204 = 0.020502 N-m, =, , Que 5.19., , Derive an expression for the acceleration of system in, , which weights are attached to the two ends of a string which passes, over a rough pulley., , Answer, 1., , Fig. 5.19.1 shows the two weights W1 and W2 attached to the two ends, of a string, which passes over a rough pulley of radius R., , 2., , As pulley is rough and having certain weight, the tensions on both, sides of the string will not be same. If W1 > W2, the weight W1 will move, downwards whereas the weight W2 will move upwards with the same, acceleration., , 3., , Let,, , a = Acceleration of the system., T1 = Tension in the string to which weight W1 is attached., T2 = Tension in the string to which weight W2 is attached., R = Radius of the pulley.
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5–24 C (CE-Sem-3), , Introduction to Kinetics of Rigid Bodies, , www.aktutor.in, , I = Moment of inertia of the pulley about the axis of rotation., = Angular acceleration., W0 = Weight of the pulley., 4., , Considering the motion of weight W1, let it is moving downwards with, an acceleration a., The net downwards force on weight W1 = (W1 – T1), Mass of weight,, , 5., , m1 =, , W1, g, , We know that,, Net force = Mass × Acceleration, , W1, a, ...(5.19.1), g, Considering the motion of weight W2, let it is moving upwards with an, acceleration a., Net upward force = (T2 – W2), (W1 – T1) =, , 6., , T2, , R, , T2, , T1, , T1, , T2, , T1, a, , W2, , a, , W1, , W2, Fig. 5.19.1., , 7., , W1, , Using, net force = Mass × Acceleration, , W2, a, ...(5.19.2), g, Now considering the rotation of the pulley, let it is rotating with an, angular acceleration ., (T2 – W2) =, , 8., 9., , If the pulley is considered as a solid disc, then moment of inertia of the, pulley is given by,, I=, , mR2, 2, , I=, , W0 R2, g 2, , (Solid disc is like a cylinder), , W0 , , m g
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Engineering Mechanics, , www.aktutor.in, , 5–25 C (CE-Sem-3), , 10. The torque on the pulley is given by,, T= I, , a, W0 R2 a , , , , R, g, 2, R , , ...(5.19.3), , 11. But torque on the pulley = Torque due to T1 – Torque due to T2, = T1 × R – T2 × R = R(T1 – T2), 12. Substituting the value of torque in eq. (5.19.3), we get, R(T1 – T2) =, T1 – T2 =, 13, , W0 R2 a, , , g, 2, R, , W0, a, 2g, , ...(5.19.4), , Adding eq. (5.19.1), eq. (5.19.2) and eq. (5.19.4), we get, W1 – W2 =, a=, , W1, W, W, a, W, a 2 a 0 a W1 W2 0 , g, g, 2g, g, 2, g(W1 W2 ), W0 , , W1 W2 , , 2 , , Que 5.20. Two weights of 8 kN and 5 kN are attached at the ends of, a flexible cable. The cable passes over a pulley of diameter 1 m. The, weight of the pulley is 500 N and radius of gyration is 0.5 m about its, axis of rotation. Find the torque which must be applied to the pulley, to raise the 8 kN weight with an acceleration of 1.2 m/sec2. Neglect, , AKTU 2013-14, (I) Marks 10, , the friction in the pulley., , Answer, Given : W1 = 8 kN, W2 = 5 kN, D = 1 m, W0 = 500 N, k = 0.5 m,, a = 1.2 m/sec2, To Find : Torque applied to pulley., 1., , As we need to raise 8 kN weight with an acceleration of 1.2 m/sec2, then, we must apply a torque on the pulley which will be given as, , 2., , Applying equilibrium equation on block of 8 kN, we get, , Torque = (T1 – T2) r + I, , T1 – 8000 =, T1 =, , 8000, a, 9.81, 8000, × 1.2 + 8000, 9.81
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5–26 C (CE-Sem-3), , Introduction to Kinetics of Rigid Bodies, , www.aktutor.in, T1 = 8978.59 N, 1m, , k = 0.5 m, , a, T1, , T2, , a, , 500 N, 8 kN, 5 kN, Fig. 5.20.1., , 3., , Applying equilibrium equation on block of 5 kN, we get, 5000 – T2 =, , 5000, a, 9.81, , 4., , 5000, 1.2, 9.81, T2 = 4388.38 N, Torque applied on the pulley = I, , 5., , a, r, 1.2, 500 , I= , (0.5)2 , 9.81 , (1 / 2), I= 30.58 N-m, Now total applied torque = (T1 – T2) r + I, , T2 = 5000 –, , I= mk2, , a, , 2, I mk , , r, , 1, = (8978.59 – 4388.38) × + 30.58, 2, = 2325.685 N-m, , Virtual Work and Energy Method, Virtual Displacements,, Principle of Virtual Work for Particle and, Ideal System of Rigid Bodies., , CONCEPT OUTLINE, Virtual Displacement : The displacement of a particle or a rigid, body in equilibrium is not at all possible. However we can assume an, imaginary displacement to occur, particularly if the system is partially, constrained, this displacement is known as virtual displacement., Virtual Work : The total work done by the system of forces causing, the virtual displacement is termed as virtual work.
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Engineering Mechanics, , www.aktutor.in, , 5–27 C (CE-Sem-3), , Questions-Answers, Long Answer Type and Medium Answer Type Questions, , Que 5.21., , Discuss in short about work done on a particle and, , work done on a rigid body., , Answer, i., , Work Done on a Particle :, , 1., , When a force acts on a particle, which is not constrained to move, it, causes a displacement of the particle. The force is then said to have, done work on the particle., , 2., , We then define work done on the particle as a product of magnitude of, the force and the displacement. Mathematically, we can write this as, W = Fs, A, , B, , F, , s, Fig. 5.21.1., , ii., , Work Done on a Rigid Body :, , 1., , We know that a rigid body is subjected to moments in addition to the, forces. Just as the forces cause linear displacements, moments cause, angular displacements., , 2., , If a moment M acting on a rigid body causes an angular displacement , then work done by the moment on the rigid body is defined as the, product of moment and angular displacement, i.e.,, W = M, , Que 5.22., , Give the principle of virtual work for a particle and a, , rigid body., , Answer, 1., , For the particle or rigid body to remain in equilibrium in the displaced, position also, we know that the resultant force acting on it must be, zero. Thus, we say that work done in causing this virtual displacement, is also zero. This is known as principle of virtual work., , 2., , For a system of concurrent forces F1, F1, .... F1, the virtual work done is, given by,, U = F1 r + F2 r + ..... + Fn r
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5–28 C (CE-Sem-3), , Introduction to Kinetics of Rigid Bodies, , www.aktutor.in, , = (F1 + F2 + ..... + Fn) r, =, , , , , , Fr, , 3., , As a system of concurrent force can be replaced by a single resultant, force, the virtual work done is equal to the work done by the resultant., , 4., , For the body to remain in equilibrium in the displaced position, we, know that the resultant must be zero. Hence, virtual work done in, causing this virtual displacement is also zero, i.e.,, U =, , F r 0, , 5., , The necessary and sufficient condition for the equilibrium of a particle, is zero virtual work done by all external forces acting on the particle, during any virtual displacement consistent with the constraints imposed, on the particle., , 6., , Similarly, for a rigid body, we can write the principle of virtual work as, U =, , Que 5.23., , F r M 0, , A uniform ladder AB of length l and weight W leans, , against a smooth vertical wall and a smooth horizontal floor as, shown in Fig. 5.23.1. By the method of virtual work, determine the, horizontal force P required to keep the ladder in equilibrium, position., A, , C, , h, , P, B, Fig. 5.23.1., , Answer, Given : Fig. 5.23.1., To Find : Horizontal force, P., 1., , Under its own weight, the ladder tries to slide down, but the horizontal, force P holds it in equilibrium. The free body diagram of the ladder is, shown in Fig. 5.23.2.
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Engineering Mechanics, , www.aktutor.in, , 5–29 C (CE-Sem-3), , A, , C, RC, h, , y, , W, , B, , P, , x, RB, Fig. 5.23.2., , 2., , Let be inclination of the ladder with respect to the horizontal. From, the geometry of the triangle, we see that the location x of the end B, and the location y of the centre of gravity of ladder with respect to the, origin are :, x=, , h, tan , , ...(5.23.1), , l, sin , ...(5.23.2), 2, The virtual displacement are obtained by differentiating eq. (5.23.1), and eq. (5.23.2) as,, y=, , 3., , l, cos , 2, From Fig. 5.23.2 we see that as decreases, y also decreases but x, increases. Hence, considering only positive virtual displacements, the, above expressions reduce to, x = – h cosec2 and y =, , 4., , l, cos , 2, Now applying the principle of virtual work, we have, x = h cosec2 and y =, , 5., , U = 0, – P x + W y = 0, 6., , It should be noted that reaction RB and RC do no work, as the virtual, displacement of the contact points B and C are perpendicular to the, direction of the forces. Therefore
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5–30 C (CE-Sem-3), , Introduction to Kinetics of Rigid Bodies, , www.aktutor.in, , l, , – P[h cosec2 ] + W cos = 0, 2, , , Wl cos , P = 2h cosec 2 , P=, , Wl, sin 2 cos , 2h, , Que 5.24. Using the principle of virtual work, determine the angle, for which equilibrium is maintained in the mechanism shown for, given values of forces P1 and P2 applied. Length of the longer links is, l and that of the shorter links is l/2., P1, , , , , P2, , Fig. 5.24.1., , Answer, Given : Fig. 5.24.1, Length of longer link = l, Length of shorter link = l/2, To Find : Angle ., 1., , Choosing the hinge point as the origin, the point of application of the, forces P1 and P2 are y and x respectively. Expressing these positions x, and y in terms of , we have, y=, , ...(5.24.1), , l, sin l sin , ...(5.24.2), 2, The virtual displacements are obtained by differentiating eq. (5.24.1), and eq. (5.24.2) as,, and, , 2., , l, l, l, 3, cos cos cos = l cos , 2, 2, 2, 2, , x= 2, , 3l, sin , 2, andy = l cos , y =
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Engineering Mechanics, , www.aktutor.in, , 5–31 C (CE-Sem-3), , P1, , y, , , , , P2, x, , Fig. 5.24.2., , 3., , From Fig. 5.24.2, we see that as increases, x increases while y, decreases. Hence, considering only positive values of virtual, displacements, the above expressions reduce to, , 3l, sin and x = l cos , 2, Applying the principle of virtual work, we get, y =, , 4., , P1y – P2x = 0, , 3l, , P1 sin P2 (l cos ) = 0, 2, , , 3 P1, sin = P2 cos , 2, 2P , = tan 1 2 , 3 P1 , , Applications of Energy Method for Equilibrium., , Questions-Answers, Long Answer Type and Medium Answer Type Questions, , Que 5.25., , State law of conservation of energy.
