Mechanical Engineering Capsule

177 Pages·2021·15.831 MB·English

by  Youth Competition Editorial Board

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
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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