Vachhani sir's Technical Gems

# Vachhani sir's Technical Gems

### This section of the site is Work In Progress. New  VTGs will be added periodically.

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### Chapter 1

Introduction to Engineering Mechanics

1. Introduction:
Object of study of Engineering Mechanics is to solve the real-life problems by making use of Physics and Mathematics. In other words we study common engineering elements in real world situations.
Thus, from now onwards you will not see any imaginary diagrams on your blackboards or in your textbooks or for that matter on this electronic white paper.
Engineering Mechanics is divided into two major parts. "Statics" and "Dynamics". Dynamics is further divided into two parts. "Kinematics" and "Kinetics".
Engineering Mechanics - 'Statics Part' deals with bodies and frames in equilibrium, no part of the given element will be allowed to move free.
Every branch of engineering has got some key words. Keywords are those words which we use quite frequently.
Major keyword in the field of statics is "Structure". How do we define it?
Just try to get the answers for the following two questions.

 Q1: How it is made? Q2: What is its function?
And to give a bit of Engineering touch.
Any frame, which is given a desirable shape (from the view point of an Architect) and is capable of supporting the design loads is called a 'Structure' .
Here you will find, the word design is added to give an Engineering touch. Structure may consist of other key words like beam, bridge truss, portal frame, gable frame and other frames.
Thus, structure is a frame built up of several members connected in such a way that no movement of a member of the joint is possible; it means it is stable. 2. Identification of a Frame:
Before solving a problem, it becomes important to know whether the given frame is a structure or a mechanism. If it is a structure, whether it is just stable or over stable.
Above three answers can be found out by calculating the degrees of freedom of the frame as a whole. 3. Degrees of Freedom of Plane Frames:
Every member in the plane has 3 degrees of Freedom.
These are:
• Horizontal Movement
• Vertical Movement
• Rotation
When this member becomes a part of a given stable frame, it loses all the 3 degrees of freedom. It means we have provided 3 restraints or so called restrictions to the movement.
3.1 Restrictions to the Movements because of ground connections.
(termed as rj- restriction because of joint)
 (a) Roller Support (only vertical movement restricted) rj = 1 (b) A pin or Hinge Joint (connection) (Vertical and horizontal movement prevented) rj= 2 (c) A fixed or Rigid Joint (vertical, horizontal and rotational movements prevented or restricted) rj = 3
3.2 Calculation of Restrictions because of connection of members amongst themselves.
(Examples are given for Pin Connections, similar arrangements can be adopted for fixed or rigid connections)
 Two members meeting at a joint rj=2 Refer table 3.1.b In other words 1 pin connection Three members meeting at a joint rj=4 In other words 2 pin connections Four members meeting at a joint rj=6 In other words 3 pin connections
All you have to do, is to count the number of members at the joint. The required pin connections will be one less than the number of members meeting at the joint. 4. Conceptual Formula:
We come to the conclusion that we have to deal with the degrees of freedom of members (∑fm) and total restrictions provided (∑rj).
Thus,
 f = Degrees of Freedom of complete frame f = ∑fm - ∑rj
So if,
 f = 0 It is a statically determinate structure. f < 0, -1, -2… It is a statically indeterminate structure. f > 0, 1, 2…… It is not a structure but it is a mechanism with “ f ” degrees of freedom.

5. Examples:
Example 1 : A Simple Bridge Beam

LEGEND: ro means roller, p means pin , r means rigid
ground is our reference member
We have only one member AB so ∑fm = 1x3=3
We have 1 pin connection at A ∑rj = 1px2+1rox1= 3
and 1 roller connection at B
(please refer above 1 pin connection provides 2 restrictions and 1 roller connection provides 1 restriction )
∑fm - ∑rj = 3-3=0.
It is a statically determinate structure.
Example 2 : A Balcony Beam or A Cantilever Beam

Above structure has one member and one rigid joint.
Therefore, ∑fm = 1x3, ∑rj = 1r x 3 = 3
f = ∑fm - ∑rj = 3-3 = 0.
It is a statically determinate structure.
Example 3 : A Bridge Truss

Total Number of Members = 9
Therefore, ∑fm = 9x3 =27
At C & E, 3 members meet, so 2p each
At D & F, 4 members meet , so 3p each
At A; 2 members and ground are connected so 2p (two pin connections)
At B, BD & BE join to give 1p, further both are connected to ground by roller support so plus 1ro
∑rj = 13p x 2 +1 rox1 = 27.
thus, ∑fm - ∑rj = 27-27 = 0
Therefore it is a statically determinate structure.
Example 4 : Gable frame (Defence Factory Security Gate )

Total Number of members = 6,
Therefore, ∑fm = 6x3 = 18
∑rj = 6r x 3+2p x2 = 22
Therefore, ∑f = ∑fm - ∑rj = 18-22 = -4
It is a statically indeterminate structure, with 4 degree of redundancy or indeterminacy.
Example 5 : A Multistorey Building Frame

Total number of members = 21
∑fm = 21x3 = 63
∑rj = 30r x 3 = 90
f = 63 - 90 = -27
It is therefore an indeterminate structure with 27 degree of indeterminacy or redundancy.
Example 6 : A motor Cycle or Scooter Mechanism

1s means one slider joint rj=2
Vertical movement & Rotation of the Piston is Prevented.
∑fm = 3x3 = 9
∑rj = 3p x 2+1sx2=8
f = 9-8 = +1
Thus it is a mechanism with one degree of freedom. Important Hint :
Any free member, if it is connected by a rigid joint will lose all the 3 degrees of freedom. So, you may count the member or not, it is immaterial.

### Chapter 2

Fundamentals of Engineering Mechanics

Fundamental Applications of Forces, their Resultant, free-body diagrams and how to transfer a force, for co-planer concurrent and non-concurrent forces.

This chapter is more or less repetition of the basics of mechanics taught at school level.

RESULTANT OF FORCES

1. Law of Parallelogram of Forces:

When two forces (F1,F2) acting on a particle are represented by the two adjacent sides of a parallelogram, the diagonal connecting the two sides represents the resultant force R in magnitude and direction.

2. Triangle Law of Forces:

If two Forces F1 and F2 acting simultaneously on a particle can be represented by the two sides of a triangle (in magnitude and direction) taken in order, then the third side (closing side) represents the resultant in opposite direction.

Thus, in above triangle, one can apply all the rules regarding trignometrical functions of the triangle as given below.