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5–32 C (CE-Sem-3), , Introduction to Kinetics of Rigid Bodies, , www.aktutor.in, , Answer, 1., , Law of conservation of energy states that the energy can neither be, created nor destroyed though it can be transformed from one form to, another form., , 2., , It can also be stated as the total energy possessed by a body remains, constant provided no energy is added to or taken from it., , Que 5.26. A body weighing 196.2 N slides up a 30° inclined plane, under the action of an applied force 300 N acting parallel to the, inclined plane. The coefficient of friction, is equal to 0.2. The body, moves from rest. Determine :, i., Acceleration of the body., ii. Distance travelled by body in four seconds., iii. Velocity of body after four seconds., iv. Kinetic energy of the body after four seconds., v. Work done on the body in four seconds., vi. Momentum of the body after four seconds., vii. Impulse applied in four seconds., , Answer, , Given : W = 196.2 N, m =, , W 196.2, = 20 kg, Applied force = 300 N,, , g, 9.81, , = 30°, = 0.2., To Find : i. Acceleration of the body., ii. Distance travelled by body in 4 sec., iii. Velocity of body after 4 sec., iv. Kinetic energy of the body after 4 sec., v. Work done on the body in 4 sec., vi. Momentum of the body after 4 sec., vii. Impulse applied in 4 sec., 1., , As body moves from rest, hence initial velocity (u) will be zero., u=0, , 2., , Fig. 5.26.1 shows the free body diagram. The net force in the direction, of motion is given by,, F = Applied force – W sin – R, = 300 – 196.2 × sin 30° – 0.2 × W cos , ( R = W cos ), = 300 – 98.1 – 0.2 × 196.2 × cos 30°, = 300 – 98.1 – 33.98 = 167.92 N
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Engineering Mechanics, , re, Di, , c ti, , W, , on, , s in, , www.aktutor.in, , m, of, , oti, , 5–33 C (CE-Sem-3), , on, , R, , ied, pl, Ap o r c e, f, of, rce R, Fo o n =, c ti, fri, , , , W cos , W = 196.2 N, Fig. 5.26.1., , 3., , We know that,, , F= m×a, , 167.92 = 20 × a, , 167.92, = 8.396 m/sec2., 20, Distance travelled in 4 sec,, a=, , 4., , s = ut +, , 1, at2, 2, , = 0×4+, , 1, × 8.396 × 42 = 67.168 m, 2, , 5., , Velocity after 4 sec, v = u + at, , 6., , The kinetic energy after 4 sec is given by,, , = 0 + 8.396 × 4 = 33.584 m/sec, , KE =, , 1, mv2, 2, , 1, × 20 × (33.584)2 = 11278.8 N-m, 2, Work done on the body in 4 sec, =, , 7., , = Net force × Distance moved in 4 sec, = 167.92 × 67.168 = 11278.8 Nm, 8., , The work done on the body is equal to the change of kinetic energy of, the body., Change of KE =, , 1, 1, mv2 –, mu2, 2, 2
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5–34 C (CE-Sem-3), , Introduction to Kinetics of Rigid Bodies, , www.aktutor.in, =, , 1, 1, × 20 × (33.5842)2 –, × 20 × 02 ( u = 0), 2, 2, , = 11278.8 – 0 = 11278.8 Nm., 9., , Momentum of the body after 4 sec, = m × v = 20 × 33.584 = 671.68 kg m/sec., , 10. Impulse applied in 4 sec, = Net force × Time = F × 4, = 167.92 × 4 = 671.68 N sec, 11. Change of momentum of the body, = mv – mu = 20 × 33.584 – 20 × 0, = 671.68 kg m/sec, = 671.68 Nsec, , Stability of Equilibrium., , Questions-Answers, Long Answer Type and Medium Answer Type Questions, , Que 5.27., , Write a short note on stability of equilibrium., , Answer, 1., , Equilibrium is a state of a system which does not change., , 2., , An equilibrium is considered stable, if the system always returns to its, initial stage after small disturbances. If the system moves away from, the equilibrium after small disturbances, then the equilibrium is, unstable., , 3., , For example, the equilibrium of a pencil standing on its tip is unstable, while the equilibrium of a picture on the wall is (usually) stable., ,
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Engineering Mechanics, , 1, , www.aktutor.in, , SQ–1 C (CE-Sem-3), , Introduction to, Engineering, Mechanics, (2 Marks Questions), , 1.1. What do you understand by a particle and a rigid body ?, Ans. Particle : A particle is a body of infinitely small volume and the, , mass of the particle is considered to be concentrated at a point., Rigid Body : A body which does not deform under the action of, external forces is known as rigid body., 1.2. Give the effect of force and moment on a body., Ans. The force acting on a body causes linear displacement while moment, , causes an angular displacement., 1.3. What are the steps in making of a free body diagram ?, , AKTU 2013-14, (I) Marks 02, Ans. The steps in making a free body diagram are as follows :, , i. A sketch of the body is drawn by removing the supporting surfaces., ii. Indicate on this sketch all the applied or active forces, which tend to, set the body in motion, such as those caused by weight of the body, or applied forces, etc., iii. Also indicate on this sketch all the reactive forces, such as those, caused by the constraints or supports that tend to prevent motion., iv. All relevant dimensions and angles, reference axes are shown on, the sketch., 1.4. Define resultant of forces., Ans. A single force which can replace a number of forces acting on a, , body and gives same effect is called resultant of forces., 1.5. The resultant of two forces 3P and 2P is R. If the first force, is doubled the resultant is also doubled, determine the, angle between the two forces., , AKTU 2013-14, (II) Marks 02, Ans., , Given : P = 3P, Q = 2P, P = 6P, R = 2R, To Find : Angle between the two forces, .
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SQ–2 C (CE-Sem-3), , www.aktutor.in, , 2 Marks Questions, , 1. From parallelogram law of forces,, R2 = P2 + Q2 + 2PQ cos , So,, , R=, , (3P)2 (2 P)2 2 3 P 2 P cos , , R = 9 P 2 4 P 2 12P 2 cos , 2. Now according to changed values,, R =, 2R =, , ...(1.5.1), , P 2 Q2 2P Q cos , (6 P)2 (2 P)2 2 6 P 2P cos , , 2R = 36 P 2 4 P 2 24 P 2 cos , 3. From eq. (1.5.1) and eq. (1.5.2), we have, , 2 9 P 2 4 P 2 12 P 2 cos =, , ...(1.5.2), , 36 P 2 4 P 2 24 P 2 cos , , 4(9P2 + 4P2 + 12P2 cos) = 36P2 + 4P2 + 24P2 cos , 12P2 + 24P2 cos = 0, 12P2(1 + 2 cos ) = 0, 4. Since, 12P20, 1 + 2 cos = 0, cos = – 1/2, = 120°, 1.6. What is static equilibrium ? Write down sufficient condition, of static equilibrium for a coplanar concurrent and, non-concurrent force system., , AKTU 2015-16, (I) Marks 02, OR, Explain condition of equilibrium of coplanar-non, concurrent forces., , AKTU 2016-17 , (II) Marks 02, , Ans. Static Equilibrium : A body is said to be in static equilibrium if all, , the forces acting on the body are balanced whether the body is at, rest or in motion., Conditions of Static Equilibrium for a Coplanar Concurrent, Force System :, Fx = 0, and, Fy = 0, Conditions of Static Equilibrium for a Coplanar NonConcurrent Force System :, Fx = 0,, Fy = 0, and, M = 0, 1.7. How do you find the resultant of non-coplanar concurrent, force system ?, , AKTU 2014-15, (II) Marks 02
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Engineering Mechanics, , www.aktutor.in, , SQ–3 C (CE-Sem-3), , Ans. The resultant of several forces in non coplanar concurrent force, , system can be found analytically by summing the components of, forces along X, Y and Z directions, i.e., resultant R can be obtained, by,, , (Fx )2 (Fy )2 (Fz )2, , R=, , 1.8. “Friction is both desirable and undesirable”. Explain, , AKTU 2014-15, (II) Marks 02, Ans. Friction helps in working of friction brakes and clutches, belt and, , rope drives, holding and fastening devices while it may also, deteriorates the working of power screws, bearing and gears, flow, of fluids in pipes. So we can say that friction is both desirable and, undesirable., 1.9. Explain the relationship between angle of friction and angle, , AKTU 2013-14, (I) Marks 02, , of repose., , Ans. Angle of friction = Angle of repose, , 1.10. A block of mass m on an inclined plane is kept in equilibrium, and prevented from sliding down by applying a force of, 500 N. If the angle of the inclination is 30° and coefficient of, friction for the contact surface is 0.35, determine the weight, , AKTU 2013-14, (II) Marks 02, , of the block., Ans., , Given : F = 500 N, = 30°, = 0.35, To Find : Weight of block., 1. Fig. 1.10.1 shows the block resting on inclined plane., R, Mo, , n, ti o, , F=, , R, , , , , W = mg, , Fig. 1.10.1., , 2. FBD of block is as shown in Fig. 1.10.2.
