Basic of ENGINEERING MECHANICS
There are two types of mechanics
1. Statics
2. Dynamics
- STATICS
It deals with the forces and their effect while body is at rest condition.
- DYNAMICS
It deals with the acting forces and their effect while body is in motion condition.
Two types of dynamics are as follows,
a. Kinetics
b. Kinematics
Kinetics
Deals with the bodies in motion due to the application of any forces.
Kinematics
Deals with the bodies in motion without any reference to the application forces.
SCALAR QUANTITIES
It is specified by Magnitude only.
This are example of scalar quantities :- Length, Area, Mass, Volume, Time, Density, Distance, Speed, Work, Temperature, Energy, Moment of inertia.
VECTOR QUANTITIES
It is specified by Magnitude and Direction.
This are the vector quantities :- Displacement, Velocity, Angular velocity, Force, Weight, Acceleration, Angular acceleration, Momentum, Moment, Impulse.
RELATIONS
1 m = 100 cm = 1000 mm
1 cm = 10 mm
1 km = 1000 m
1 cm2 = 100 mm^{2 }
1 m2 = 10^{6} mm^{2}
1 N = 1 kg.m/s^{2 }
1 kgf = 9.81 N
1 MN = 10^{6} N = 10^{3} KN
1 GN = 10^{9} N
1 Pa = 1 N/m^{2 }
1 MPa = 1 N/mm^{2 }
1 GPa = 103 N/mm^{2 }
1 N.m = 1 Joule = 10^{7} Arg
1 watt = 1 N.m/s = 1 J/s
1 hp = 746 watt
1° = π/180 rad
FORCE
The agent which produces or tends to produces, destroy or tends to destroy motion of any body.
S.I. Unit = N.
Vector quantity.
Nature = Tensile or Compressive type.
WEIGHT
The force by which the body is attracted towards the centre of the earth.
W = mg.
S.I. Unit = N.
M.K.S. Unit = kgf.
Vector Quantity.
MASS
Matter contained in the body.
S.I. Unit = kg, Tonne.
Scalar Quantity.
Coplanar Forces :- It is the forces whose line of action lie on the same plane.
Concurrent Forces :- It is the forces which meet at one point.
Co-linear Forces :-It is the forces whose lines of action lie on the same line.
Coplanar Concurrent Forces :- It is the forces which meet at one point and their lines of action also lie on the same plane.
Coplanar Non-current Forces :- It is the forces whose lines of action lie on the same plane but they do not meet at one point.
Non-coplanar Concurrent Forces :- Forces whose lines of action do not lie on the same plane, but they meet at one point.
Non-coplanar Non-concurrent Forces :- It is the forces whose lines of action do not lie on the same plane and they do not meet at one point.
Like parallel Forces :- It is the forces whose lines of action are parallel to each other and all of them act in the same direction.
Unlike parallel Forces :- It is the forces whose lines of action are parallel to each other but all of them do not act in the same direction.
RIGID BODY
The body which retain its actual shape and size, even if subjected to some external forces.
No body is perfectly rigid.
PARALLELOGRAM LAW OF FORCES
R = √(P^{2}+Q^{2}+2 PQ cosθ)
Tan α = Q sinθ/(P+Q cosθ)
LAMI’S THEOREM
P/sinα = Q/sinβ = R/sin γ
RESULTANT OF TWO OR MORE FORCES
R =√ (ƩH ^{2}+ ƩV ^{2})
Tan θ = ƩV/ƩH
Resultant of two concurrent tensile force is maximum, when the angle between them is 0℃.
Resultant of two concurrent tensile force is minimum, when the angle between them is 180℃.
The condition of equilibrium for coplanar concurrent forces are as follows ƩH = 0, ƩV = 0 & ƩM = 0.
PRINCIPLE OF TRANMISSIBILITY
“ The effect of a force upon the body is the same at every point on its line of action.”
This principle is applicable to the Rigid body.
MOMENT
Moment = Force × Perpendicular Distance
S.I Unit = N.m or kN.m.
COUPLE
A system of two equal and opposite forces separated by the distance.
Resultant force of couple is also zero.
Arm of Couple
It is the distance between two forces of a couple.
Concentrated Load :- Unit = N, kN.
Uniformly Distributed Load :- Unit = kN/m.
Support | Reaction |
Simple support | One Reaction |
Hinged support | Two Reaction |
Roller support | One Reaction |
Fixed support | Three Reaction |
Roller support is provided at the end of a bridge structure.