3. Law of Polygon of Forces:

If a number of concurrent forces acting simultaneously on a particle are represented in magnitude and direction by the sides of a polygon taken in order, then the resultant of the system of forces is represented by the closing side of the polygon in the opposite order.

Explanation:

Polygon of forces abcda' has been drawn by adopting well known Bow's Notation. In the given Force diagram spaces between two forces are denoted as space A, B, C & D
In the polygon drawn aa' is the resultant R in magnitude and direction. In case a' falls upon a, the Resultant will obviously be zero (Above diagram shows all the three tensile in nature).

4. Lami's Theorem:

If three forces acting on a particle are in equilibrium, then each force is proportional to the Sine of the angle included between the other two forces.

It may be noted that all the three forces should be either tensile or compressive in nature. Diagram above shows tensile.

5. Funicular Polygon:

If the co-planar forces are non-concurrent, then the resultant is obtained by drawing the funicular polygon as given below. In this, polygon of forces diagram is drawn and a suitable pole O is chosen. Rest is self explanatory from the diagrams.

Polygon of forces give us magnitude and direction of the Resultant of Forces F1,F2,F3 and F4. Funicular Polygon gives us the location of the Resultant.

6. Varignon's Theorem:

This theorem states that algebraic sum of the moments of two or more number of concurrent forces about any point in their plane is equal to the moment of their resultant about the same point.
Here, the proof is given for two forces. Similar one will be valid for any number of concurrent forces.

Let there be two forces F1 and F2 represented by OA and OB.
Complete parallelogram of forces and find Resultant R as shown.
Choose point D for moments.
∴ Moment of F1 about D =2(Area of △AOD)
Moment of F2 about D = 2(Area of △BOD)
∴ Moment of F1 & F2 about D = 2[Area of △AOD + Area of △BOD]
= 2[Area of △AOC + Area of △BOD]
= 2[Area of △BOC + Area of △BOD]
= 2 Area of △DOC = Moment of R about D

7. Moment of a Force, Couple and Effect of the Couple:

Moment of a force is always calculated at a given point. It is equal to the force multiplied by the perpendicular distance measured from the point. In the figure given below the moment at O is Fxd anticlockwise i.e Mo = -F.d

where as MA = 0
Couple - Two forces "F" of same magnitude, parallel lines of action and opposite sense are said to form a couple.

It may be seen that effect of the couple is a moment and this moment is same about any point O in the plane of forces.

Example: 8. Free Body Diagrams:

Definition: It is the sketch of the isolated body on which external forces acting on the body and the reactions of the removed elements are shown.

Simple examples:
(a) Ball resting on a floor.

(b) Two balls of different weights resting inside a smooth channel.

It may be noted that horizontal forces are denoted by H and vertical forces by V and inclined forces by R.
(c) Ladder resting on rough floor and rough wall

9. Transfer a force from one point to another point:
Let there be a force P at Point A as shown. Suppose for some reason it can not be applied at A and we wish to apply at B, what will happen can be explained as given below.

It shows a force can be transferred as it is, but with a moment of p.a.

10. Resultant of Coplaner Forces:

The analytical method of finding the resultant of a given force system is based on the principle of resolution of forces along horizontal and vertical axes.
Let F1,F2,F3. . . . . be the forces inclined to x axis (horizontal) at α123. . . .Then the various force components will be

where Φ is the angle R makes with the x-axis (horizontal)
11. Examples:
Example 1 : Find the resultant (magnitude and location) of the force system shown below.

i) Transfer 20KN force from point B to point A, we will have

ii) Let us suppose equivalent single force of 60KN is applied at C at a distance of x from A, we have

it may be noted that all the three diagrams are equivalent system.

Examples 2 : A rigid bar is subjected to a system of parallel forces as shown below. Reduce this system to a single force-moment system at A.

First transfer 20kN from B to A,
we will have (30-20) kN+60 kN.m
i.e.

Now transfer 30kN.m couple from C to A. It will go as it is, Therefore we will have at A

Example 3 : Reduce the given system of forces to an equivalent force-couple system at B.

Given System Up Force +ve , i.e. ↑ +ve

Net force at B = +60 + 30 - 80 - 40 = -30 kN ↓
Net Moment at B = +30 X 3 - 80 X 2 + 60 X 0 + 40 X 2 = +10 kN.m Answer ▼

Example 4 : Find the values of the forces P and Q if the system of forces in (a) & (b) are equivalent.

 system (a) Vertical forces +20 - 10 = +Q - P system (b) Vertical forces ∴ Q - P = 10 .................① Mat B = -10 x 2 = -Q x 2 -20 = -2Q ∴ Q = 10 .................② Q-P = 10 from ① Substituting the value of Q 10 - P = 10 ∴ P = 0 Answer P = 0, Q = 10kN ▼

Example 5 : A square plate of dimensions 4m x 4m is subjected to forces as shown. Determine magnitude, direction and location of the resultant single force.

Reduce the entire system at A

Therefore magnitude and direction of the resultant will be 28.28 KN and Φ = +135°
The direction of the Resultant is shown in the diagram by double arrows to distinguish from the given forces .
Determination of Location:
Suppose R crosses line AD at "y" distance below A. Moment at A due to R =(∑H) y = 20y (vertical component will vanish because it will pass through A)
∴ 20y = 230 ∴ y = 11.5m ▼ (Below A)
It is obvious that point E will not be between A and D but will be 11.5-4 = 7.5m below D

Example 6 : In each of the following three problems, give necessary equations for the solution of the problem.

(i) Apply Lami's Theorem to point O.

(ii) Two balls of same radius "r" and same weight W in a smooth trough

Consider FBD of LHS Ball,

Consider FBD of RHS Ball,

Solving (1) & (2),
we will get HA,HB & Vc
(iii) Distance between walls is 2b. Weight of each ball is W. Radius of each ball is r.

Example 7 : Two cylinders, 'A' of weight 4000N and 'B' of weight 2000N rest on smooth inclined planes. They are connected by a bar of negligible weight hinged in each cylinder at their geometrical center by smooth pins. Find the force P to be applied as shown in the figure such that the system is held in the position shown.

FBD of B :
 Σ along the inclined plane of 45° = 0 (Forces up the plane +ve. Plane will be parallel and pass through Centre B and N2 will be normal to this plane) = -P Cos15°- 2000 Cos45°+ N3 Cos60° = 0 Therefore P Cos15° = 1035.5 or P = 1071.8 N Answer ▼ . Note: In above equation N2 does not appear because N2 cos90° = 0.

Note: Force Representation Convention Used in VTG.