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SQ–4 C (CE-Sem-3), , www.aktutor.in, , 2 Marks Questions, , R, , R, mg sin 30°, 30° mg cos 30°, 500, , N, , mg, Fig. 1.10.2., , 3. Equation of equilibrium along plane,, 500 + R = mg sin 30°, 500 + 0.35 × mg cos 30° = mg sin 30°, ( R = mg cos 30°), 500 = mg (sin 30° – 0.35 cos 30°), 500 = mg (0.5 – 0.30) 500 = 0.2 mg, mg = 2500 N, 1.11. Write any four engineering applications of friction., , AKTU 2015-16, (I) Marks 02, Ans. Following are the engineering applications of friction :, , i., ii., iii., iv., , In producing relative motion between bodies., In transmitting power., In braking system to stop the vehicle., In lifting the heavy blocks, machinery etc., over wedges., , 1.12. State Varignon’s theorem of moments., , AKTU 2016-17, (I) Marks 02, Ans. Varignon’s theorem of moments states that the algebraic sum of, , the moments of a system of coplanar forces about a moment centre, in their plane is equal to the moment of their resultant force about, the same moment centre., 1.13. Define the principle of transmissibility., , AKTU 2016-17 , (II) Marks 02, Ans. Principle of transmissibility states that the state of rest or of motion, , of a rigid body is unchanged if a force acting on the body is replaced, by another force of same magnitude and same direction but acting, anywhere on the body along the line of action of the replaced force., 1.14. Explain free body diagram with example., , AKTU 2016-17 , (II) Marks 02, Ans. Free Body Diagram : A body may consist of more than one element, , and supports. Each element or support can be isolated from the
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Engineering Mechanics, , www.aktutor.in, , SQ–5 C (CE-Sem-3), , rest of system by properly incorporating the effect of forces. The, diagram of the isolated element or a portion of the body along with, the net effect of forces is known as free body diagram (FBD)., Example :, FBD, , Body, , W, W, R, Fig. 1.14.1., , 1.15. Define parallelogram law of forces., , AKTU 2016-17 , (II) Marks 02, Ans. Parallelogram law of forces states that if two forces acting, , simultaneously on a body at a point are represented in magnitude, and direction by two adjacent sides of a parallelogram, their, resultant will be represented in magnitude and direction by the, diagonal of the parallelogram which passes through the point of, intersection of two sides representing the forces., ,
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SQ–6 C (CE-Sem-3), , 2, , www.aktutor.in, , 2 Marks Questions, , Centroid and Centre, of Gravity, (2 Marks Questions), , 2.1. What is the difference between centroid and centre of, gravity ?, , AKTU 2014-15, (II) Marks 02, , Ans. The term centre of gravity applies to bodies with weight while, , centroid applies to lines, planes, areas and volumes., 2.2. Define axis of symmetry., Ans. The line about which the figure can be cut into equal halves is, , known as axis of symmetry., 2.3. Determine the centroid of a circular arc having radius, 20 mm and central angle 180°. AKTU 2013-14, (I) Marks 02, Ans., , Given : 2 = 180°, = 90° = /2 rad, R = 20 mm, To Find : Centroid of circular arc., 1. Position of the centroid for circular arc is given as,, R sin 20 sin ( / 2), x =, , , , 2, 40, =, = 12.73 mm, , y = 0 (due to symmetry), 2.4. What is the centroid of segment of a circular disc of radius, 5 cm and subtended angle of 120° ?, , AKTU 2013-14, (II) Marks 02, Ans., , Given : R = 5 cm, 2 = 120°, = 60° = /3 rad, To Find : Centroid of segment of a circular disc.
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Engineering Mechanics, , www.aktutor.in, , SQ–7 C (CE-Sem-3), , 1. Centroid of circular lamina is given as,, , 2R, sin , 3, 25, 10, sin 60° =, sin 60°, x=, , , 3, 3, x = 2.75 cm, y = 0 (due to symmetry), , x=, , ( = /3), , 2.5. Explain polar moment of inertia., , AKTU 2013-14, (II) Marks 02, Ans. Moment of inertia about an axis perpendicular to the plane of an, , area is known as polar moment of inertia., 2.6. Find the polar moment of inertia of a circular area of, diameter 5 mm., , AKTU 2013-14, (I) Marks 02, , Ans., , Given : D = 5 mm, To Find : Polar moment of inertia., 1. Polar moment of inertia of a circular disc is given as,, J=, , D4 54, = 61.36 mm4, , 32, 32, , 2.7. What do you understand by radius of gyration ?, , AKTU 2015-16, (I) Marks 02, Ans. Radius of gyration is the distance which is when squared and, , multiplied by area gives the moment of inertia of that area., 2.8. State perpendicular axis theorem., , AKTU 2015-16, (I) Marks 02, OR, State and explain perpendicular axis theorem., , AKTU 2014-15 , 2016-17, (II) Marks 02, Ans. Perpendicular axis theorem states that the moment of inertia of an, , area about an axis perpendicular to its plane (polar moment of, inertia) at any point O is equal to the sum of moments of inertia, about any two mutually perpendicular axis through the same point, O and lying in the plane of the area., Mathematically, IZZ = IXX + IYY
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SQ–8 C (CE-Sem-3), , www.aktutor.in, , 2.9. State parallel axis theorem., , 2 Marks Questions, , AKTU 2016-17, (I) Marks 02, , Ans. Parallel axis theorem states that the moment of inertia about any, , axis in the plane of an area is equal to the sum of moment of inertia, about a parallel centroidal axis and the product of area and square, of the distance between the two parallel axis., 2.10. Define mass moment of inertia., Ans. Mass moment of inertia of a body about an axis is defined as the, , sum total of product of its elemental masses and square of their, distance from the axis., ,
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Engineering Mechanics, , www.aktutor.in, , 3, , SQ–9 C (CE-Sem-3), , Basic Structural, Analysis, (2 Marks Questions), , 3.1. Write the different types of support., Ans. Following are the different types of support :, , i., ii., iii., iv., v., , Simple support or knife edge support,, Roller support,, Pin joint or hinged support,, Smooth surface support, and, Fixed or built-in support., , 3.2. List the various types of loads to which the beam can be, subjected., , AKTU 2016-17, (I) Marks 02, , Ans. Following are the different types of loads to which the beam can be, , subjected :, i. Concentrated or point load,, ii. Uniformly distributed load (UDL), and, iii. Uniformly varying load (UVL)., 3.3. Differentiate between perfect and imperfect truss., , AKTU 2015-16, (I) Marks 02, Ans., , S. No., , Perfect Truss, , Imperfect Truss, , 1., , Prefect trusses always, retain their shape., , Imperfect trusses cannot retain, their shape when loaded and, get distorted., , 2., , Number of members in perfect N umbe rs o f me mbers are, truss are equals to (2j – 3), where either more or less than 2j – 3., j is number of joints., , 3.4. What do you understand by point of contraflexure ?, , AKTU 2015-16, (I) Marks 02
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SQ–10 C (CE-Sem-3), , www.aktutor.in, , 2 Marks Questions, , Ans. The point of contraflexure is a point which represents the section, , on the beam where bending moment is zero or bending moment, changes its sign., 3.5. A truss structure is made up of five members. If the number, of joints in the truss is four then state the nature of truss., , AKTU 2013-14, (II) Marks 02, Ans., , Given : m = 5, j = 4, To Find : Nature of truss., 1. Nature of truss can be determine by the following formula,, m = 2j – 3, m = 2j – 3, = 2×4–3=5, LHS = RHS, So, the given truss is a perfect truss., 3.6. What are the different methods of analysing a frame ?, Ans. A frame is analysed by the following methods :, , i. Method of joints,, ii. Method of section, and, iii. Graphical method., 3.7. What assumptions are made while determining stresses in, a truss ?, Ans., , i., ii., iii., , AKTU 2014-15, (II) Marks 02, , Following are the assumptions made while determining stresses, in a truss :, The frame should be a perfect frame., The frame carries load at the joints., All the members are pin-joined., , 3.8. Discuss the conditions under which the method of section, is preferred over method of joints in analysis of truss., , AKTU 2013-14, (I) Marks 02, Ans. Under the following two conditions the method of section is, , preferred over the method of joints :, i. In a large truss in which forces in only few members are required., ii. In the situation where the method of joints fails to start/proceed, with analysis., 3.9. Define zero force members., Ans. The members of a truss in which net force is zero are known as, , zero force members.
