ENGINEERING MECHANICS MADE SIMPLE

COMPILED BY

Prof.S.RAJENDIRANFormer Scientist in DRDO

R2 = d2+g2

= (m+n)2+g2

= m2 +n2+2mn+g2

= F12+ 2F1F2cosά+F22-n2+n2 hence R = F12+ F2+2F1F2cosά

Pl note that if F1=F2= R to find angle between F1andF2 apply triangular formula

i.e. =

Parallelogram lawIf two forces acts on a particle ,are represented as adjacent sides of the parallelogram, then the resultant is thediagonal of the parallelogram passing through thepoint of application of the forces

ProofForce F1andF2 acts at point o(fig)Let the resultant is Rlet us extend force vector as shownAnd all construction can be as per fig

R = F12+ F2 - 2F1F2cosά

F2

F1

R

m n

d= (m+n)

g

ά ά

F2

o

Then We get

Equilibrium

Is a force which brings the body to state of equilibrium. It is equal in magnitude but opposite in direction to the resultant force.

Centre of Gravity

Centre of gravity can be defined as a point through which resultant force of gravity of the body acts.

Centroid

The centre point obtained by dividing first moment of line segment by the total length of the line or first moment of area about reference axis with a total area or first moment of volume about reference axis by total volume is termed as centroid.

Moment of inertia of area.

Second moment of area is sum of area times the square of the distance from the axis is considered, which is the plane of the area. This is also termed as second moment of area or it is moment of a moment of an area.

Polar Moment

It is second moment of area about an axis normal to the area is

termed as polar moment of area.

I.e. IZZ =IXX+IYY

Moment

Moment of a force is a measure of its tendency to cause a body

to rotate about a specific point or axis. This is different from the

tendency for a body to move, or translate, in the direction of the

force. In order for a moment to develop, the force must act upon

the body in such a manner that the body would begin to twist. This

occurs every time a force is applied so that it does not pass

through the centroid of the body. A moment is due to a force not

having an equal and opposite force directly along it’s line of action.

F

d

Where

‘d' is perpendicular distance between force ‘F’ and to the end of cantilever beam

‘F’ Force act on the diameter of the beam where beam cross section is circular

M = Moment

Moment of force M = F X d

Z

Y XF

d

O

Moment of force “F” about “O” is given by M = F X d

Where “d” is perpendicular distance between force “F” and point “O”

M = Moment, (force F creates rotary motion about O)

Example.1

Example.2

Moment

F

F

d

Couple

• Two parallel forces equal in magnitude but opposite in direction separated by particular distance acts on a body form a Couple

• System of forces whose total of force F = ∑ F i is zero.

• Couple is a pure moment. Please note that moment of a couple is independent of reference point.

• Couple is a system of forces with a resultant moment but no resultant force.

• Magnitude of couple is F X d (which is also called moment of couple if d perpendicular distance between forces)

• Where d is the perpendicular distance between forces of F

• Since the forces are equal and oppositely directed, the resultant force is zero. But the displacement of the force couple create a couple moment

F

d

Where

‘d' is perpendicular distance between force ‘F’ and to the end of cantilever beam

‘F’ Force act on the diameter of the beam where beam cross section is circular

M = Moment

Moment of force M = F X d

Z

Y XF

d

O

Moment of force “F” about “O” is given by M = F X d

Where “d” is perpendicular distance between force “F” and point “O”

M = Moment, (force F creates rotary motion about O)

Example.1

Example.2

Moment

Torque

F

R

Torque = F X R

Where

“F” is a force

“R” is a radius

It is true that force F act radially as shown. Torque is a resultant of couple. Torque is a moment of force. Force applied on diameter like moment will not make it rotate whereas force applied at radius as shown in above fig will make it rotate.

The forces have a turning effect or moment called a torque about an axis which is normal to the plane of force.