 +ve -ve

### Chapter 3

Reactions

1. Reactions- Definition, Calculations and Transfer of Loads:
So far we have observed the effect of freedom and restrictions on the structure. It is seen that any type of freedom of a member will make the structure useless and it will collapse.
In our life, we are required to face several restrictions (e.g. laws of the institution where we study, city and state laws and finally national and international laws. This may include the religious laws.)and these restrictions make us good citizens.
Every structure is to face actions (given design loads) and these actions are transmitted through man made connections (already explained roller support, hinged support and fixed support). Thus entire load of the structure is transferred to the ground through the supporting connections.
Religiously speaking, we all are supposed to merge into good earth, only in a different form depending upon religion. Well I am sorry to mention about the religious philosophy but that is the truth.
So far only Actions (given loads) have been described. What about Reactions.
Let us consider a Hinged support A as shown below.

This support will not allow any vertical movement and any horizontal movement.
In order to prevent vertical movement, a vertical force has to be applied. This vertical force is called a Vertical Reaction.

Similarly to prevent horizontal movement , a horizontal force is applied by the ground and this is called a Horizontal Reaction.
Now, we can define what is a Reaction.
Reaction is a force required to prevent the movement.
Moment Reaction is moment required to prevent the rotation at the end or connection of the structure.
Important note: Reactions are under the control of a Design Engineer. He can put the supports anywhere he likes.
2. Type of Beams, Type of Loads and Calculation of Reactions:
Figures of Beams and Loads are given below-
 (i) A Simple Beam (ii) An Overhanging Beam (iii) A Cantilever Beam (iv) A Fixed Beam

3. Type of Loads and their Mathematical Notations:

4. General Conditions of Equilibrium of Rigid Bodies or Structures:
The structure or rigid body subjected to loads P1, P2, P3, P4 & P5 is shown here.
Every force has two components horizontal and vertical.
In case we do not want the body to move horizontally, we should see that algebraic sum of all the horizontal components of forces(loads and reactions) should be zero.
Therefore, ΣH = 0, → +ve ---------- (1)
Similarly, if we want no vertical movement,
ΣV=0, ↑+ve ---------- (2)
In order to maintain the equilibrium, we must see that the body should not rotate about any point in the plane.
ΣM=0, +ve ---------- (3)
Mathematically, following are the general conditions of Equilibrium.
 ΣH = 0, →+ve ΣV = 0, ↑+ve ΣMat any point=0 , +ve
It is important to show the sign conventions.

5. Calculation of Reactions-Examples:
Example 1.

Step 1
Locate the Reactions and show them on the diagram. Since there is a hinge at A, we will have 2 reactions HA & VA to prevent horizontal and vertical movement.
There is a Roller Support at B which prevents only vertical movement, hence show VB at B.
Step 2
Apply the conditions of equilibrium in such a manner that one reaction is obtained easily. Since HA & VA are both acting at A, it will be better to apply the condition ΣM = 0, +ve at A.
 ΣMat A=0, +ve, Therefore HA × 0 + VA × 0 + (2×6)(3)+4×9- VB×12=0 Therefore VB = +6KN ↑ ▼ Answer 1 ΣH=0,→+ve , + HA=0 ▼Answer 2 ΣV=0, ↑+ve, + VA + VB –(2×6)-4=0 or + VA+6-12-4 = 0 ∴ VA=+10KN↑ ▼ Answer 3

Example 2. Cantilever Beam

Since only end A is fixed. It will prevent all the three movements, hence all the three reactions HA , VA & MA will act at A. B is completely free and therefore nothing should be done at B. Students normally make a mistake and apply VB at B which is wrong.
 ΣH = 0, →+ve , ∴ HA = 0 ▼(1st answer) ΣV=0, ↑+ve, + VA -2×6-4 = 0 ∴ VA = +16KN ▼(2nd Answer) ΣMat A = 0, +ve, HA × 0 + VA × 0 + MA+(2×6)(5)+4×10+6=0 ∴ MA= -106KN.m ▼ (3rd Answer)

Example 3 An Overhanging Beam.

2-uniformaly distributed loads 2KN/m & 4KN/m
Reactions VA & VB to be determined
Applying the general condtions of Equilibrium,
We have,
 ΣMat A=0, +ve, +6-(2×4)(2)+ VA×0 +(4×10)(5)- VB×10+2×14+4=0 ∴ VB = (6−16+200+28+4)/10 = +22.2 ↑KN ▼ ΣV = 0 , ↑+ve, +VA +VB-(2×4)- (4×10)-2 = 0 ∴ VB = +27.8KN ↑▼ & VA=+22.2KN↑▼
Example 4. A determinate Portal Frame.

End A is hinged. It will prevent two movements, so there will be 2 Reactions HA & VA . Assume +ve direction.
End B is a Roller Support.it will prevent vertical movement, so assume VB +ve direction i.e. upwards.
There are 3 concentrated loads. Two horizontal 6kN→ and 2kN ← plus one vertical concentrated load ↓10kN
Apply conditions of Equilibrium
 ΣH=0,→ +ve , +6+HA-2 = 0 ∴ HA = -4kN← ▼ ̆ ΣMat A=0, +ve, HA×0+VA×0+6×4+10×4 -2(8-3)- VB×8=0 ∴ VB= +6.75KN ↑ ▼ Σ V = 0 ,↑+ve , VA + VB -10 = 0 ∴ VA =+2.25kN ↑ ̆▼

### Chapter 4

Friction

(Friction involving horizontal plane, inclined plane, and Wedges)

1.What is Friction
Friction is the resistance to motion of one object moving relation to an-other. It takes place on the surface because of inter penetration of irregularities of two surfaces in contact and due to molecular attraction.
1.1 Friction on a Horizontal Plane