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Engineering Mechanics, , www.aktutor.in, , SQ–11 C (CE-Sem-3), , 3.10. With neat sketches describe in brief different types of beams., , AKTU 2014-15, (II) Marks 02, Ans. Following are the different types of beams :, , S. No. Type of beam, , Diagram, , i., , Cantilever beam, , ii., , Simply supported beam, , iii., , Overhanging beam, , iv., , Fixed beam, , v., , Continuous beam, , 3.11. Determine the maximum bending moment in a simply, supported beam having span of 5 m and carrying a, uniformly distributed load of 10 kN/m throughout its span., , AKTU 2013-14, (I) Marks 02, Ans., , Given : l = 5 m, w = 10 × 103 kN/m, To Find : Maximum bending moment., 1. We know maximum bending moment for simply supported beam, carrying uniformly distributed load is given as,, (BM)max =, , wl2 10 103 52, = 31250 N-m, , 8, 8, , 3.12. Determine the maximum bending moment in a simply, s upported beam of s pan 5 m, carrying uniformly, distributed load of 2 kN/m over its entire span., , AKTU 2013-14, (II) Marks 02, Ans., , Given : w = 2 kN/m, l = 5 m, To Find : Maximum bending moment.
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SQ–12 C (CE-Sem-3), , www.aktutor.in, , 2 Marks Questions, , 1. Maximum bending moment for a simply supported beam carrying, uniformly distributed load is given by as,, 2, 2, (BM)max = wl = 2 5 25 = 6.25 kN-m, 8, 8, 4, ,
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Engineering Mechanics, , www.aktutor.in, , 4, , SQ–13 C (CE-Sem-3), , Review of Particle, Dynamics, (2 Marks Questions), , 4.1. Define rectilinear motion., Ans. The motion of a body along a straight line is known as rectilinear, , motion., 4.2. A mass of 3 kg is dropped from a height from rest. Find the, distance travelled in 5 seconds., , AKTU 2013-14, (I) Marks 02, Ans., , Given : Mass = 3 kg, t = 5 sec,, To Find : Distance travelled in 5 sec., , But,, , 1 2, gt, 2, u = 0 (body is initially at rest), , , , s=, , 1. We know that, s = ut +, , 1 2 1, gt =, × 10 × 52, 2, 2, = 125 m, , (Using, g = 10 m/sec2), , 4.3. The equation of motion for motion of a particle is given by, s = 18t + 3t2 – 2t3. Find acceleration and velocity at t = 2 sec., , AKTU 2014-15, (II) Marks 02, Ans., , Given : s = 18t + 3t2 – 2t3, To Find : Acceleration and velocity at t = 2 sec., , ds, = 18 + 6t – 6t2, dt, t = 2 sec, v = 18 + 6 × 2 – 6 × 4, = 6 m/sec, , 1. We know that, v =, Velocity at,
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SQ–14 C (CE-Sem-3), , www.aktutor.in, , 2. Also, acceleration, a =, , 2 Marks Questions, , d2 s, , = 6 – 12t, dt2, Acceleration at, t = 2 sec, a = 6 – 12 × 2 = – 18 m/sec2, , 4.4. What do you understand by plane curvilinear motion ?, Ans. The motion of a body in a plane along a circular path is known as, , plane curvilinear motion., 4.5. Define relative motion., Ans. The motion of a moving body with respect to another moving body, , is known as the relative motion of the first body with respect to, second body., 4.6. Define work., Ans. Work is defined as the product of force and displacement. Its unit is, , joule (J)., 4.7. What do you mean by energy ?, Ans. The capacity of doing work is known as energy. It is the product of, , power and time., 4.8. Define kinetic energy and potential energy., Ans. Kinetic Energy : The energy possessed by a body by virtue of its, , motion is known as kinetic energy. It is given by,, , 1, mv2, 2, Potential Energy : The energy by virtue of position of a body with, respect to any given reference or datum is known as potential, energy. It is given by,, PE = mgh, KE =, , 4.9. Define impulse and momentum., Ans. Impulse : The product of force and time is known as impulse., , Momentum : The product of mass and velocity of a body is known, as momentum., 4.10. What do you understand by angular momentum ?, Ans. The product of mass moment of inertia and angular velocity of, , rotating body is known as angular momentum., 4.11. State the law of conservation of energy., Ans. Law of conservation of energy states that the energy can neither, , be created nor destroyed, though it can be converted from one, form into another form., ,
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Engineering Mechanics, , www.aktutor.in, , SQ–15 C (CE-Sem-3), , Introduction to, Kinetics of Rigid, Bodies, (2 Marks Questions), , 5, , 5.1. State Newton’s second law of motion., Ans. Newton’s second law of motion states that the rate of change of, , momentum of a body is proportional to the external force applied, on the body and takes place in the direction of the force., 5.2. Define instantaneous centre of rotation., Ans. The point about which motion of a body having both translational, , and rotational motion is assumed to be pure rotational is known as, instantaneous centre of rotation., 5.3. State and explain D’Alembert’s principle., , AKTU 2014-15, (II) Marks 02, OR, State D-Alembert’s principle. AKTU 2015-16, (I) Marks 02, Ans. D’Alembert’s principle states that the net external force acting on, , the system and the resultant inertia force are in equilibrium., 5.4. What do you understand by work-energy principle ?, , AKTU 2015-16, (I) Marks 02, Ans. Work-energy principle states that the change in kinetic energy of a, , body during any displacement is equal to the work done by the, body., 5.5. Write D’Alembert’s principle for rotary motion., Ans. According to D’Alembert’s principle, when external torques acts on, , a system having rotating motion, then the algebraic sum of all the, torques acting on the system due to external forces and reversed, active forces including the inertia torque is zero., 5.6. Give the expression for the kinetic energy of rotating bodies., Ans., , KE =, , 1 2, I, 2
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SQ–16 C (CE-Sem-3), Where,, , www.aktutor.in, , 2 Marks Questions, , = Angular velocity, and, I = Moment of inertia., , 5.7. Define virtual displacement., Ans. The displacement of a partially constrained body which is occurring, , only in imagination but not in reality is known as virtual, displacement., 5.8. Write principle of virtual work for a particle and for a rigid, body., Ans. For Particle :, U = Fr = 0, For Rigid Body :, U = Fr + M = 0, 5.9. Write the different types of motion., , AKTU 2015-16, (I) Marks 02, Ans. Following are the different types of motion :, , i. Translation,, ii. Rotation, and, iii. General plane motion (combined motion of translation and rotation)., ,
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Engineering Mechanics, , www.aktutor.in, , SP–1 C (CE-Sem-3), , B.Tech., (SEM. III) ODD SEMESTER THEORY, EXAMINATION, 2019-20, ENGINEERING MECHANICS, Time : 3 Hours, Note : 1., , Max. Marks : 100, , Attempt all section. If require any missing data; then choose, suitably., Section-A, , 1. Attempt all questions in brief., a. Define shear force and bending moment., , (2 × 10 = 20), , b. How does a rigid body differ from an elastic body ?, c. Define center of mass and write down the co-ordinates of, center of gravity of trapezoid., d. Define work and power. Write the mathematical relation, and SI unit., e. State and prove law of conservation of momentum., f. Enlist different types of supports and loading system., g. Explain with the help of neat diagram, the concept of, limiting friction., h. Write down D’Alembert’s Principle., i. Differentiate between stable and unstable equilibrium., j. State parallel axis theorem. Define radius of gyration., Section-B, 2. Attempt any three of the following :, (10 × 3 = 30), a. State and prove Lami’s theorem., The greatest and least resultant of two forces acting on, body are 35 kN and 5 kN respectively. Determine the, magnitude of the forces. What would be the angle between, these forces if the magnitude of the resultant is stated to, be 25 kN ?
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SP–2 C (CE-Sem-3), , www.aktutor.in, , Solved Paper (2019-20), , b. Calculate the centroid of a semi-circular ring of radius ‘r’,, using method of moments., c. Find moment of inertia of the Fig. 1 about X-X axis,, thickness of member is 20 mm., Y, B, , C, , 20 mm, , A, , F, , G, , E, , H, , 20 mm, , 160 mm, Fig. 1., , D, 20 mm, , 150 mm, , X, , d. Differentiate between rectilinear and curvilinear motion., Also derive the expression for the horizontal range, Time, of flight and maximum height of a projectile with initial, velocity u and inclined at an angle “” with the horizontal., e. State Work Energy principle., A uniform cylinder of 125 mm radius has a mass of 0.15 kg., This cylinder rolls without slipping along a horizontal, surface with a translation velocity of 20 cm/sec. Determine, its total kinetic energy., Section-C, 3. Attempt any one part of the following :, (10 × 1 = 10), a. Explain how a wedge is used for raising heavy loads. Also, mention the principle., A body resting on a rough horizontal plane required a pull, of 24 N inclined at 30° to the plane just to move it. It was, also found that a push of 30 N at 30° to the plane was just, enough to cause motion to impend. Make calculations for, the weight of body and the coefficient of friction., b. A ladder 5m long rests on a horizontal ground and leans, against a smooth vertical wall at an angle 70° with the, horizontal. The weight of the ladder is 900N and acts at its, middle. The ladder is at the point of sliding, when a man, weighing 750N stands 1.5m from the bottom of the ladder., Calculate coefficient of friction between the ladder and, the floor.