Cantiliver beam

Moment of a couple

F

F

D

d

Couple is formed when two forces equal in magnitude act at a distance of D in opposite directionMagnitude of couple is FD where as moment of couple is Fd

Varignon’s principle (Theorem)

θ

θ1

θ2 F1

FF2

d1

d

d2

Fx1

Fx2

Fx

O

A

x

Y

θ

θ1

θ2

Proof

By principle of resolving forces about X axis we know that

Fx = Fx1 + Fx2

Multiplying OA the above equation is equal to

OAFx = OAFx1 + OAFx2 ------------------- (Eq 1)

We also know that Fx is equal to F cos θ, similarly Fx1 is equal to F1 cos θ1 and Fx2 is equal to F2 cos θ2

therefore rewriting Eq 1

OAFCos θ = OA F1 cos θ1 + OA F2 cos θ2 -------------- (Eq 2)

Statement

Varignon’s theorem states that the moment of a force F about an axis is equal to the sum of moments of its compliments (F1 & F2) about the same axis.

i.e, Fd = F1d1 + F2d2

We know that

θ1

θ1

O x

Y

A

d1= OA Cos θ1

F1

OA Cos θ1 = d1 from above fig

Substituting the above d, d1 and d2 values the Eq 2 can be rewritten as

Fd = F1d1 + F2d2

Hence Varinon’s theorems is proved

O

A

x

Y

θ2

d2 =θ2

F2

OA Cos θ2 = d2 from the above figOA Cos θ = d from above fig

θ

O

A

F

θ

Y

x

d = OA Cos θ

OA Cos θ2

Principle of transmissibility

AB is line of action

Looking at the above fig we know that when an object acted upon by

a force F, when line of action (AB) remains the same whereas changing

the point of application of force without changing the magnitude of

force. The condition of the object remains same i.e. motion, or

equilibrium.

A

BObject

A

B

Object Point of application

Point of application

=

F

F

Definition

When an object subjected to rotation and translation motion, then at an instant a point exists (which may inside the object or out side the object) which is the plane of other points may looked as pure rotation about this point and that point has the velocity is zero, which is called instantaneous centre. Axis passing through and directed at right angles to the plane of motion is called instantaneous axis of rotation

Points to remember

• The instantaneous center is not a fixed point its location keeps on changing at every instant and path traced or locus by it is called centrode

• The instantaneous center may within the body or outside the body if it is within the body then it is called body centrode if it is outside the body it is called space centrode

• The velocity at instantaneous center is zero

Instantaneous center (Instantaneous centre of rotation)

Proof Let us consider a disc rotates with a constant angular velocity on the surface of the floor as shown in the fig below which will have translation velocity shown in Fig 1 and rotation velocity shown in Fig 2. Superimposing Fig 1 & 2 we get Fig 3, where at point M the velocity equal to zero since translation velocity cancels rotation velocity.

From the fig 1, 2 & 3 the instantaneous centre is well defined where M is the instantaneous centre and at that point velocity is zero

M

V

Translation velocity

Fig 1

V

VM

V

Rotation velocity

Fig 2

M

2VFig 3

Rotation velocity Translation velocitycancels

-V +V

Velocity at point M = 0

VωR

O

Object

V1

V2

A

B

O

Object

Different methods of locating instantaneous centre

Since V = ωR we can find out R = V/ ω on the velocity line make a perpendicular line as shown above and mark the distance = R then O is the instantaneous centre

Similarly, when an object subjected by rotary and translation motion if we know the velocity of points A & B as shown above (which are not parallel) are drawn and we can draw perpendicular lines from A & B as shown above (angular velocity is perpendicular to velocity) both the lines will intersect at point O. which is called instantaneous centre

Let us assume that an object subjected to rotary and translation motion and if we know the angular acceleration equal to ω and translation velocity equal to V at an instant shown below.

II Method

I Method

III Method

From the above fig we know that an object subjected to rotary and translation motion point A1 occupies the position of A2 and point B1 occupies the position of B2. Join A1 A2 and B1 B2 draw a perpendicular bisector for both the lines which will intersect at point O then point o is called instantaneous centre.