In above figure, there is a body of Weight "W" kN lying an a rough Horizontal Plane. Our aim is to move the body by applying a horizontal force "P" kN. P will be increased from zero to a value till the body starts moving.
In the beginning, the applied force "P" will be small and it will be opposed by equivalent frictional force so that body will remain at rest.
When P is increased to a value such that body starts moving. It means the limiting static frictional force Fs has been achieved. As soon as the body starts moving Fs gets reduced to "Fk". The reduction in "Fs" to "Fk" is because of irregularities get lifted up (which were interlocked in rest position).
Fs is called Limiting Static Frictional Force
Fk is called Kinematic Frictional Force.
Fk = µk.N & Fs = µs.N = µs.W , or µs = Fs/N = tanΦ
where as Φ is Angle of Friction.
µk is usually between 0.75 to 0.85 of µs.
P = W tanΦ ----------------------
1.2 Friction on an inclined plane
∝ = Angle of inclined plane with respect to horizontal plane
W = Weight of the body lying on the inclined plane.
µs = Coefficient of static friction
= tanΦ = Fs/N

from the Free body Diagram, we get, by using Lami's theorem
W/Sin(90+∝+Φ) = P/Sin[180-(∝+Φ)]
∴ P = W Sin(∝+Φ)/Cos(∝+Φ) = Wtan(∝+Φ)----------------------
Valid for Motion up the Inclined Plane
For Motion down the plane, frictional force will be in the oposite direction, therefore in place of +Φ, substitute -Φ
Accordingly P = W tan(∝-Φ)
When body slides down without P being applied,
We will have P = W(tan(∝-Φ) = 0
or tan(∝-Φ) = 0 ∴ ∝ = Φ
i.e When Angle of Friction is equal to Angle of Inclined plane the body will slide down.

1.3 Frictional Forces on Wedges
1.3.1 What are Wedges
Any triangular or trapezoidal piece of metal or hard wood which can be used to make the small adjustments in heavy pieces of machinery.
1.3.2 General Formula for Wedges
µ1 = Coefficient of Friction between wall and Weight W = tanΦ1
µ2 = Coefficient of Friction between Weight W and Wedge = tanΦ2
µ3 = Coefficient of Friction between Wedge and bottom floor = tanΦ3

1.3.3 Single Formula Concept of Friction Formula (Explanation and Justification)
 P = W tan(∝+2Φ)
(1) Horizontal Plane
∝ = 0, Friction taken place on a single surface so we will have "Φ" instead of 2Φ.
i.e P = W tanΦ----------------------
(2) Inclined Plane
"∝" remains as it is.
Similarly, single surface of contact and as such we will have "Φ" instead of 2Φ.
i.e P = W tan(∝+Φ)----------------------
(3) Wedge
"∝" remains as it.
Wegde suffers friction on two surfaces so we will have 2Φ
i.e P = W tan(∝+2Φ) ----------------------
General Formula when three surfaces including wall have angle of friction as Φ12and Φ3 .
Above Formula is not difficult to remember.
All the angles should appear in numerator and denominator.
For no Friction on the wall, Φ1 = 0

### Chapter 5

Wedges - using The Wedge Formula

Problems on wedges are normally solved by drawing free body diagrams of the Weight and the Wedge. Using conditions of equilibrium or Lami’s theorem, the desired results are obtained.
Here, the problems on the Wedge and the Inclined Plane are solveed by using Wedge formula given above,
 i.e. P = W tan(α+2Φ) when Φ1 = Φ2= Φ3= Φ

Otherwise, we use
Where α is the angle of the wedge and Φ1, Φ2 and Φ3 are the angle of friction on three diferent surfaces.
Example 1 : Determine the force P to be applied on the wedge to lift a weight of 600N. The coefficient of friction(µs) is 0.3 for all surfaces in contact.
The angle of wedge is 30°.

Since,
 ∝ = 30° and µs = 0.3, ∝ = 30° ∴ Φ1 = Φ2 =Φ3 = Φ 16.7° P = W tan(∝ +2Φ) = 600 tan(30+2×16.7) = 600 tan63.4° = 1200N =1.2kN (Answer)
Example 2 :Determine the Weight of Block B that will keep at rest when a force P start Block A up the plane on inclined surface of B( ∝ = 30°). The weight of A is 10kN. Angle of friction for all surfaces in contact is 15°( µs =0.268)

P is pushing wt. A =10KN up on the inclined plane.
Inclined plane formula
 P = W tan(∝ + Φ) = 10 tan(30 + 15) = 10KN N =WA +WB ∴ Fs = µsxN For WB to be at rest , Fs ≥ P i.e. µs (WA +WB) ≥ P or 0.268(10+WB) = 10

Example 3 : The block A shown in the figure, supports a load W and is to be raised by the wedge B under it as shown. If the angle of friction is 10° for all surfaces in contact, determine maximum wedge angle ”∝” that will give the wedge a mechanical advantage, i.e. make P less than the weight W of the block.

For mechanical advantage P < W at the most P will be equal to W
i.e. P = W,
Using Wedge Formula
 P = W tan(∝ + 2Φ) or W = W tan(∝ + 2×10) i.e. tan(∝ + 20) = 1 =tan45° ∴ ∝ + 20 = 45 or ∝ = 25° (Answer)
Example 4 : What force P must be applied to the wedges in order to lift the weight of W = 1000N. Angle of friction for all surfaces in contact is Φ =10°. The angle of the wedge is 15°.

Since there is no wall support , Φ1 = 0, Φ2 = Φ3 = Φ = 10°
We will use original wedge formula,
i.e
Because of symmetry each wedge will be required to lift 500N.
We have therefore Φ1 = 0, Φ2 = Φ3 = 10°, ∝ = 15° , W = 500

Example 5 : In the figure shown, Block A of Weight W =100kN is to be raised by pushing the wedge B of weight 40kN. What will be the value of force P on the wedge to raise the load. Also if Wedge is to be removed, what force P will be applied in the opposite direction.

Angle of wedge ∝ = 20°. Angle of friction = 15°
For all the surfaces in contact
 Let P1 be the force required to raise the load, assuming Weight of B as zero. & Let P2 = Force to overcome friction of Weight of wedge P2 = µs x40 P2 = tan-1 15° x 40 P2 = 0.268 ×40 = 10.72kN P1 = W tan(∝ + 2Φ) P1= 100 tan(20+2×15) = 100 tan50° = 119.17kN ∴ P = P1 + P2 = 119.17 + 10.72 = 129.89 kN (Answer)
Case 2 : When the wedge is to be pulled out
We can again write P = P1 + P2
P2 remains to be 10.72 but in opposite direction.
 P1 = W tan(∝ - 2Φ) because friction will at an opposite direction and such Φ will be -ve P1 = 100 tan(20 – 2 ×15) = 100 × tan(-10) = 100(-0.1763) P1 = -17.63 P2 = -10.72 ∴ P = -(17.63 + 10.72) = 28.35 kN (Answer)

Example 6 : In order to adjust the vertical position of a column supporting 200kN load , two 5° wedges are used as shown in the figure. Determine the force P necessary to move the load upwards. Angle of friction for all surfaces in contact is 25°. Neglect friction at the rollers.