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Engineering Mechanics, , SP–3 C (CE-Sem-3), , www.aktutor.in, , 4. Attempt any one part of the following :, (10 × 1 = 10), a. Draw the SF and BM diagram for the simply supported, beam loaded as shown in Fig. 2., 2 kN, 4 kN, 2 kN, A, B, C, D, E, 1m, 1m, 1m, 1m, RA, RB, Fig. 2., , b. Define and explain the term imperfect truss., Fig. 3 shows a framed of 4 m span and 1.5 m height subjected, to two point loads at B and D. Find the forces in all the, members of the structure., B, , 8 kN, , 1.5 m, A, , , D, , 2m, , C, 2m, , 12 kN, , Fig. 3., , 5. Attempt any one part of the following :, (10 × 1 = 10), a. Explain the principle of virtual work. An overhanging beam, ABC of span 3 m is loaded as shown in Fig. 4. Using the, principle of virtual work, find the reactions at A and B., 1 kN, , 2 kN/m, , C, , A, B, 2m, , 1m, Fig. 4., , b. In a reciprocating pump, the lengths of connecting rod, and crank is 1125 mm and 250 mm respectively. The crank, is rotating at 420 rpm. Find the velocity with which the, piston will move, when the crank has turned through an, angle of 40° from the inner dead centre., 6. Attempt any one part of the following :, (10 × 1 = 10), a. Derive an equation for moment of inertia of triangle, centroidal axis and about its base., b. An I-section is made up of three rectangles as shown in, Fig. 5. Find the moment of inertia of the section about the
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SP–4 C (CE-Sem-3), , www.aktutor.in, , Solved Paper (2019-20), , horizontal axis passing through the centre of gravity of, the section., 60 mm, 20 mm, , 2, , 100 mm, , 20 mm, , 1, , 3, , 20 mm, , 100 mm, Fig. 5., , 7. Attempt any one part of the following :, (10 × 1 = 10), a. A body of mass 20 kg moving towards with a velocity of, 16 m/sec strikes with another body of 40 kg mass moving, towards left with 50 m/sec. Determine, i. Final velocity of the two bodies., ii. Loss in kinetic energy due to impact., iii. Impulse acting on either body during impact., Take coefficient of restitution as 0.65, b. A particle start with velocity u and the accelerationvelocity relationship is prescribed as a = – kv where k is a, constant. Set up an express ion that prescribes the, displacement time relation for the particle., ,
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Engineering Mechanics, , www.aktutor.in, , SP–5 C (CE-Sem-3), , SOLUTION OF PAPER (2019-20), Note : 1., , Attempt all section. If require any missing data; then choose, suitably., Section-A, , 1. Attempt all questions in brief., a. Define shear force and bending moment., , (2 × 10 = 20), , Ans., , 1. Shear Force : It is the force that tries to shear off the section of, a beam. It is obtained as algebraic sum of all forces acting normal, to axis of beam, either to the left or to the right of section., 2. Bending Moment : It is the moment that tries to bend the beam, and it is obtained as algebraic sum of moment of all forces about, the section, acting either to left or to the right of section., b. How does a rigid body differ from an elastic body ?, Ans., , S. No., , Elastic Body, , Rigid Body, , 1., , On applying lo ad it, undergoes deformation., , On applying load it do es not, deform., , 2., , On removal of load, comes, back to its original size and, shape., , On removal of load shape and size, remains unchanged., , 3., , Deformation is temporary., , No deformation., , c. Define center of mass and write down the co-ordinates of, center of gravity of trapezoid., Ans., , 1. Center of Mass : It is the point at which the whole weight of the, body acts. A body is having only one centre of gravity for all positions, of the body., 2. Co-ordinates of Centers of Gravity : From Fig. 1., Y, , h, , a, , x, CG, , y, X, , b, Fig. 1.
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SP–6 C (CE-Sem-3), , www.aktutor.in, x b / 2, y , , Solved Paper (2019-20), , 2a b h, , a b 3, , d. Define work and power. Write the mathematical relation, and SI unit., Ans., , 1. Work : Work is defined as the product of force and displacement., Its unit is joule (J)., 2. Power : In SI system of units, the unit of power is Joule per second, (J/sec), also called watt (W)., 3. Relation :, i. If W is the total work done in a time interval t, then average power, is given by,, , Total work done W, , ...(1), T ime taken, t, ii. The instantaneous power, i.e., power at a particular instant of time, is given by,, Pavg =, , dW d( Fs), , ...(2), dt, dt, iii. The force can be assumed to be constant over this infinitesimally, small time interval dt. Hence, we can write the above expression, as :, P=, , P=, , Fds, Fv, dt, , ...(3), , e. State and prove law of conservation of momentum., Ans., , 1. Conservation of Linear Momentum : When no external forces, act on bodies forming a system, the momentum of the system is, conserved i.e., the initial momentum of the system is equal to final, momentum of the system., 2. Proof :, i. Let,, F = Net force acting on a rigid body in the direction, of motion through CG of the body., m = Mass of the rigid body., a = Acceleration of the body., ii. We know that,, F = ma = m, , dv, dt, , Fdt = mdv, iii. Integrating the above equation, we get, , , , t2, , t1, , Fdt =, , , , v2, , v1, , mdv, , = m(v2 – v1), , d v, , a , , dt
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Engineering Mechanics, , www.aktutor.in, , SP–7 C (CE-Sem-3), , Impulse = mv2 – mv1, Impulse = Final momentum – Initial momentum, f. Enlist different types of supports and loading system., Ans., , 1., i., ii., iii., iv., v., 2., , Types of Support : Following are the different types of support :, Simple support or knife edge support,, Roller support,, Pin joint or hinged support,, Smooth surface support, and, Fixed or built-in support., Types of Load : Following are the different types of loads to which, the beam can be subjected :, i. Concentrated or point load,, ii. Uniformly distributed load (UDL), and, iii. Uniformly varying load (UVL)., g. Explain with the help of neat diagram, the concept of, limiting friction., Ans. The force of friction has a remarkable property of adjusting its, magnitude, so as to become exactly equal and opposite to the, applied force, which tends to produce motion. There is, however,, a limit beyond which the force of friction cannot increase. If the, applied force exceeds this limit, the force of friction cannot balance, it and the body begins to move in the direction of the applied force., This maximum value of frictional force, which comes into play,, when a body just begins to slide over the surface of the other, body, is known as limiting friction., , F (Friction), , Condition of, impending motion, , Limiting, friction, Static, condition, , Kinetic, condition, , P (Applied force), , Fig. 2., , h. Write down D’Alembert’s Principle., Ans. D’Alembert’s principle states that the net external force acting on, , the system and the resultant inertia force are in equilibrium., i. Differentiate between stable and unstable equilibrium.
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SP–8 C (CE-Sem-3), , www.aktutor.in, , Solved Paper (2019-20), , Ans., , S. No., , Stable Equilibrium, , Unstable Equilibrium, , 1., , A body is said to be in stable, equilibrium, if it returns, back to its original position,, after it is slightly displaced, from its position of rest., , A body is said to be in an unstable, equilibrium, if it does not return, back to its original position and, heels farther away, after slightly, displaced from its position of rest., , 2., , This happens when some, additional force sets up due, to displacement and brings, the body back to its original, position., , This happens when the additional, force moves the body away from, its position of rest., , 3., , A smooth cylinder, lying in a, concave surface, is in stable, equilibrium., , A smooth cylinder lying on a, convex surface is in unstable, equilibrium., , j. State parallel axis theorem. Define radius of gyration., Ans., , 1. Parallel Axis Theorem : Parallel axis theorem states that the, moment of inertia about any axis in the plane of an area is equal, to the sum of moment of inertia about a parallel centroidal axis, and the product of area and square of the distance between the, two parallel axis., 2. Radius of Gyration : Radius of gyration is the distance which is, when squared and multiplied by area gives the moment of inertia, of that area., Section-B, 2. Attempt any three of the following :, (10 × 3 = 30), a. State and prove Lami’s theorem., The greatest and least resultant of two forces acting on, body are 35 kN and 5 kN respectively. Determine the, magnitude of the forces. What would be the angle between, these forces if the magnitude of the resultant is stated to, be 25 kN ?, Ans., , A. Statement of Lami’s Theorem : Lami’s theorem states that if, three forces acting at a point are in equilibrium, then each force, will be proportional to the sine of the angle between the other two, forces., B. Proof of Lami’s Theorem :, 1. The three forces acting on a point are in equilibrium and hence, they can be represented by the three sides of the triangle taken in, the same order.
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Engineering Mechanics, , SP–9 C (CE-Sem-3), , www.aktutor.in, , 2. Now draw the force triangle as shown in Fig. 3(b)., 3. Now applying sine rule, we get, , P, Q, R, , =, sin(180 ), sin(180 ) sin(180 ), 4. This can also be written as,, P, Q, R, , =, sin , sin sin , Q, , , , , O, , P, , (a), , R, , (1, 8, , R, , 0°, –, , ), , , , Q, , (180° – ), (180° – ), , O, , , , P, , , (b), Fig. 3., , C. Numerical :, Given : Maximum resultant = 35 kN, Minimum resultant = 5 kN, To Find : Angle between two forces, magnitude of forces., 1. Let P and Q be the two forces and be the angle of indication, between them. According to the parallelogram law of forces, the, resultant R is, R2 = P2 + Q2 + 2PQ cos , 2. The resultant will be maximum when the forces are collinear and, in the same direction, i.e., = 0°. The gives, R2 =, , , P 2 Q2 2 PQ cos 0 P 2 Q 2 2 PQ P Q, 35 = P + Q, , ...(1)
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SP–10 C (CE-Sem-3), , www.aktutor.in, , Solved Paper (2019-20), , 3. The resultant will be minimum when the forces are collinear and, act in the opposite direction, i.e., = 180°. That gives, R = P 2 Q 2 2PQ cos 180 P 2 Q 2 2 PQ P Q, , 5= P–Q, ...(2), 4. From the eq. (1) and eq. (2), we get, P = 20 kN and Q = 15 kN, 5. Let be the angle between the force P = 20 kN and Q = 15 kN, when their resultant is 25 kN, then, 252 = 202 + 152 + 2 × 20 × 15 cos , 625 = 400 + 225 + 600 cos , cos = 0; = 90°, Thus the given system of forces is at right angles to each other, when the resultant is 25 kN., b. Calculate the centroid of a semi-circular ring of radius ‘r’,, using method of moments., Ans., , 1. Consider a semicircular arc of radius r as shown in Fig. 4., Y, dL, d, , r, , , (– r, 0), , y, , X, , (r, 0), Fig. 4., , 2. Let us take an elemental strip of thickness dL at a distance ‘y’ from, the X-axis., 3. Solving the problem using polar co-ordinates., Integral to be evaluated is yc =, , ydL, dL, , 4. From Fig. 4. y = r sin and dL = r d, , , (r sin )rd, yc =, , 0, , , rd, , =, , r 2[– cos ]0, r[ ]0, , 0, , 2r, r 2[cos cos 0], r 2[ 1 1], =, =, =, , r, r, 5. Thus the centroid of a semicircle of radius R is at a distance 2r/, from the base., c. Find moment of inertia of the Fig. 5 about X-X axis,, thickness of member is 20 mm.