Centrode

The locus of the instantaneous centre is called centrode. If it lies inside the object, then it is called body centrode, if it lies out side the object it is space centrode.

O

A1

B1

A2

B2

Object

Area Theorem

If a line or profile rotates about an axis then

the surface area formed by the line about the

axis is equal to product of the line length and

distance traveled by the C.G of the line.

Let us consider that cone of height h and

base radius R is being formed by the rotation of

the line AB (slant length square root of h2 + R2))

about the axis MN .

Then surface area of the cone as per Pappus

theorem is AB multiplied by distance traveled

by the C.G

M

N

B

A

R

R/2

R/2C.G

h

Slant length AB=√ h2 + R2

Distance traveled by C.G = 2π R/2 = πR

Therefore surface area = πR √ h2 + R2

Theorems of Pappus - Guldinus.

M

N

B

A

R

C.GR/3

h

R/3

Pappus Theorems on Volume

If a cross section of a solid revolves

about an axis then, volume formed by

generation of the cross section is

equivalent to product of the area of

cross section of area and distance

traveled by the C.G while cross section

revolves

Let us consider the cone cross section

is a right angle triangle shown below

rotates about axis MN then volume of the

cone generated equal to product of

cross section area of triangle (R h/2) and

distance made by the C.G equal to2πR/3

Then Volume of cone = π R2h

3

Resolution of force to couple and a force

FF

F F

F

Fig A Fig B Fig C

= =

Object

M M

N

Let us assume that a single force F act on an object (at point M) as shown in Fig A.

And Let us also introduce equivalent opposite force at point N (shown in Fig B).

By introduction of the forces at N the condition of the object remains same (since we have introduced force equal in magnitude but opposite in direction at point N and net force at Point N is zero)

Now if we look at Fig c. we were able to understand that the right side force at N and left side force at M forms a couple. The net result is a force and a couple as shown in Fig C. Thus the single force is resolved into a couple and a force.

Mt

ObjectF1F2

F4F5

F3

C.G=

Before D'Alembert's principle

When an object subjected to multiple forces (F1, F2,…..etc) then all forces reduces to one resultant = R and one net torque = Mt which produces translation and

rotary motion.

As per D'Alembert's principle

R – ma = 0

Mt – Iά = 0

D'Alembert's principle

D'Alembert's principles introduces that

body will be in dynamically equilibrium such that

inertia force –ma acts opposite to the translation

motion produced by resultant and – Iά inertia force

due to rotation in the opposite direction of net

torque Mt

Mt

– Iά

– ma

R

C.G

= 0

The force system consisting of external forces and inertia force can be considered to keep the object in equilibrium. Since the resultant force externally acting on the object is not zero, the object is said to be in dynamic equilibrium. This known as D'Alembert's principle.

Centre of Area or area center

The centre point obtained by dividing first moment of area about the reference axis by the total area termed as centre of area.

Centre of mass or mass center

The centre point obtained by dividing first moment of masses about the reference axis by the total mass termed as centre of mass.

Centre of mass, Centre of volume, Centre of area

Xi, Yi, are the coordinates of ΔMi/ΔAi/ΔVi

ΔMi/ΔAi/ΔVi

Object

Xi

Yi

Mass Centre mass centre x = ∑ΔMi Xi

M

mass centre y = ∑ΔMi Yi

M

WHERE M=∑ΔMi

The centre point obtained by dividing first moment of area about the reference axis by the total area termed as centre of area.

The centre point obtained by dividing first moment of volume about the reference axis by the total volume termed as centre of volume.

Volume centre x = ∑ΔVi Xi

V

Volume centre y = ∑ΔVi Yi

V

Centroid or Centre of Area or area center

Centroid x = ∑ΔAi Xi

A

Centroid y = ∑ΔAi Yi

A

Centre of Volume or volume center

Volume centre

Centroid

WHERE V=∑ΔVi

WHERE A=∑ΔAi

Drawing a FBD, steps

1). Decide which body to analyze.