Wall Friction(rollers ) = Φ1 = 0
Φ2 = Φ3 = 25°
First imagine wedge A fixed to the ground. Only wedge B acts and by applying force P, it will raise the load by suffering friction at two places, with wedge A and load W.

 P = 200 [ (Sin(05+25 + 25) )/( Cos(25 + 0+25))]( Cos0° / Cos25°) ∴ P = 200 [ (Sin(55°) )/( Cos30xCos25)] = (200x0.819)/(0.866x0.906) = 208.77kn
There will be no difference in value of P if both wedges start working as shown.

Example 7 : Two blocks A & B, each weighing 20 KN and resting on a horizontal surface are to be pushed apart by a 30° wedge. The angle of friction is 15° for all surfaced in contact. What value of P is required to start movement of blocks. How this will changed if the weight of one block A is increased to 40kN.

As done in previews example, first consider block A fixed to the horizontal surface.
Since load is always vertical and horizontal force is applied on Wedges.
Thus we will call block “C” as load and block A & B as Wedges. When A is fixed, push on Block “B” wedge will be (P+20)µ and load W of C is P
 Φ = 15° , ∴ µ = tan-1 15 = 0.268 and ∝ = 15° Applying Wedge Formula P = W tan(∝+2Φ) Push on the wedge B will be (P+20)µ and load W = P ∴ (P+20)(0.268) = P tan(15+2×15) = P×1 = P ∴ 0.732 P = 20×0.268 ∴ P = 7.32kN (Answer) Now if Block A has weight 20kN or more, say 40kN P = Push on Block B or A will not change.

It may be noted that examples 1 to 7 were solved by using wedge formula. Examples 1 to 5 were straight forward but examples 6 and 7 required some imagination and common sense.
Most of the real life problem on wedges are straight forward as given in the derivation of Wedge- Formula. Therefore solution can be obtained in two minutes.
An attempt has been made to tabulate the value of horizontal force “P” on the Wedge in terms of the Load “W” to be lifted or adjusted for popular values of co-efficient of static friction µs and the angle of the Wedge “∝”.
 Angle of Wedge µs Φ ∝ = 5° ∝ = 7° ∝ = 10° ∝ = 12° ∝ = 15° µs = 0.2 Φ =11.31 P = 0.5232W P = 0.5685W P = 0.640W P = 0.6904W P= 0.770W µs = 0.25 Φ =14.04 P = 0.6512W P= 0.7021W P=0.7833W P=0.841W P=0.935W µs = 0.30 Φ =16.70 P= 0.7925W P= 0.8510W P=0.9456W P=1.014W P=1.126W µs = 0.35 Φ =19.29 P=0.9516W P=1.021W P=1.1335W P=1.2166W P=1.355W
There is no point in discussing the results of above table since the value obtained are obvious.
Solve again
Example 7. using standard method of applying Lami’s Theorem to Free Body diagrams of Load and the Wedge.
Given Diagram
 i).Free Body diagram of wedge “C” ii).Free Body Diagram of load B Apply Lami’s Theorem P/(Sin(180-30-30)) = R2/(Sin(90+30)) Apply Lami’s Theorem R2/(Sin(180-15)) = P 20/(Sin(90+30+15)) ∴ R2 P Cos30/Sin60 .................① ∴ R2 = 20 x(Sin15/Cos45) .................② Comparing Equation ① and ②, we have, R2 = P Cos30/Sin60 = P = 20Sin15/Cos45 = 20 x (0.2588/0.707) = 7.32kN (Answer)

Summary and Conclusion
Frictional force is independent of the area of contact Frictional force acts in tangentical direction to the surface in contact and opposite to he direction of motion.
Coefficient of static friction always greater than coefficient of dynamic friction.The volume of static friction is known as limiting friction.
Angle of friction is the angle between normal reaction and the resultant of normal reaction and limiting friction.
The relation between the different parameters “µ”, “Φ”, “∝” and N are as follows:
µ=tanΦ, µ = Coefficient of friction, Φ = Angle of friction.
F= µN, F = Friction Force
∝ = Φ here ∝ = Angle of repose
"∝" is also used as an Angle of the inclined plane.
A body of weight “W” on an inclined plane of angle “ ∝ ” is being pulled up by a horizontal force “P”. The motion impends when P=W tan(∝+Φ ),

For N threaded screw the “Lead” of the screw is L = n times the pitch and angle of the screw jack is tanΦ= L / 2πr
Wedges are used to lift or adjust the heavy blocks or machinery by applying a force P which is usually horizontal.
Wedges formula in general form is

In most of the cases it is given that when Φ1 = Φ2= Φ3= Φ
Hence
 P = W tan(α+2Φ)
This Formula is valid for horizontal plane and inclined plane also
Horizontal Plane ∝ = 0, single surface for friction, therefor 2Φ to be replaced by Φ , P = W tanΦ
Inclined plane P = W tan(α+Φ)

### Chapter 6

Analysis of Trusses

(Method of Joints And Method of Sections)

What is a Truss
Truss is a structure made up of straight members joined together at their ends by pin joints.
There are two categories of Trusses.
1. Plane Trusses
2. Space Trusses
Examples of Plane Trusses are Roof Trusses and Bridge Trusses.
Example of space Truss is a Transmission Line Tower.
Types of Plane Trusses
1. Roof Trusses
2.Bridge Trusses 3.Stadium or Pavilion Truss