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Engineering Mechanics, , www.aktutor.in, , SP–11 C (CE-Sem-3), , Y, B, , C, , 20 mm, , A, , F, , G, , E, , H, 160 mm, Fig. 5., , 20 mm, , D, 20 mm, , 150 mm, , X, , Ans., , Given : Thickness of member = 20 mm, To Find : Moment of inertia about XX axis., 1. We know that, moment of inertia of rectangular section about its, base = bd3/3, 2. Moment of inertia of hatch portion = Moment of inertia of rectangle, ABCD – Moment of inertia of rectangle EFGH about its base., =, , 200 1503 160 1303, , 107.83 106 mm 4, 3, 3, , d. Differentiate between rectilinear and curvilinear motion., Also derive the expression for the horizontal range, time, of flight and maximum height of a projectile with initial, velocity u and inclined at an angle “” with the horizontal., Ans., , A. Difference :, S. No., , Rectilinear Motion, , Curvilinear Motion, , 1., , The motion of the body along, a straight line is called, rectilinear motion., , The motion of the body along a, curved path is called curvilinear, motion., , 2., , It is also kno wn as one, dimensional motion., , It is also kno wn as multi, dimensional motion., , 3., , Equatio ns of motio n for, rectilinear motion are given, by,, v = u + at, s = ut + (1/2) at2, v2 = u2 + 2as, , Equatio ns o f mo tio n fo r, curvilinear motion are given by,, w = w0 + t, = w0t + (1/2) t2, w2 = w02 + 2, , 4., , Example : A ball thrown, vertically upward, a car, travelling on a straight road., , Example : A golf ball hit from the, ground, a motion travelling on a, curved road.
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SP–12 C (CE-Sem-3), , www.aktutor.in, , Solved Paper (2019-20), , B. Expression :, 1. Equation of Motion for Projectile Motion :, i. Motion along the X-direction (Uniform Motion) :, ax = 0, ...(1), vx = v0 cos , ...(2), x = (v 0 cos ) t, ...(3), ii. Motion along the Y-direction (Uniform Accelerated, Motion) :, ay = – g, ...(4), vy = v0 sin – gt, ...(5), vy2 = (v 0 sin )2 – 2gy, y = (v 0 sin ) t –, , ...(6), , 1 2, gt, 2, , ...(7), , Y, vy, , v 0 sin , , v, , vx, , vx, , v0, , v x, v y, , , , v, X, , O v cos , 0, Fig. 6. Projectile motion., , 2. Derivation of Various Terms :, i. Time Taken to Reach Maximum Height and Time of Flight :, a. When the particle reaches the maximum height, we know, that the vertical component of velocity i.e., v y is zero., Therefore, from the eq. (5), we have, 0 = v0 sin – gt, b. Hence, the time taken to reach the maximum height is,, , v0 sin , ...(8), g, Since the time of ascent is equal to the time of descent, the, total time taken for the projectile to return to the same level, of projection is,, t=, , c., , 2 v0 sin , g, ii. Maximum Height Reached :, a. Substituting the value of time of ascent in the eq. (7), we get, T=, , v sin 1 v 0 sin , y = v 0 sin 0, 2 g , , , g, g, , 2
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Engineering Mechanics, , www.aktutor.in, , SP–13 C (CE-Sem-3), , v02 sin 2 1 v02 sin 2 , v2, g, = 0 sin 2 , 2, , g, 2 , g, , 2g, Hence, the maximum height reached is,, =, , b., , hmax =, , v02 sin 2 , 2g, , iii. Range :, a. The horizontal distance between the point of projection and, point of return of projectile to the same level of projection is, termed as range., b. Hence, range is obtained by substituting the value of total, time of flight in the eq. (3),, , c., , 2 v sin , R = (v 0 cos )T = (v0 cos ) 0, , g, , , Since, sin 2 = 2 sin cos , we can write,, v2 sin 2, R= 0, g, , e. State Work Energy principle., A uniform cylinder of 125 mm radius has a mass of 0.15 kg., This cylinder rolls without slipping along a horizontal, surface with a translation velocity of 20 cm/sec. Determine, its total kinetic energy., Ans., , A. Statement of Work-Energy Principle : Work-energy principle, states that the change in kinetic energy of a body during any, displacement is equal to the work done by the net force acting on, the body or we can say that work done is equal to change in kinetic, energy of the body., B. Numerical :, Given : Translation velocity, v = 20 cm/sec = 0.20 m/sec,, Mass of cylinder, m = 0.15 kg, Radius of cylinder, R = 0.125 m., To Find : Kinetic energy., 1. Total KE of rotating body is given by,, , 1, 1, I 2 m v 2, 2, 2, 2. Moment of inertia of solid cylinder,, MR 2, 0.1252, I=, = 0.15 , 1.172 10 3, 2, 2, v, 0.20, 3. Angular velocity, =, , 1.6 rad/ sec, R 0.125, Total KE =, , ...(1)
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SP–14 C (CE-Sem-3), , www.aktutor.in, , Solved Paper (2019-20), , 4. Substituting these values in eq. (1), we get, , 1, 1, 1.172 10 3 1.62 0.15 0.22, 2, 2, = 4.5 × 10– 3 Joule, , Total KE =, , Section-C, 3. Attempt any one part of the following :, (10 × 1 = 10), a. Explain how a wedge is used for raising heavy loads. Also, mention the principle., A body resting on a rough horizontal plane required a pull, of 24 N inclined at 30° to the plane just to move it. It was, also found that a push of 30 N at 30° to the plane was just, enough to cause motion to impend. Make calculations for, the weight of body and the coefficient of friction., Ans., , A. Wedge :, 1. A wedge is, usually, of a triangular or trapezoidal in cross-section., It is, generally, used for slight adjustments in the position of a, body i.e., for tightening fits or keys for shafts. Sometimes, a wedge, is also used for lifting heavy weights as shown in Fig. 7., F, , E, , Body, B, D, , G, , A, , Wedge, , C, , P, , Fig. 7., , 2. It will be interesting to know that the problems on wedges are, basically the problems of equilibrium on inclined planes., 3. Thus these problems may be solved either by the equilibrium, method or by applying Lami’s theorem., 4. Now consider a wedge ABC, which is used to lift the body DEFG, Let,, W = Weight fo the body DEFG., P = Force required to lift the body., µ = Coefficient of friction on the planes AB, AC, and DE such that, tan = µ., 5. A little consideration will show that when the force is sufficient to, lift the body, the sliding will take place along three planes AB, AC, and DE will also occur as shown in Fig. 8(a) and (b).
