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Youth Competition Times MECHANICAL ENGINEERING CAPSULE Useful for All Competitive AE/JE Exam : (cid:1) UPPSC AE (cid:1) UKPSC AE (cid:1) BPSC AE (cid:1) CGPSC AE (cid:1) MPPSC AE (cid:1) RPSC AE (cid:1) UPSSSC JE (cid:1) SSC JE (cid:1) DMRC JE (cid:1) LMRC JE (cid:1) JMRC JE (cid:1) BMRC JE (cid:1) DSSSB JE (cid:1) UP Jal Nigam JE (cid:1) UKSSSC JE (cid:1) RSMSSB JE (cid:1) MPPEB SUB ENGINEER (cid:1) HPSSC JE (cid:1) HSSC JE (cid:1) Punjab JE (cid:1) CGPEB SUB ENGINEER (cid:1) BSSC JE (cid:1) DRDO JE (cid:1) ISRO JE (cid:1) UPPCL AE/JE (cid:1) UPRVUNL AE/JE (cid:1) JVUNL JE (cid:1) SAIL JE (cid:1) GAIL JE (cid:1) BHEL JE (cid:1) NTPC JE (cid:1) DFCCIL (cid:1) COAL INDIA LTD. JE (cid:1) RRB JE etc. Chief Editor A.K. Mahajan Compiled & Written by Er. Brijesh Kuamr Computer Graphics by Balkrishna, Charan Singh Editorial Office Youth Competition Times 12, Church Lane Prayagraj-211002 Mob. : 9415650134 Email : [email protected] website : www.yctbooks.com Publisher Declaration Edited and Published by A.K. Mahajan for YCT Publications Pvt. Ltd. and printed by Laxmi Narayan Printing Press, Prayagraj. In order to publish the book, full care has been taken by the editor and the publisher, still your suggestions and queries are welcomed. Rs. : 150/- In the event of any dispute, the Judicial area will be Prayagraj. INDEX (cid:1) Mechanics ...................................................................................................... 3-8 (cid:1) Strength of Material ................................................................................... 9-17 (cid:1) Theory of Machine ................................................................................... 18-34 (cid:1) Machine Design ........................................................................................ 35-43 (cid:1) Material Science ....................................................................................... 44-50 (cid:1) Production Engineering ........................................................................... 51-71 (cid:1) Metrology .................................................................................................. 72-75 (cid:1) Industrial ................................................................................................... 76-86 (cid:1) Engineering Drawing ............................................................................... 87-99 (cid:1) CAD-CAM, NC & CNC Machine ...................................................... 100-106 (cid:1) Robotics & Mechatronics .................................................................... 107-111 (cid:1) Fluid Mechanics & Hydraulic Machine ............................................. 112-126 (cid:1) Hydraulic Machinery ........................................................................... 127-136 (cid:1) Thermodynamics ................................................................................. 137-146 (cid:1) Thermal Power Plant ........................................................................... 147-156 (cid:1) IC Engine .............................................................................................. 157-163 (cid:1) Refrigration & Air Conditioning ........................................................ 164-169 (cid:1) Heat and Mass Transfer ...................................................................... 170-176 Mechanical Engineering Capsule 2 YCT Mechanics (cid:1) Newton's law of motion (cid:1) Principle of transmissibility of force– First law of It states that everybody continues in When a force acts on a body, this force may be motion the states of rest or of uniform motion, assumed to be acting on all particles of the body in a straight line, unless it is acted which lie on the line of action of the force. upon by some external force to change (cid:1) Parallelogram law of forces– that state. Second law dP R = P2+Q2+2PQcosθ F∝ of motion dt F=ma Third law The forces of action and reaction of motion between bodies in contact have