4. Tower Truss
Stability of Truss and Idealised Truss
When we examine the stability of the truss, we come across the terms
(i) Perfect Frame
(ii) Redundant Frame
(iii) Deficient Frame
An idealized truss is said to be “just-rigid ”. If we remove any of its member, the frame so called truss will collapse and the rigidity of the truss is lost.
A basic stable structure (which is just rigid ) is a structure with only three members and three joints it forms a triangle.
Thus , when we make a simple plane truss, we begin with an elementary stable structure of three members and three joints (a triangle ).It may be seen from the sketch given below that to make it stable we have to add two members with a joint. For every additional two members, one joint has to be provided
Let m = Total number of members in a given Plane Truss &
j = Total numbers of joints in the same truss under consideration
If we proceed from the basic elementary truss
(a) which is a triangle having 3 members and 3 joints
m = 3 +2(j-3) or m = 2j-3
If the number of members are more then 2j-3, the truss is said to be over-rigid. Thus, we say,
 m = 2j-3 , perfect frame m > 2j-3 , redundant frame m < 2j-3 , deficient frame.
A simple example is given below
Assumptions in the analysis of trusses
1. All the joints are assumed to be friction less pins.
2. The members of the truss are connected at their ends.
3. All the given loads are applied at the joints only.
4. The weight of the members may be neglected.
5. C.G line of all the members meet at one point at the joint.
6. The section of the members are assumed to be uniform.
Free Body Diagrams(FBD) of a Member and the Joint Joint A
FBD of Joint A
Assuming tension in members AB and AC (for tension arrows will be away from the Joints). VA is assumed to be +ve reaction upwards and P is assumed to be load at the joint .
In all practical applications, we normally consider FBD of joints only.
Identification of Zero force members
Rule 1. When two members meeting at the joint are not in one straight line and there is no load or reaction at the joint, each of the two members will have Zero Force .
i.e FAB = FAC = 0
Rule 2. When three members meet at the joint, if two members are in one straight line, then the only inclined member will have zero force (provided there is no load or reaction at the joint)
In the diagram FAD = 0,
because its vertical components cannot be balanced.
Examples
1. 2 3 4 Step-wise Procedure for Method of Joints :
1. Consider the equilibrium of entire truss and determine the reactions.(VA,VB and HB).
2. Choose a joint where there are only two unknown member forces(say joint A)
3. Sketch the Free Body Diagram of the joint(A) so chosen.
4. Assuming Tension in the unknown members forces,(FAC & FAD – for tension arrow away from the joint A). Apply conditions of equilibrium ∑H = 0,→ +ve, ∑V = 0 ↑ +ve.
This will give us the magnitude of these two unknown member forces (FAC & FAD)
5.Proceed to next joint and repeat the procedure (Joint C or D). This will give the values for next two member forces
6.In this manner, we can determine the forces in all the members of the truss.
Method of Joints is used to determine forces in all the members of the truss.

Example 1 : Determine the forces in all the members of the truss given below

Reactions :-
 ∑ Mat A= 0, +ve VA × 0 + 16×4+4×3+HBx 0 -VB ×8 = 0 ∴ VB = + 9.5kN ↑ ∑V= 0, ↑ +ve, +VA +VB-16 = 0,or VA + 9.5 -16 = 0 ∴ VA = +6.5kN ↑ ∑H = 0 , →+ve, +HB + 4 = 0 ∴ HB= -4kN ←
i)FBD of joint A :
 ∑V = 0 , ↑ +ve, +6.5 + FAC SinΘ = 0 ∴ FAC = -6.5/SinΘ = -6.5/3/5= -10.83kN (comp). ∑H = 0, → +ve, FACCosΘ +FAD ∴ FAD = +8.67 (T)
ii)FBD of joint D
 ∑V = 0 , ↑ +ve, +FCD -16= 0 ∴ FCD = +16kN (T) ∑H = 0 , → +ve, +FDB -8.67= 0 ∴ FDB= + 8.67kN(T)
iii)FBD of Joint B
 ∑V= 0, ↑+ve FBC SinΘ +9.5 = 0 ∴ FBC = -15.83(C)
Example 2 : Determine the Forces in all the members of the cantilever truss given below

Since its is a cantilever truss, the reaction need not be determined.
i). Joint E
 ∑V = 0, ↑ +ve -30 - FDE SinΘ = 0 ∴ FDE = -30×5/3 = -50kN (C) ∑H = 0, → +ve, -FCE-FDE CosΘ = 0 ∴ -FCE + 50CosΘ = 0 ∴ FCE = +40kN(T)
ii). Joint C
Assume tension in unknown members AC and CD.

 ∑H = 0, → +ve, +40- FAC = 0 ∴ FAC = +40 kN(T) ∑V = 0, ↑+ve, -30-FCD = 0 ∴FCD = -30kN (C)

iii) Joint D
Assume Tension in unknown members AD and BD.
 ∑V = 0, ↑+ve, FADSinΘ -30-50 SinΘ = 0 ∴ FAD = 60/SinΘ = +100kN(T) ∑ H = 0, →+ve -FBD-FADCosΘ-50CosΘ = 0 ∴ -FBD -100 x 4/5-50 x4/5 = 0 ∴ FBD = -120kN(C)
It may be noted that each joint gives us answers for two unknown member forces.

Example 3 : Determine forces in all the members of the truss shown in the figure

i). joint E
 ∑ H = 0, →+ve, -FEF = 0 ∴ FEF = 0 ∑ V = 0, ↑+ve +FDE -10 = 0 ∴ FDE = 10kN (T)
ii). Joint D

 ∑V = 0, ↑ +ve, -10-FDF CosΘ = 0 ∴ FDF = -10/CosΘ = -10/4/5= -12.5kN (C ) ∑H = 0, → +ve, -FCD-5-FDFSinΘ = 0 ∴ -FCD -5 +12.5x3/5 = 0 ∴ FCD = +2.5kN(T)
iii). Joint C FCF = 0 and forces in BC & CD will be equal
 ∴ FCF = 0, FCD = FBC = +2.5 kN(T)(Can be Solved orally)
iv). Joint F Can be solved orally by looking at the joint from, ∑V = 0, FBF will be equal and opposite to FDF
 ∴ FBF = +12.5kN(T) since FDF is compressive
FGF automatically works out as follows Horizontal components of both FDF &
 FBF is add to 7.5 kN +7.5 kN = 15kN → ∴ FGF = 15kN → (to the joint)( C )

v). Joint B
 ∑V = 0, ↑+ve, -5-12.5×4/5 -FBH x 4/5 = 0 ∴ FBH = -15×5/4 = -18.75 ( C) ∑H = 0, →+ve, -FAB +2.5 + 12.5×3/5 +18.75 ×3/5 = 0 ∴ FAB = +21.25 kN(T)

Example 4 : Determine the forces in all the members of the truss shown in the figure.

Reactions:
Most of the students find it difficult to calculate the reactions because of determining the horizontal distance of 12kN load from A or B.
Horizontal distance AE = 2.5+(7.5/2)=6.25 m.
 ∑ M at A = 0, +ve, VA x 0 +10x2.5+12x6.25-VBx10 =0 ∴ VB =10 kN ↑ and ∴ VA= +12kN ↑
i). Joint A: Assume tension in AC & AD as shown in FBD,
 ∑V=0, ↑ +ve, +12+FACSin60=0 Therefore, FAC= -13.86kN (c) ∑H = 0, → +ve, +FAD+FAC cos60=0 Therefore, FAD=+6.93kN (T)
ii). Joint B:
 ∑V=0,↑ +ve,+10+FBE Sin30=0 Therefore, FBE= -20kN(c) ∑H=0, → +ve, -FBD- FBE cos30=0 Therefore, -FBD +(20 x 0.866) =0 Therefore, FBD = +17.32kN (T)
iii). Joint E:
 ∑along DE=0, -FDE -12cos30 = 0 Therefore, FDE= -10.39 kN (c) ∑along perpendicular to DE=0, △ +ve -FCE -20+12cos60 = 0 Therefore, FCE = -14kN(c)
iv). Joint D: ∑V=0, ↑+ve, this will obviously give us FCD = +10.39kN (T).