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Engineering Mechanics, F, , E, R1, , , R2 , , W, D, A, , SP–15 C (CE-Sem-3), , www.aktutor.in, , G, , , B, , B, A, , P, , C, , R2, , R3 , (a) Forces on the body DEFG, (b) Forces on the wedge ABC, Fig. 8., , 6. The three reactions and the horizontal force (P) may now be, found out either by graphical method or analytical method as, discussed below :, Analytical Method :, i. First of all, consider the equilibrium of the body DEFG and resolve, the forces W, R1 and R2 horizontally as well as vertically., ii. Now consider the equilibrium of the wedge ABC, and resolve the, forces P, R2 and R3 horizontally as well as vertically., B. Numerical :, Given : Pull = 24 N, Push = 30 N, Angle inclination with horizontal, plane (α) = 30°, To Find : Weight of body and the coefficient of friction., 1. Let,, , 2., , 3., 4., 5., , W = Weight of the body., R = Normal reaction., = Coefficient of friction., Firstly we consider a pull of 24 N acting on the body. We know, that in this case, the force of friction (F1) will act towards left as, shown in Fig. 9(a)., Resolving the forces horizontally,, F1 = 24 cos 30° = 24 × 0.866 = 20.785 N, Resolving the forces vertically,, R1 = W – 24 sin 30° = W – 24 × 0.5 = W – 12, We know that the force of friction (F1),, 20.785 = R1 = (W – 24), ...(1), 24 N, , R1, , 30 N, , R2, , 30°, , 30°, , F1, , F2, W, (a), , W, (b ), , Fig. 9., , 6. Now consider a push of 30 N acting on the body. We know that in, this case, the force of friction (F2) will act towards right as shown, in Fig. 9(b)., 7. Resolving the forces horizontally,, F2 = 30 cos 30° = 30 × 0.866 = 25.98 N
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SP–16 C (CE-Sem-3), , www.aktutor.in, , Solved Paper (2019-20), , 8. Resolving the forces horizontally,, R2 = W + 30 sin 30° = W + 30 × 0.5 = W + 15, 9. We know that the force of friction (F2),, 25.98 = R2 = (W + 15), 10. Dividing eq. (1) and eq. (2), we get, , ...(2), , (W 24) W 24, 20.785, , =, (W 15) W 15, 25.98, 20.785 W + 311.775 = 25.98 W – 623.52, 5.195 W = 935.295, Weight of body, W = 935.295/5.195 = 180.04 N, 11. Now substituting the value of W in eq. (1), we get, 20.785 = (180.04 – 24) = 156.04 , Coefficient of friction, = 20.785/156.04 = 0.133 N, b. A ladder 5m long rests on a horizontal ground and leans, against a smooth vertical wall at an angle 70° with the, horizontal. The weight of the ladder is 900N and acts at its, middle. The ladder is at the point of sliding, when a man, weighing 750N stands 1.5m from the bottom of the ladder., Calculate coefficient of friction between the ladder and, the floor., Ans., , Given : Length of the ladder (l) = 5 m; Angle which the ladder, makes with the horizontal () = 70°; Weight of the ladder, (w1) = 900 N; Weight of man (w2) = 750 N and distance between, the man and bottom of ladder = 1.5 m., To Find : Calculate coefficient of friction between ladder and the, floor., 1. Forces acting on the ladder are shown in Fig. 10,, B, , 750 N, , 5m, , 1.5 m, 900 N, Ff, , A, , 70°, , C, , RA, Fig. 10., , 2. Let,, , f = Coefficient of friction between ladder and floor., RA = Normal reaction at point A., 3. Resolving the forces vertically, RA = 900 + 750 = 1650 N, ...(1), 4. Force of friction at A, Ff = f × RA = f × 1650, ...(2)
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Engineering Mechanics, , SP–17 C (CE-Sem-3), , www.aktutor.in, , 5. Now taking moments about B, and equating the same,, RA × 5 sin 20° = (Ff × 5 cos 20°) + (900 × 2.5 sin 20°), + (750 × 3.5 sin 20°), = (Ff × 5 cos 20°) + (4875 sin 20°), 6. Now substituting the values of RA and Ff from eq. (1) and eq. (2),, we get, 1650 × 5 sin 20° = (f × 1650 × 5 cos 20°) + (4875 sin 20°), 7. Dividing both sides by 5 sin 20°,, 1650 = (f × 1650 × cot 20°) + 975, = (f × 1650 × 2.7475) + 975 = 4533.375 f + 975, , f = 0.15, 4. Attempt any one part of the following :, (10 × 1 = 10), a. Draw the SF and BM diagram for the simply supported, beam loaded as shown in Fig. 11., 2 kN, 4 kN, 2 kN, A, B, C, D, E, 1m, 1m, 1m, 1m, RA, RB, Fig. 11., Ans., , Given : Load on beam as shown in Fig. 11., To Find : Draw SFD and BMD., 2 kN, , 4 kN, , 2 kN, , D, , E, , B, , A, C, RA = 4 kN, , 1m, , 1m, , 1m, , 1m, , (a) Loading diagram, , 4 kN, +, 2 kN, , 2 kN, , 2 kN, , –, 4 kN, , 4 kN, , 4 kN-m, , RB = 4 kN, , (b) SFD, 6 kN-m, 4 kN-m, , +, , (c) BMD, Fig. 12. SFD and BMD.
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SP–18 C (CE-Sem-3), , www.aktutor.in, , Solved Paper (2019-20), , 1. Calculate the Support Reaction :, i. To determine the support reactions taking moments about A, we, get, RB × 4 = 2 × 1 + 4 (1 + 1) + 2(1 + 1 + 1) = 2+ 8 + 6 = 16, RB = 16/4 = 4 kN, ii. Fy = 0 R A + RB = 2 + 4 + 2 = 8 kN, , RA = 8 – RB = 8 – 4 = 4 kN, 2. Shear Force Calculations :, SB – E = + 4 kN, SE – D = 4 – 2 = 2 kN, SD – C = 2 – 4 = – 2 kN, SC – A = – 2 – 2 = – 4 kN, SF at point A, SA = – 4 + 4 = 0 kN, SF diagram is shown in Fig. 12(b)., 3. Calculation of Bending Moment :, MB = 0, ME = 4 × 1 = 4 kN-m, MD = 4(1 + 1) – 2 × 1 = 8 – 2 = 6 kN-m, MC = 4(1 + 1 + 1) – 2 (1 + 1) – 4 × 1 = 12 – 4 – 4 = 4 kN-m, MA = 4(1 + 1 + 1 + 1) – 2(1 + 1 + 1) – 4(1 + 1) – 2 × 1, = 16 – 6 – 8 – 2 = 0, BM diagram is shown in Fig. 12(c)., b. Define and explain the term imperfect truss., Fig. 13 shows a framed of 4 m span and 1.5 m height, subjected to two point loads at B and D. Find the forces in, all the members of the structure., 8 kN, , B, 1.5 m, , A, , , 2m, , D, , C, 2m, , 12 kN, , Fig. 13., Ans., , A. Imperfect Truss :, 1. A frame in which number of members and number of joints are not, given by n = 2j – 3 is known as imperfect frame. This means that, number of members in an imperfect frame will be either more or, less than (2j – 3)., 2. If the number of members in a frame are less than (2j – 3), then the, frame is known as deficient frame., 3. If the number of members in a frame are more than (2j – 3), then, the frame is known as redundant frame.
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Engineering Mechanics, , www.aktutor.in, , SP–19 C (CE-Sem-3), , B. Numerical :, Given : Span = 4 m, Horizontal load, H = 8 kN, Vertical load,, V = 12 kN., To Find : Forces in all the members., 1. Horizontal reaction, Fx = 0,, 2. Vertical reaction, Fy = 0,, , HA = 8 kN (), , VA + VC = 12 kN, , ...(1), , 3. Taking moments about A and equating the same,, VC × 4 = (8 × 1.5) + (12 × 2) = 36, VC = 36/4 = 9 kN (), 4. From eq. (1), we get VA = 12 – 9 = 3 kN (), 5. From the geometry of the Fig. 14, we get, tan = 1.5/2 = 0.75 or = 36.9°, Similarly, sin = sin 36.9° = 0.6 and cos = cos 36.9° = 0.8, 6. Consider the equilibrium at joint A,, i., Fx = 0, FBA, 8 kN, , A, , B, , B, , B, FBC, , , FAD, , D, , FBD, , , , D, , C, , FCD, , A, , 12 kN, , 12 kN, , FDA, , D F, DC, , 3 kN, 9 kN, (a), , (b ), Fig. 14., , 12 kN, (c ), , FAD = 8 kN + FBA cos = 8 + cos 36.9° FBA, ii., , C, , ...(2), , Fy = 0, FBA sin = 3 kN, FBA = 3/sin 36.9° = 5 kN, , iii. From eq. (2) we get, FAD = 8 + 5 × 0.8 = 12 kN, 7. Consider the equilibrium at joint C,, i., Fx = 0, FCD = FBC cos = FBC cos 36.9°, ii., , Fy = 0, FBC sin = 9 kN, FBC = 9/sin 36.9° = 15 kN, , iii. From eq. (3) we get, FCD = 15 × 0.8 = 12 kN, 8. Considering the equilibrium of joint D, Fy = 0, FDB = 12 kN, 9. Now tabulate the results as given below :, , ...(3)
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SP–20 C (CE-Sem-3), , www.aktutor.in, , Solved Paper (2019-20), , S. No., , Member, , Magnitude of force in kN Nature of Force, , 1., , AB, , 5.0, , Compression, , 2., , AD, , 12.0, , Tension, , 3., , BC, , 15.0, , Compression, , 4., , CD, , 12.0, , Tension, , 5., , BD, , 12.0, , Tension, , 5. Attempt any one part of the following :, (10 × 1 = 10), a. Explain the principle of virtual work. An overhanging beam, ABC of span 3 m is loaded as shown in Fig. 15. Using the, principle of virtual work, find the reactions at A and B., 1 kN, , 2 kN/m, , C, , A, B, 2m, , 1m, Fig. 15., , Ans., , A. Principle :, 1. For the particle or rigid body to remain in equilibrium in the, displaced position also, we know that the resultant force acting on, it must be zero. Thus, we say that work done in causing this, virtual displacement is also zero. This is known as principle of, virtual work., 2. For a system of concurrent forces F1, F1, .... F1, the virtual work, done is given by,, U = F1 r + F2 r + ..... + Fn r, = (F1 + F2 + ..... + Fn) r, =, , , , , , Fr, , 3. As a system of concurrent force can be replaced by a single resultant, force, the virtual work done is equal to the work done by the, resultant., 4. For the body to remain in equilibrium in the displaced position, we, know that the resultant must be zero. Hence, virtual work done, in causing this virtual displacement is also zero, i.e.,, , , , , , U = F r 0, 5. The necessary and sufficient condition for the equilibrium of a, particle is zero virtual work done by all external forces acting on, the particle during any virtual displacement consistent with the, constraints imposed on the particle.