CENTROID
The point at which the whole area of the figure is concentrated is centroid.
Two dimensional.
Example :- Square, Rectangle, Triangle, Circle etc.
CENTRE OF GRAVITY
The point at which whole mass of the body is concentrated.
Third dimensional.
Example :- Cube, Cuboids, Cone, Sphere etc.
Geometrical Shape | Length | ẍ | ȳ |
Straight wire | L | L/2 | L/2 |
Wire ring | 2πr | r | R |
Semicircular wire | πr | r | 2π/r |
Quarter circular wire | Πr/2 | 2π/r | 2π/r |
Arc of circle | 2αr | r sin∝/∝ | On axis of symmetry |
Shape | Area | ẍ | ȳ |
Rectangle or Square | B.D | B/2 | D/2 |
Right angle triangle | ½ (bh) | (1/3) b | 13 h |
Trapezium | (a + b)h/2 | b/2 | h3(b+2ab+a) |
Circle | π/4 d^{2} | r | r |
Semi-circle | π/2 r^{2 } | r | 4r/3π |
Quarter | π/4 r^{2 } | 4r/3π | 4r/3π |
Circular sector | αr^{2} | 2/3 (r sin∝)/∝ | On axis of symmetry |
Shape | Volume | ẍ | ȳ |
Cylinder | πr^{2}h | r | h/2 |
Cone | π /3r^{2}h | r | h/4 |
Sphere | 4/3 πr^{3} | r | r |
Hemisphere | 2/3 πr^{3} | r | 3r/8 |
L-section is not symmetrical about any axis.
C.G. of Isosceles triangle = √(4p^{2}−q^{2})/6
C.G. of equilateral triangle with each side (a) = a/2√3 from any of the three sides.
FRICTION (F)
Force act opposite to the direction of motion is called friction.
If the contact surface is smooth, F will less.
If the contact surface is rough, F will more.
If the external force becomes greater than friction force, body will move.
Friction does not depends upon the area of surface between two surface but depends upon the roughness of the surface.
Limiting Friction
Maximum friction force that can be developed at the surface, when body is just on the point of moving.
Magnitude of limiting friction bears a constant ratio to the normal reaction between the two surfaces.
Static Friction
Friction when experienced by a body when it is at rest.
External force > Friction force.
No motion.
Static friction is slightly less than the Limiting friction.
Dynamic Friction
Friction experienced by a body when it is in motion.
External force < Friction force.
Dynamic friction is always less than static friction.
Sliding Friction
Friction experienced by body when it slides over another body.
Rolling Friction
Friction experienced by body when it rolls over another body.
Dry Friction
Friction between dry surfaces is called dry friction.
Also known as Coulomb Friction.
ANGLE OF FRICTION (∅)
Angle between normal reaction and resultant force is called angle of friction.
Also called Limiting angle of friction.
COEFFICIENT OF FRICTION
Ratio of limiting friction and normal friction.
μ = F/N
Relation between ∅ & μ is μ = tan∅.
ANGLE OF REPOSE (∝)
Maximum angle at which body starts sliding down is called angle of repose.
∝ = ∅.
MECHANICAL ADVANTAGES
Ratio of load lifted to effort required.
Maximum M.A. = 1/m.
VELOCITY RATIO (V.R)
Ratio of distance moved by effort to the distance moved by load.
V.R of simple wheel and axle = D/d
V.R of differential pulley block = 2D/(D−d)
V.R of single purchase crab increased by increasing the length of the handle.
V.R of simple screw jack = 2πl/p
V.R of differential wheel and axle = 2D/(d1−d2)
EFFICIENCY
Ratio of work done by machine to work done on machine.
Ratio of output to input.
Ratio of M.A to V.R.
Maximum Efficiency = 1/(m x V.R )
IDEAL MACHINE
Ideal machine have 100% efficiency.
Friction is zero.
Output = Input.
M.A = V.R.
In actual machine, M.A < V.R.
REVERSIBLE MACHINE
If machine is capable of doing some work in the reverse direction after the effort is removed.
Ƞ ≥ 50%.
NON-REVERSIBLE MACHINE
Also known as Self-locking machine.
If machine is not capable of doing work in reverse direction after the effort is removed.
Ƞ < 50%.
SYSTEM OF PULLEY
System | Velocity Ratio |
First | 2^{n } |
Second | n |
Third | 2^{n} – 1 |
WORK
Whenever force act on body, body undergoes a displacement, some work is said to be done.