Example 5 : Determine forces in all the members of the truss shown in diagram

Reactions:
Because of symmetry;
VA=VB=10kN ↑ each
i). Joint A Assume tension in AD & AC

 ∑V=0, → +ve, +10+FAD cos60+FACcos30=0 +10+0.5FAD+0.866 FAC=0 ....① ∑H=0, → +ve FAD cos30+ FACcos60=0 Therefore, 0.866 FAD+0.5FAC=0 ....② FAC = -1.732FAD Substituting FAC in ① above, we get FAC = -17.32 kN (c), FAD = +10kN (T)
ii). Joint B Because of symmetry,
 FBD = +10kN (T) and FBC =-17.32 kN (c)
iii). Joint C Assume tension in CD

 ∑V=0, ↑ +ve , -20-FCD +17.32 cos30+17.32 cos30 = 0 Therefore, FCD = -20+(2x17.32cos30x0.866) = +10 Therefore, FCD = +10kN (T)

### Chapter 7

Method of Sections

This method is used to find the forces in one or a few members of the truss. It is used as a design check. The method consists in dividing the given truss into two parts, by passing a curved section through the members of interest. Free body diagram may be drawn for the left hand or right hand side portion of truss and the forces in the members cut are determined.

1. Step-by-step Procedure for Method of section :

1. Consider the equilibrium of the complete truss and determine the reactions.
2. Take a section through the members of interest(taking care that not more than 3 members are cut).
3. This will result in having two portions of the truss, left hand side and right hand side. In case of vertical truss, lower portion or upper portion trusses may be obtained.
4. Depending upon the convenience, we may consider the FBD of either LHS truss portion or RHS truss portion. Do not forget to replace forces of the members cut as external forces.And these three forces may be arounded tensile
5. The three members cut and replaced by unknown forces will act like three unknown reactions.
6. Consider the FBD of LHS portion of the truss. Applying conditions of equilibrium ∑H=0, ∑V=0 and ∑M=0, we can find the unknown members forces to be removed.
2. Conclusion :
It is concluded that one has to apply the conditions of equilibrium to two trusses.
Firstly, the main given truss is considered to find the reactions.
Secondly, the left hand side or right hand side portion truss is considered to determine unknown member forces.
3. Use of components of a force :
When there are two or more inclined unknown member forces, it may be easier to find its horizontal or vertical component.

 If H is determined If V is determined F = H x (r/h) F = V x (r/v)

4. Examples:
Example 1 : Determine forces in the members CD and DH of the truss shown

(i) Consider equilibrium of complete truss
 ∑H=, → +ve , will give HA=-1.5 kN ← ∑M at A = 0, +ve , will give us VB=6.5 kN ↑ ∴ VA=5.5 kN
(ii) Consider equilibrium of LHS truss as shown above,

a. Member DH
 ∑V=0, ↑+ve +5.5-4.0+FDH Sinθ =0 ∴ FDH=(-1.5/sinθ)=-1.875 kN (c)
b. Member CD
 ∑Mat H=0, +ve (only LHS truss) (+5.5x3)+(1.5x0)+(4x0)+(FDHx0)+FGH x 0 +FCDx4 =0 ∴ FCD=(-5.5x3/4)=-4.125 kN (c)
Example 2 : Determine the forces in the members CE, CD, and DB of the vertical truss shown in the figure.

(i) Reactions : (these may not be required because of canti-lever truss)
 ∑H = 0,→ +ve ∴ HA +1+2+2 = 0 ∑HA=-5 kN ← ∑Mat A=0, +ve, HAx0+VAx0+8x0+8x4+1x9+2x6+2x3-VBx4= 0 ∴ VB=+14.75 kN , hence VA=+1.25 kN

(ii) Consider FBD of upper portion of the truss by taking section through the members in which we are interested.
Assume tension in members cut,then:
a. Member CD:
 ∑H = 0, ←+ve ,+1+2+2-FCD = 0 ∴ FCD = +5 kN (T) ▼
b. Member CE:
 ∑M at D = 0, +ve, (-8x4)-(FCEx4)+(8x0)+(1x6)+(2x3)+(FCDx0)+(FDBx0)+(2x0)=0 ∴ FCE = (-20/4) = -5 kN (c) ▼
C. Member DB:
 ∑V=0,(-8-8+5-FDB) = 0 ∴ FDB=-11 kN (c) ▼

Example 3 : Determine the forces in the members GE, EH, and FH for truss shown in figure.

(i) Reactions :
Consider FBD of complete given Truss
 ∑M at A=0, +ve, VAx0+4x4+6x8+8x12+10x16+12x20-VBx24 = 0 ∴ VB=+23.33 kN ↑, hence VA=16.67 kN ↑

(ii) Take a section through members of interest EG, EH, and HF.
Assuming tension in these members, FBD of LHS Truss is shown

(iii) Member FH :
Choose a point where remaining two members EG and EH meet i.e. point E.
 ∑M at E = 0, +ve, [(+16.67x8)-(4x4)+(6x0)+(FGE x 0)+(FEH x 0)-(FFH x 4)]=0 ∴ FFH=+29.34 kN (T)▼
(iv) Member EG:
Choose a point where remaining two members EH amd FH meet i.e. point H.
 ∑M at H = 0, +ve, [(+16.67x12)-(4x8)-(6x4)+(FEH x 0)+(FFH x 0)+(VGE x 0)+(HGE x 5)=0

{please note that VGE & HGE are components of “FGE” at G}

 ∴ HGE = -28.08 Hence, FGE = (HGE/h)x r = (-28.08/4)x (√17) = -28.94 kN(C) ▼
5.Member EH:
Remaining two members GE & FH meet at 0.
 ∑M at O =0, +ve, [(-16.67x8)+(4x12)+(6x16)+(FGE x0)+(FFH x0)+(HEH x0)+(VEH x 20)=0

{please note that HEH &VEH are components of FEH at H}

 ∴ VEH=(-10.64/20)=-0.532 & FEH =(0.532/4)x (4√2)=-0.752 kN (C) ▼

Example 4: Determine the forces in the members CE,DE and DF using method of sections figure of the truss is shown below

FBD of upper portion or LHS portion of truss

(i) Reaction
FBD of given Truss, ∑M at E = 0 , +ve ,-100×10VE×0-VFx5 =0
∴ VF = -200 kN ↓, Hence VE = 300kN ↑
In the method of sections for a cantilever Truss these are not required.