same magnitude, same line of action but opposite in direction. Newton's Every particle of matter attracts every law of other particle of matter a force directly gravitation proportional to the product of the masses and inversely proportional to Qsinθ Psinθ tanα= tanβ= the square of the distance between P+Qcosθ Q+Pcosθ then. Case Resultant m m F=G 1 2 I. If two forces are like parallel R = P + Q r2 θ = 0o G = 6.67 × 10–11 N-m2/kg2 II. If forces are unlike parallel R = P – Q (cid:1) Types of forces θ = 180o Coplanar Line of action of all forces lying on III. If forces are perpendicular forces single plane R = P2+Q2 θ = 90o None- Line of action of all forces are not coplanar lying on a single plane. IV. If magnitude of two forces α = θ/2 forces are same Concurrent Line of action of all forces passes (cid:1) Law of triangle of forces– forces through a single point. If three forces acting a point are in equilibrium, then None Line of action of all forces do not pass their magnitude & directions can be represented by concurrent through a single point. successive sides of a triangle. forces (cid:1) Lami's theorem– Collinear Line of action of all forces passes forces through a single line. Parallel Line of action of all forces are parallel forces to each other. (a) Like Line of action of all forces are parallel P Q R parallel to each other in same direction = = forces sinα sinβ sinγ (b) Unlike Line of action of all forces are parallel (cid:1) Law of polygon of forces– If all the forces acting at parallel to each other in different direction. a point can be represented by successive sides of a forces closed polygon, then forces will be in equilibrium. Mechanical Engineering Capsule 3 YCT (cid:1) Resolution of forces (cid:1) Geometrical representation of moment of a force– • If magnitude & direction of forces are known then there will be only one resultant of definite Suppose a force 'P' is acting along AB on a body. Body magnitude & direction. is free to rotate about a fixed point. Rsinβ Rsinα Moment of a force = 2 × area of ∆OAB P= Q= sin(α+β) sin(α+β) (cid:1) Resolution of force in two perpendicular Varignon's Algebric sum of moment of two direction– theorem coplanar forces about a point is equal to the moment of resultant force about that point. Principle If algebric sum of moments of all of moment forces acting on a body about a point is zero, then body will be in state of P=Rcosα Q=Rsinα rotational equilibrium ∑M = 0 (cid:1) Resolution of concurrent coplanar forces– (cid:1) Lever Principle of lever– An ideal lever works on principle of moments when the lever is in equilibrium. Load Effortarm M.A. of lever ⇒ = Effort Loadarm Class I lever • Fulcrum is between effort & load MA≥1 • {maybe} MA<1 Ex. ⇒ Scissors, see-saw, claw ∑Fx =F1cosθ1+F2cosθ2+F3cosθ3+F4cosθ4+... hammer etc. ∑Fy =F1sinθ1+F2sinθ2+F3sinθ3+F4sinθ4+... Class II lever • Load is in between effort & fulcrum Resultant force, (R) = (∑F )2+(∑F )2 x y • MA > 1 (cid:2) Direction of resultant– Ex. ⇒ Wheel barrow, lemon • If resultant is inclined at θ angle with X axis crusher, nut cracker, paper cutter. ∑F Class III lever • Effort is in between fulcrum & y tanθ= ∑F load x (cid:1) Moment– • MA < 1 • Vector quantity Ex.⇒ Sugar tongs, forearm used • Moment of force = Force × Perpendicular distance for lilting a load. Mechanical Engineering Capsule 4 YCT (cid:1) Friction– • Path of projectile is parabola. Time of flight 2usinθ (T) = g Range u2sin2θ (R) = g Maximum height u2sin2θ (H) = 2g Condition for maximum α = 45o range u2 Law of static friction Law of kinetic friction R = max 2g • Frictional force (f ) ∝ • µ >µ S s k Normal reaction (R ). N • Force of dynamic Body projected upward Body projected • Frictional force is friction is to inclined plane downward inclined independent of surface independent of plane area of contact. relative motion. • Frictional force depends • Force of friction is upon surface roughness. opposite to relative • Friction force depends motion. upon materials of surfaces in contact. • T= 2u sin(α−β) • T= 2u sin(α+β)     