2). Separate this body from everything else and sketch the contour,

3). Draw all applied forces (weight).

4). Draw all reactions.

5). Include any necessary dimensions and coordinate axis.

If you don't know a direction assume a direction and let the sign of the answer tell you if the direction is correct or not.

Rules:

The magnitude and direction of known forces should be clearly indicated (usually applied forces)

Indicate the direction of the force exerted on the body, not the force exerted by the body.

Unknown forces are usually the reactions (constraining forces).

Reactions are supports and connections in 2-D.

Free - body diagrams

Friction

Dry Friction(Coulomb Friction)

Fluid Friction

W

NF

Fig 1

P

Friction

• Occurs un-lubricated surfaces are in contact during sliding

• Friction force always oppose theθ

• sliding motion

• Occurs when the adjacent layers in a fluid (liquid, gas) are moving at different velocities

• This motion causes friction between fluid elements

• Depends on the relative velocity between layers

• No relative velocity –no fluid friction

• Depends on the viscosity of fluid – (measure of resistance to shearing action between the fluid layers)

From the above fig it is seen that an object having weight W rest on a floor is pushed by a force F. the contact area of the floor and the area of the object which in contact with the floor experience in frictional force (F) which is against the motion of the object.

θ

Limiting Friction

The maximum value of frictional force which comes into play, when the motion is impending, is known as limiting friction. In this case, applied force is less than the friction force.

Static Friction

The frictional force for which the body at rest is called static friction.

Dynamic Friction

If the value of applied force is exceeding the limiting friction, the body starts moving over the floor and the frictional resistance are experienced by the body is known as dynamic friction.

Sliding Friction

It is the friction experienced by a body when it slides over the other body

Rolling Friction

It is the friction experienced by the body when it rolls over a surface.

Coefficient of friction

The limiting friction bears a constant ratio to the normal reaction between the two forces is called coefficient of friction. = F/N

Where,

F – Limiting friction

N – Normal reaction

Which is otherwise Coefficient friction denoted as μ,

Hence, μ = F/N or = tan θ

Laws of Coloumb Friction

• The force of friction always acts in a direction opposite to that in which the body tends to move.

• Till the limiting value is reached, the magnitude of friction is exactly equal to the force which tends to move the body.

• The magnitude of the limiting friction bears a constant ratio to the normal reaction between the two surfaces

• The force of friction depends upon the roughness/smoothness of the surfaces

• The force of friction is independent of the area of contact between the two surfaces

• After the body starts moving, the dynamic friction comes into play, the magnitude of which is less than that of limiting friction and it bears a constant ratio with normal force. This ratio is called coefficient of dynamic friction.

P

F

F limit

Static state

Dynamic State

The maximum inclination of the plane on which a body free from external applied force can repose (sleep) is called angle of repose.

θN

F

θ

W

From the above fig we conclude that

N = W cos θ and F= W sin θ

Since,

F/N = W sin θ / W Cos θ

F/N = tan θ = μ

F/N = tan ά (ά – angle of friction)

θ = ά

The value of angle of repose same as the value of limiting friction

Angle of repose

Cone of Friction

W

F

P

ά ά

N

F

R

When a body having impending motion in the direction of P, the frictional force will

be the limiting friction and the resultant reaction R will make limiting friction angle with

the normal as shown in the fig if body is having impending motion with some other

direction, again the resultant reaction makes limiting frictional angle ά with the normal

in that direction. Thus, when the direction of force P is gradually changed through 3600

the resultant R generates right circular cone with the semicentral angle equal to ά.

If the resultant reaction is on the surface of this inverted right circular cone whose

semi central angle in limiting frictional angle ά, the motion of the body is impending. If

the resultant is within the cone the body is stationary. This inverted cone with semi

central angle, equal to limiting frictional angle ά, is called cone of friction.