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Engineering Mechanics, , www.aktutor.in, , SP–21 C (CE-Sem-3), , 6. Similarly, for a rigid body, we can write the principle of virtual, work as, U =, , F r M 0, , B. Numerical :, Given : Span, AB = 2 m and span, BC = 1 m, Concentrated load,, W = 1 kN, Intensity of UDL, w = 2 kN/m, To Find : Reaction at A and B., 1. From the geometry of the Fig. 16, we find that when the virtual, upward displacement of the beam at B is y, then the virtual upward, displacement of the beam at C is 1.5 y as shown in Fig. 16., 2. Total virtual work done by the two reactions RA and RB, = +[(RA × 0) + (RB × y)] = + RB × y, (Plus sign due to reactions acting upwards), 1 kN, , 2 kN/m, , C, , A, B, 2m, , 1m, , 1.5 y, , y, A, , B, RA, , C, RB, , Fig. 16., , 3. Total virtual work done by the point load at C and uniformly, distributed load between A and C, , , 0 1.5 y, , = (1 1.5 y) 2 , 3 = – (1.5y + 4.5y) = – 6y, , , 2, , (Minus sign due to loads acting downwards), 4. We know that from the principle of virtual work, the algebraic, sum of the total virtual works done is zero, therefore, RB × y – 6y = 0, RB = 6 kN, 5. Fy = 0,, , RA + RB = 2 × 3 + 1 = 7 kN,, , RA = 7 – 6 = 1 kN, , b. In a reciprocating pump, the lengths of connecting rod, and crank is 1125 mm and 250 mm respectively. The crank, is rotating at 420 rpm. Find the velocity with which the, piston will move, when the crank has turned through an, angle of 40° from the inner dead centre., Ans., , Given : Radius of the crank (r) = 250 mm = 0.25 m; Length of, connecting rod (l) = 1125 mm = 1.125 m; Angular rotation of, crank (N) = 420 rpm and Angle () = 40°, To Find : Velocity of piston.
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SP–22 C (CE-Sem-3), , Solved Paper (2019-20), , www.aktutor.in, , 1. We know that angular velocity of the crank,, , 2 420, = 14 rad/sec, 60, 2. From the geometry of the Fig. 17, we get, 1 =, , BM AB sin 40 0.25 0.643, = 0.143 or = 8.125°, , , BC, BC, 1.125, , sin =, , vB, B, m, m, , , vC, , C, , 0, 25, , 1125 mm, , 40°, A, D M, , Fig. 17., , 3. We know that velocity of the piston, vc = 1(l sin + r cos tan ), = 14[1.125 sin (8.22°) + 0.25 cos 40° tan (8.22°)] = 8.286 m/sec, 6. Attempt any one part of the following :, (10 × 1 = 10), a. Derive an equation for moment of inertia of triangle, centroidal axis and about its base., Ans., , 1. Consider an elemental strip at a distance y from the base AA. Let, dy be the thickness of the strip and dA its area. Width of this strip, is given by,, b1 =, , y, , (h – y), b = 1 – b, h, h, , , h, , dy, b1, , y X, , X, h/3, , A, , A, , b, Fig. 18., , 2. Moment of inertia of this strip about AA, = y2dA, = y2 b1 dy, , y, 2 , = y 1 – bdy, h, , 3. Moment of inertia of the triangle about AA,, h, , IAA =, , y, 2, 0 by 1 – h dy =, , h, , , , b y, 0, , 2, , –, , y3 , dy, h
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Engineering Mechanics, , www.aktutor.in, , SP–23 C (CE-Sem-3), , h, , y3, y4 , = b –, , 4h 0, 3, bh3, 12, 4. By parallel axis theorem,, IAA = IXX + Ay2, IXX = IAA – Ay2, IAA =, , =, , bh3 1 h , – bh , 12 2 3 , , 2, , ( y = h/3), , bh3 bh3, , 12, 18, bh3, =, 36, , =, IXX, , b. An I-section is made up of three rectangles as shown in, Fig. 19. Find the moment of inertia of the section about the, horizontal axis passing through the centre of gravity of, the section., 60 mm, 20 mm, , 2, , 100 mm, , 20 mm, , 1, , 3, , 20 mm, , 100 mm, Fig. 19., Ans., , Given : I-section is shown in Fig. 19., To Find : Moment of inertia about its centroidal axis., 1. As the section is symmetrical about Y-Y axis, therefore its centre, of gravity will lie on this axis. Let bottom face of the bottom flange, be the axis of reference., 2. Rectangle-1 :, Area,, a1 = 60 × 20 = 1200 mm, , y1 = 20 + 100 + 20/2 = 130 mm, and, 3. Rectangle-2 :, Area,, a2 = 100 × 20 = 2000 mm2
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SP–24 C (CE-Sem-3), , www.aktutor.in, , Solved Paper (2019-20), , y2 = 20 + 100 / 2 = 70 mm, and, 4. Rectangle-3 :, Area,, a3 = 100 × 20 = 2000 mm2, and, , y3 = 20 / 2 = 10 mm, , 5. Centre of gravity of the section from bottom face,, , y =, =, , a1 y1 a2 y2 a3 y3, a1 a2 a3, (1200 130) (2000 70) (2000 10), mm, 1200 2000 2000, , y = 60.8 mm, Y, 1, , 2, X, , X, y, 3, , X, Fig. 20., , 5. Moment of inertia of rectangle (1) about an axis through its centre, of gravity,, , 60 (20)3, = 40 × 103 mm4, 12, Distance between centre of gravity of rectangle (1) and X-X axis, of whole section,, h1 = 130 – 60.8 = 69.2 mm, 6. From parallel axis theorem moment of inertia of rectangle (1), about X-X axis,, = IG1 + a1h12 = (40 × 103) + [1200 × (69.2)2] = 5786.37 × 103 mm4, 7. Similarly, moment of inertia of rectangle (2) about an axis through, its centre of gravity,, IG1 =, , 20 (100)3, = 1666.67 × 103 mm4, 12, Distance between centre of gravity of rectangle (2) and X-X axis, h2 = 70 – 60.8 = 9.2 mm, 8. Moment of inertia of rectangle (2) about X-X axis,, IG2 =, , = IG2 + a2 h22 = (1666.67 × 103) + [2000 × (9.2)2], = 1836.95 × 103 mm4, 9. Moment of inertia of rectangle (3) about an axis through its centre, of gravity,
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Engineering Mechanics, , SP–25 C (CE-Sem-3), , www.aktutor.in, , 100 (20)3, = 66.67 × 103 mm4, 12, Distance between centre of gravity of rectangle (3) and X-X axis,, h3 = 60.8 – 10 = 50.8 mm, Moment of inertia of rectangle (3) about X-X axis,, = IG3 + a3h32 = (66.67 × 103) + [2000 × (50.8)2], = 5227.95 × 103 mm4, Now moment of inertia of the whole section about X-X axis,, IXX = 5786.37 × 103 + 1836.95 × 103 + 5227.95 × 103, = 12851.27 × 103 mm3, IG3 =, , 7. Attempt any one part of the following :, (10 × 1 = 10), a. A body of mass 20 kg moving towards with a velocity of, 16 m/sec strikes with another body of 40 kg mass moving, towards left with 50 m/sec. Determine, i. Final velocity of the two bodies., ii. Loss in kinetic energy due to impact., iii. Impulse acting on either body during impact., Take coefficient of restitution as 0.65, Ans., , Given : m1 = 20 kg, u1 = 16 m/sec, m2 = 40 kg, u2 = 50 m/sec,, e = 0.65, To Find : i. Find velocities of both bodies., ii. Loss in kinetic energy., iii. Impulse acting on either body during impact., A. Final Velocity of the Two Bodies :, (, , ) u1, , (, , m1, , ) u2, m2, , (, , ) v1, m1, , (, , ) v2, m2, , After collision, Before collision, Fig. 21., , 1. Applying movement conservation of system, m1u1 + m2u2 = m1v1 + m2v2, 20 × 16 + 40 × (– 50) = 20(– v1) + 40(– v2), v1 + 2v2 = 84, ...(1), v1 v2, 2. We know that, – e =, u1 u2, , v 1 ( v 2 ), 16 ( 50), – v1 + v2 = – 42.9, v1 – v2 = 42.9, 3. Solving the eq. (1) and eq. (2), we get, v1 = 56.6 m/sec, and v2 = 13.7 m/sec, – 0.65 =, , ...(2)
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SP–26 C (CE-Sem-3), , www.aktutor.in, , Solved Paper (2019-20), , B. Loss in Kinetic Energy :, Loss in kinetic energy = Kinetic energy before collision – Kinetic, energy after collision, 1, 1, 1, 1, , = m1u12 m2u22 m1 v12 m2 v 22 , 2, , 2, 2, 2, , 1, 1, 1, 1, 20 (16)2 40( 50)2 20 ( 56.6)2 40 ( 13.7)2, 2, 2, 2, 2, = 2560 + 50000 – 32035.6 – 3753.8 = 16770.6 J, C. Impulse :, 1. Impulse of first body,, I1 = m1v1 = m1(v 1 – u1), = 20(– 56.6 – 16) = – 1452 kg-m/sec, 2. Impulse of second body,, I2 = m2v2 = m2(v 2 – u2) = 40[– 13.7 – (– 50)] = – 1452 kg-m/sec, , =, , b. A particle start with velocity u and the accelerationvelocity relationship is prescribed as a = – kv where k is a, constant. Set up an express ion that prescribes the, displacement time relation for the particle., Ans., , Given : Acceleration, a = – kv, Initial velocity = u., To Find : Expression for displacement and time., 1. Acceleration of particle is given by, a =, , dv, kv, dt, , ...(1), , 2. Upon rearranging, we get, , dv, = – kdt, v, 3. Integrating both side, ln v = – kt + C1, ...(2), Since v = u at t = 0, we have C1 = ln u. Therefore, the eq. (2) can be, written as, v, ln = – kt, u, v, or, = e– kt, u, , v = ue – kt, ...(3), 4. Further, we can write velocity as :, dx, ue kt, dt, On rearranging, dx = ue – ktdt, 5. Integrating both side, we get, v=, , x= , , u kt, e C2, k, , ...(4)
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Engineering Mechanics, , www.aktutor.in, , 6. At t = 0, x = 0, we get, , u, k, 7. Put the value of C2 in eq. (4), we get, C2 =, , x=, , u, [1 e kt ], k, , , , SP–27 C (CE-Sem-3)