Unit = S.I. = Joule. ( 1 Joule = 1 N.m )
POWER
Rate of doing work.
S.I. Unit = Watt.
Unit = N.m/s or Joule/s.
1 H.P. = 746 watt.
ENERGY
Capacity to do work.
Unit = N.m or Joule.
Two types of energy
1. Potential energy
2. Kinetic energy
Potential energy
Energy stored in body = mgh.
Kinetic energy
Energy possessed by body = ½ mv^{2}.
MOMENT OF INERTIA (I)
Second moment of Area.
Second moment of Mass.
Second moment of Force.
Unit = mm^{4}, cm^{4} or m^{4}.
Unit of Mass moment of inertia = kg.m^{2}.
SECTION MODULUS (Z)
Z = I/y_{max }
Unit = mm^{3}, cm^{3} or m^{3}.
RADIUS OF GYRATION (k)
k = √(I/A )
M.I. OF SOME SECTIONS
Rectangular
I_{xx }= bd^{3}/12 Z_{xx} = bd^{2}/6 I_{base} = bd^{3}/3
I_{yy} = bd^{3}/12 Z_{yy} = bd^{2}/6
Square (side a)
I_{c.g. }= a^{4}/12 I_{base }= a^{4}/3
Circular
I_{xx} = I_{yy} = π/64 D^{4 }
Z_{xx} = Z_{yy} = π/32 D^{3}
Triangular
I_{base} = bh^{3}/12
I_{c.g. }= bh^{3}/36
I_{top} = bh^{3}/4
Semicircular
I_{xx} = 0.11 r^{4}
I_{yy} = π/128 d^{4}
Quarter circle
I_{xx} = I_{yy }= 0.055 r^{4}
IOA = IOB = π/16 r^{4}
Greater the M.I of section, greater would be the load carrying capacity.
Strength of section is increase with increase in radius of gyration.
Resultant of two equal forces acting in opposite direction will be zero.
Resultant of two forces P & Q in opposite direction will be P – Q.
Resultant of two equal forces at angle 90° will be √2 P.
Resultant of two equal forces is equal to either of them, angle between them is 120°.
Resultant of two equal forces at angle θ is 2P cos𝜃/2.
SCREW JACK
The efficiency of screw jack is increased by increasing the pitch.
The efficiency of screw jack = tan∝/tan(∝ + ∅)
The maximum efficiency of screw jack if ∅ is angle of friction = 1−sin∅/1+sin∅
The efficiency of screw jack is maximum if helix angle is 45° − ∅/2
The screw thread is imagined to be unwound then, tan ∝ = p/πd
LAW OF MACHINE
P = mW + C
Where, P = effort required
W = Load lifted
m = Coefficient of friction
C = Machine friction
COUPLE
Couple does not produce translatory motion but it produce a motion of rotation of a body.
Translatory motion = Motion in straight line.
M.I of thin disc of mass m and r about axis of c.g and perpendicular to plane of disc = 0.5 mr^{2}
M.I of thin disc of mass m and r about its diameter = 0.25 mr^{2 }
M.I of thin rod of mass m and l about axis of c.g & perpendicular to length = ml^{2}/12
M.I of thin rod of mass m and l about parallel to length = ml^{2}/3
M.I of solid cylinder = mr^{2}/2
M.I of thin spherical shell = 2mr^{2}/3
M.I of sphere about axis tangential to it = 7mr^{2}/5
M.I of sphere = 2mr^{2}/5
M.I of solid cone = 3mr^{2}/10
Median :- Line joining the apex and the middle point of opposite side in triangle is called median.
Minimum force which required to slide a body on Rough horizontal plane is = P_{min }= W sin θ.
EFFORT REQUIRED TO MOVE THE BODY UP AN INCLINED PLANE
P = Wsin(∝ + ∅)/sin[θ− (∝ + ∅ )]
When friction is neglected, ∅ = 0
P = Wsin∝/sin(θ− ∝)
When θ = 90°, P applied horizontally
P = W tan ( ∝ + ∅ )
When P applied parallel to plane θ = 90° + ∝
P = W ( sin∝ + μ cos∝)
EFFORT REQUIRED TO MOVE THE BODY DOWN AN INCLINED PLANE
P = Wsin(∝ − ∅)/sin[θ− (∝ − ∅) ]
When friction is neglected, ∅ = 0
P = Wsin∝/sin (θ− ∝ )
When θ = 90°, P applied horizontally
P = W tan ( ∝ − ∅ )
When P applied parallel to plane θ = 90° + ∝
P = W ( sin∝ − μ cos∝)
The acceleration of body sliding down inclined plane = g sinθ.