Impulse momentum = Final Momentum – Initial Momentum

We know R is the resultant force acting on a body of mass m, then from Newton’s second law,

R=ma

But acceleration a = dv/dt

R=m dv/dt

Rdt = mdv

∫ Rdt = ∫ mdv

If initial velocity is u and after time interval t it becomes v, then

t v

∫ Rdt = m [v] = mv - mu

0 u

Impulse Momentum – mv – mu

The impulse momentum principle can be stated as the component of the resultant linear impulse along any direction is equal to the component of momentum in that direction.

Impulse Momentum

If mv- mu is equal to zero then, momentum is conserved

The principle of conservation of momentum may be stated as the momentum is

conserved in a system in which resultant force is zero. In other words in a system

the resultant force is zero, initial momentum will remain equal to the final

momentum.

Coefficient of restitution may be defined as the ratio between relative velocity of separation and relative velocity of approach along the line of impact, when two bodies collide. v2-v1

u1-u2

Conservation momentum

Coefficient of restitution

e =

The displacement of the particle in rotation is measured in terms of angular displacement θ, where θ is in radins. Thus when a particle moves from position A to B, the displacement is θ as shown in the Fig. 1. This displacement has the direction – clockwise or counter clockwise.

S

B

Aθ Reference Axis

r

The distance traveled by a particle A to B is S which is equal to r θ

S = r θ

ds/dt = r dθ/dt

v = r ω, (since dθ/ dt = ω)

dv/dt = r dω/dt

at = r ά (where dω/dt = ά)

ά = at/r

we know that an = v2 /r = (r ω )2 /r = r ω2

Relation between ω, v, r and ά

Span

A B

Supports

Beam

Span

A B

Supports

Beam

OverhangingOverhanging

Pl note that If there is no inclined or horizontal load on the beam, then the reaction at the hinged support will be vertical, otherwise the reaction will be inclined to the vertical.

Beams are classified on the basis of supports.

Simply supported beam

When supports are provided at both ends of the beam is called simply supported beam.

When the beam is projecting beyond the supports is called overhanging beam

Beams

Cantilever beam, Fixed beam and Continuous Beam

Fig A Fig B Fig C

When one end is fixed other end is free it is called cantilever beam (Fig A)

When both ends are fixed it is called fixed beam (Fig B)

When beam is supported more than two positions it is called Continuous Beam (Fig C)

Condition for skidding (for vehicle)

W- Weight

F- Friction of force

Inertia force = (W/g) (V2/r)

F = μ W

(W/g) (V2/r) = μ W

V = μ g r

tan θ = μ

Super Elevation

Let ά be super elevation angle, then tan ά = v2 / g r Velocity of vehicle at super

elevation condition v = g r (μ+ tan ά)

(1- μ tan ά )

Condition for Overturning

V = g r B

2 h

Where,

B – distance between inner and outer wheels

h – height of C.G from the ground level

B

h

W V2

g r

w

ά

μ= coefficient of friction

θ = angle of motor cycle with ground

In case of rotation

ω = d θ , dt

ά= d2

dt r

θ, v= r ω , a = r ά , a= v2 = r ω2

θ =ωt ωv = ω u + ά t

ωv2 - ω u

2 = 2άθ

kinetic energy = Iω2 2

impulse momentum = I( ω2-ω1)

Mt = Iά

I1ωu1+ I2ωu2 = I2ωv1+I2ωv2

m1u1 +m2 u2 =m1v1+m2v2

θ = ωut + 1 άt2

2

m = I a = ά u = ωu

V = ωv

S = θ

Pl note

linear rotation

150

θ

PR

600300

P = R = W

Sin (180-60) sin (90+θ) sin (180 - 30 - θ)

P = R = W Sin (120) sin (90+θ) sin (150 - θ)

P = W Sin (120) sin (90-( θ- 60) )

P will be least when θ= 60 sinceWhen sin 90= 1300

Sin (θ)= 150/300 θ1 = 30o

w

θ1

Wheel

block

D

O

E

O

DE