gcosβ gcosβ • Coefficient of friction (µ) = u2 u2 f • R = • R = f ∝ R , f = µ R , µ= gcos2β gcos2β N N R N sin(2α−β)−sinβ f = Friction force RN = Normal reaction sin(2α+β)+sinβ   • Limiting friction– • Condition for maximum range– flim =µS×RN • For maximum range β α=45o + • Kinetic friction– 2 α=45−β 2 f =µ ×R f >f K K N S K Maximum height Motion of an object µ >µ S K obtained by an object falling freely under • Angle of friction– thrown in upward gravity tanθ=µ θ=tan−1(µ ) S s • Angle of repose– It is angle of inclination of the plane to the horizontal, at which the body just begins to move down the plane. α=φ • Maximum height • Velocity before Angle of inclination of plane = Angle of friction. u2 hitting the ground (cid:1) Projectile motion– H= 2g v= 2gH • Time taken by object to • Time taken by object 1 gx2 reach the ground. to reach the ground. y=x.tanθ− 2 u2cos2θ 2u 2H t= t= g g Mechanical Engineering Capsule 5 YCT (cid:1) System of pulleys– Semicircle 1 r 4r πr2 First system of pulleys Velocity ratio (VR) = 2n 2 3π Second system of pulleys VR = n Quadrant 1 4r 4r πr2 Third system of pulleys VR = 2n – 1 circle 4 3π 3π n = number of pulleys Three 3 4r 4r (cid:1) Motion of a lift– quadrant 4πr2 9π 9π Lift is moving upward Lift is moving circle downward (cid:1) Centre of gravity for given area– a x +a x +a x +........ x= 1 1 2 2 3 3 a +a +a +....... 1 2 3 a y +a y +a y +........ y= 1 1 2 2 3 3 a1+a2+a3+...... • R–mg = ma • mg – R = ma • Centre of gravity for remains part after cut out a • R =m(g+a) • R =m(g−a) lamina– (cid:1) Screw jack– p tanα= πd x= a1x1−a2x2 y= a1y1−a2y2 a −a a −a Effort required 1 2 1 2 (cid:1) Mass moment of inertia– •For raising the load P = W tan (α + φ) Shape Name I • For lowering the load P = W tan (α – φ) Note– Rod mL2 1. When friction is neglected then φ = 0 12 P =Wtanα o Rod mL2sin2θ 2. The efficiency of screw jack– 12 tanα η= tan(α+φ) Ring mR2 3. The efficiency of screw jack is maximum– φ α=45− 2 1−sinφ Disc mR2 η = max 1+sinφ 2 i(cid:1) Centroid of regular plane figure Lamina Area x y Right angle 1 b h Hollow mR2 Triangle 2b.h 3 3 cylinder Rectangle b.h b h Solid mR2 cylinder 2 2 2 Mechanical Engineering Capsule 6 YCT Spherical 2 u = Initial velocity of body, v = Final velocity, mR2 shell 3 t = Time, a = Uniform acceleration, s = Distance covered, ω = Final angular velocity, ω = Initial angular velocity, o Solid 2 θ = Angular displacement, α = Angular acceleration. mR2 sphere 5 • Distance covered in nth second– a S = u + (2n−1) n 2 Rectangular m(a2+b2) (cid:1) Motion of particle in a plane (2D motion)– plate Velocity Acceleration 12 dx d(v ) Square ma2 • (vx) = dt • (ax) = dtx plate dy ( ) 6 • vy = dt • a = d vy y dt (cid:1) Area moment of inertia– • vresultant = v2x +v2y • a = a2 +a2 resultant x y bd3 • About x axis I = Rectangular XX 12 (cid:1) Momentum– P=mv KE= P2 section db3 2m • About y axis IYY = 12 (cid:1) Impulse momentum theorem– Impulse = change in momentum BD3 bd3 Hallow • About x axis IXX = 12 − 12 Impulse(J)=∫Fdt=∆P=Pf-Pi rectangular section • About y axis I = DB3 −db3 (cid:1) Law of conservation of momentum– YY 12 12 If F = 0 then, ext. Circular πD4 initial momentum = final momentum I = I section XX YY 64 m u +m u =m v +m v 1 1 2 2 1 1 2 2 • About an axis (cid:1) Angular momentum– L=Iω passing (cid:1) Conservation of angular momentum– through its bh3 If T = 0 then, ext. centre of I = Triangular G 36 initial angular momentum = final angular momentum gravity and section parallel to the I1ω1 =I2ω2 base (cid:1) Simple harmonic motion– • About the bh3 From From I = base B 12 origin extreme (cid:1) Equation of motion– position Displacement x A sin ωt A cos ωt For linear motion For circular motion Velocity v=±ω A ω cost ωt –A ω • v = u + at • ωw = ω + αt o • v2 = u2 + 2as • ω2 = ω2+2αθ A2−x2 sin ωt o • s=ut+1at2 • θ=ω t+1αt2 Acceleration –ω2x –Aω2 sin ωt –Aω2 2 o 2 cos ωt Mechanical Engineering Capsule 7 YCT (cid:1) Time period for different pendulum– (cid:1) Principle of transmissibility of force– When a Type of pendulum Time period force acts on a body, this force may be assumed to be acting on all particles of the body which lie on Simple pendulum ℓ T=2π the line of action of the force. g (cid:1) Spring-mass system m Principle of • It states that for a body to be in T=2π k virtual equilibrium, the virtual work should be Compound pendulum work zero. k2 +h2 T=2π G If, P , P ........ P = force g.h 1 2 n δ , δ .......... δ = corresponding 1 2 n Conical pendulum ℓcosθ displacement T=2π g M , M ..........M = moment 1 2 n δθ , δθ ........... δθ = Corresponding • Time period of second's pendulum is 2 second. 1 2 n angular displacement • Equivalent length of compound pendulum is– P δ + P δ ...+ M δθ + M δθ + ... = 0 k2 +h2 1 1 2 2 1 1 2 2 L = G (cid:1) Centre of percussion– h (cid:1) Truss– • Point at which a blow may be struck on a suspended body on a suspended body so that the reaction at the Plane truss If all members lie in a single plane support is zero. Space truss Consists of members joined together at their ends to form a stable 3D structure. Plane truss Space truss Statically m = 2j – 3 m = 3j – 6 determinate • Centre of percussion is always below the centre of Statically m > 2J – 3 m > 3j – 6 indeterminate k2 gravity. (ℓ) = G Unstable truss m < 2j – 3 m < 3j – 6 h (cid:1) Collision between two bodies • The distance between the centre of suspension (O) & Perfectly • Initial kinetic energy = Final kinetic the centre of percussion (C) is equal to equivalent elastic energy length (L) of simple pendulum collision • e = 1 L=ℓ+h • Velocity of approach = Velocity of • Centre of suspension (O) and centre of percussion separation (C) are interchangeable • u – u = v – v 1 2 2 1 (cid:1) D'-Alembert's principle– Perfectly • e = 0 • It is used for analyzing the dynamic problem which inelastic • (KE) = (KE) – (KE) loss initial final collision can reduce it into a static equilibrium problem. Partially • 0 < e < 1 • It is an alternative form of Newton's second law of elastic • Velocity of separation = e (velocity of motion. approach) • F = ma (Newton's second law) • v – v = e (u – u ) • F + (–ma) = 0 (D' Alembert's principle) 2 1 1 2 • Coefficient of restitution (e) = Where, Velocityofseparationalonglineofimpact F = Real force, Velocityofapproachalonglineofimpact (–ma) = Inertia force or Fictitious force Mechanical Engineering Capsule 8 YCT Strength of Material Types of Material Relation between E, G, K & µ Homogeneous A material which have same • E = 2G (1+µ) Material elastic properties at any point in a • E = 3K(1−2µ) given direction. 9KG Isotropic Material This material has same • E= identical properties in all 3K+G direction at a point. 3K−2G Anisotropic It has different properties in all • µ= 6K+2G Material direction at a point in the body. Orthotropic A material which has different Material properties in three mutually Types of Total number of No. of perpendicular planes. Material Elastic Independent Material Poission's Ratio Constants Elastic Constant Cork - 0 Homogeneous 4 2 Glass - 0.02 - 0.03 and Isotropic Cast Iron - 0.23 - 0.27 Elastic Material - 0.25 - 0.40 Orthotropic 12 9 Steel - 0.27 - 0.33 (wood) Rubber - 0.50 Anisotropic ∞ 21 Human Tissues - –1 Axial Elongation in Different Types of Bar– Wrought Iron - 0.30 Type of bar Elongation due to Concrete - 0.10 - 0.20 Elastic Constant external load Elastic Constant Formula Prismatic bar Young's Modulus or LongitudinalStress Pl σl E= δl = = Modulus of Elasticity LongitudinalStrain AE E σ Fl = = ε δl×A Modulus of Rigidity/ Shear stress τ Circular tapered bar δl = 4Pl G= = Shear Modulus πdd E Shear strain φ 1 2 Poisson's Ratio LateralStrain δd/d µ=− =− LinearStrain δl/l Rectangular tapered bar b  Bulk Modulus K= Direct stress = σd δl= Pllogeb21 VolumetricStrain ε (b −b )Et v 2 1 Load with respect time t = thickness Type of load Stress Composite bars P = P + P 1 2 Gradual P Change in length σ = load A PL PL δ1=δ2 = A1E = A2E 1 1 2 2 Impact W 2hAE load σi = A 1+ 1+ Wℓ  P1 = A EA+1EA1 E ×P 1 1 2 2 A E P = 2 2 ×P Sudden load σ =2σ 2 A1E1+A2E2 sudden Mechanical Engineering Capsule 9 YCT