Negative acceleration is called Retardation.
FRAME
Perfect frame :- Satisfy the equation n = 2j – 3.
Imperfect frame :- Not satisfy above equation.
If n < 2j – 3, then deficient frame.
If n > 2j – 3,then redundant frame.
EQUATION OF LINEAR MOTION
v = u + at^{2}
s = u + ½ at^{2 }
v^{2} = u^{2} + 2as
When body fall from height h, velocity with which it will hit the ground V = 2gh.
For downward motion, g = + 9.81 m/s2.
For upward motion, g = − 9.81 m/s2.
If the Lift moving upwards direction, then Reaction R = m (g + a).
If the Lift moving downwards direction. Then Reactions R = M (g − a).
MOTION OF TWO BODIES CONNECTED BY STRING
Acceleration
a = g(m_{1}− m_{2})/m_{1}+ m_{2} m/s2
Tension
T = 2 m_{1}m_{2}g/m_{1}+ m_{2} N
PROJECTILE
Particle moving under combined effect of vertical and horizontal forces.
Trajectory
Path traced by a projectile in space.
Shape of trajectory is parabola.
Velocity of projection
Velocity with which a projectile is projected.
Angle of projection
Angle with the horizontal at which projectile is projected.
Time of flight
Total time taken by any projectile to reach maximum height and return back to the ground.
Time of flight on horizontal plane = t = 2usin∝/g
Range
Distance between the point of projection and the point where the projectile strikes the ground.
R = u^{2}sin^{2}∝/g
Range is maximum when ∝ = 45°.
Height of projectile
Maximum height of projectile on horizontal plane
H = u^{2}sin^{2 }∝/2g
ANGULAR VELOCITY
Denoted by ω (omega).
Expressed in revolution per minute (r.p.m.) or rad/s.
Vector quantity.
ω = 2πN/60 rad/s
Body rotating at ω rad/s along circular path of radius r then velocity = ω.r.
ANGULAR ACCELERATION
Rate of change of angular velocity.
Unit = rad/s^{2}.
Denoted by α (alpha).
Vector quantity.
Body move along circular path radius r with α then, acceleration = α.r.
Body move along circular path, linear acceleration have two component i.e Normal or Centripetal & Tangential component.
Normal Component = ac = ω2.r.
Body moves along circular path with uniform velocity, no tangential acceleration acts only centripetal acceleration.
Body moves along the straight path, no centripetal acceleration acts only tangential acceleration.
AMPLITUDE
Maximum displacement of body from its mean position .
Always equal to the radius of circle.
PERIODIC TIME
Time taken for one complete revolution of the particle.
t_{p} = 2π/ω seconds
FREQUENCY
Number of cycles per second.
Reciprocal of time period.
f = ω/2π Hz
S.I. Unit = Hz (one cycle per second)
SIMPLE PENDULUM
- Periodic time = t_{p} = 2π √(L/g)
- Frequency of oscillation = f = 1/t_{p} = 1/2π √(g/L)
SECOND’S PENDULUM
The motion of any body from one extremity to the other is known as beat.
Pendulum executes one beat/second is second’s pendulum.
Length of second’s pendulum = 99.4 cm.
SUPERELEVATION
Super elevation = s = Gv2/gr
Tan θ = v2/gr = Angle of inclination = Angle of banking = Centrifugal ratio.
FOR ANGULAR MOTION
- Work = T.θ
- Power = T.ω
- Kinetic energy of rotation = ½ Iω2
For both cases linear & angular motion
- Kinetic energy = ½ mv2 + ½ Iω2
The process of finding out the resultant force is called the Composition of forces.
The resultant of the two forces P & Q, if Q is doubled then new resultant is perpendicular to the P, then R = Q.
D’Alembert’s principle basically depends the upon Newton’s second law of the motion.
Body falls under the gravitational force it possesses No weight or weight less condition.
Motion of wheel of any car is translation & rotation type motion.
Bodies which rebound after the impact are called Elastic bodies [ e = 1 ].
Bodies which does not rebound at all after the impact are called the inelastic bodies [ e = 0 ].
e = coefficient of restitution = Velocity of the two bodies after impact / Velocity of the two bodies before impact