(ii) Member CE : Remaining Two members meet at point D.(DF & DE)
 ∴ Mat D = 0, +ve , -100×10- VCE ×10 +FDF ×0+FDE ×0 =0 ∴ VCE & HCE are components of FCE at A. ∴ VCE = -100kN , hence FCE = VCE/ϑ×r = (-100×√(52+2.52 ))/2.5 = (-223.61)/((c))kN (C) ▼

iii). Member DF :
Remaining two members meet at point E.(DE & CE)
 ∴ ∑ M at E = 0 , +ve , -100×10+FCE × 0 +FDE×0+HDF×0+VDF×5 = 0 ∴ VDF = +200kN ∴ FDF = VDF/ϑ×r = (-200×5√2)/5 = 282.8 kN(T)▼
Please note that HDF & VDF are components of FDF at F.

iv) Member DE:
Remaining Two members DF and CE meet at o
 ∴ ∑ M at O = 0 , +ve , -100×20-FDE × 10 +FDF×0+FCEx0= 0 ∴ FDE = -200 kN(C) ▼

Determine the forces in the members of the truss shown in Example 4 by the method of Joints. Forces to be determined only in three members CE,DE and DF
Reactions to be determined as given in Example 4 .

 VE = 300kN ↑, VF = 200kN ↓
Joint F
 ∑V=0 , ↑ +ve FDF sin45-200 = 0 ∴ FDF = 282.8kN( T)▼ ∑H= 0, +ve -FDF cos45-FEF = 0 ∴ FEF = -200kN(C ) ▼
Joint E
 ∑H=0 ,→ +ve -200-FCECosθ = 0 ∴ FCE = -223.6kN( C) ▼ ∑V = 0 , ↑ +ve , +FDE +300+FCE Sinθ = 0 ∴ FDE = -300-FCESinθ = 0 = -300+223.6×2.5/5.59 -200kN( C)▼

### Chapter 8

Centre of Gravity

1. Centre of Gravity :
Centre of gravity is defined to be a point through which the weight of the body is assumed to act. A rigid body consists of a large number of particles. The force of gravity acts on these particles. These gravitational forces acting on these particles form a system of parallel forces. The resultant of these parallel forces act in all the positions of the body through a point called centre of gravity.  If all the particles of a rigid body lie in one plane, it is called as a plane figure. The centre of gravity of plane figure is termed as centroid because, plane figures do not have any weight.

2. Centroid Of Plane Figure & Axis of Symmetry :
A plane figure has no thickness and hence it has no weight. Hence the centre of the gravity of the plane figure is called Centroid of the area or just a centre. It may be defined as a point through which entire area is assumed to act. In most of the problems, in Engineering Mechanics, it is observed that the rigid bodies or plane figures are symmetrical about an axis. This axis is termed as an Axis of Symmetry”. Therefore, it is obvious that the  C.G of the rigid body or the centroid of the area necessarily lies on the axis of symmetry.

3. Location Of the Centroid :
Since we are dealing with plane figures, their areas will lie in one plane XY as shown.
Let, ͞x = Unknown distance of C.G from Y-axis.
“For lines and areas, C.G is known as centroid and depends only on the geometry of body”.
The centroid of the area can be located by calculating the values of ͞x  by simple Mathematics.
Moment of the entire area about Y-axis:

4. Location Of Centroid – General method and procedure :
When a given area is built up of number of known areas (rectangle, triangle, circles, semi-circles etc.) with known centroids as shown, the following procedure may be adopted. i). Choose a reference axis, usually bottom most point of the section passes through the reference axis.
ii). Identify the known areas and assign a number to each of the areas (Semi-circle 1, Rectangle 2 & 3, triangle 4).
iii). Calculate the areas A1,A2,A3, and A4 and also calculate the sum of the areas. A or ∑A= A1+A2+A3+A4
iv). Calculate the moment of all these areas about the reference axis, the distance of the C.Gs of these areas from reference axis will be y1,y2,y3 & y4.
We will get, ∑Ay=A1y1+A2y2+A3y3+A4y4.
v). Finally as per theory above, ?⋅ ̅y = ∑?? =?1 ?1+?2 ?2+?3 ?3+?4 ?4 It will be convenient to choose the reference axis in such a way that given area lies in the positive co-ordinates, which is shown in the diagrams below:
Example 1: A square plate of metal has a square of one quarter cut from one corner. calculate the position of the centroid of remaining portion of the plate.

let the dimensions of the given square plate be a x a
Area of given plate is
Example 2: Determine the centroid of the area shown in the figure given below

Total Area of the section is:

### Chapter 9

Moment Of Inertia

Moment of inertia is defined as the second moment of an area about the given axis. The section shown in the figure has two axes passing through the centroid.
The moments of inertia about the two axes are:
 IXX = ∑ y2.dA IYY = ∑ x2.dA

This appears to be a mathematical expression for a given section.
Its physical significance lies in the fact that it is a sort of measure of resistance to bending. Greater the Moment of Inertia, greater will be resistance to bending.
Similarly mass Moment of Inertia is used to calculate the resistance to rotation of the fly wheels and the kinetic energy of the rotation.
In order to determine the Moment of Inertia of any section of the beam or a member, we will need the application of two theorems described below.
(i) Parallel-Axis Theorem:
It states that the Moment of Inertia of an area about an axis is equal to the sum of Moment of Inertia passing through the centroid and parallel to given axis, plus the product of area and square of the distance between the two parallel axis.
Proof:
 By Definition of Moment of Inertia
Important: Students normally make mistake in using this theorem. The theorem is not valid for any two parallel axes as its name suggests. One of the two parallel axes,must be a centroidal axes.
Example: For the given rectangular section the moment of inertia about the horizontal centroidal axis is 216mm4. Determine I?1?1, I?2?2 & I?3?3
Given IXX (C.G.) = 216mm4 Using parallel axis theorem.
Important: After obtaining the value of I?1?1, attempt is made to apply parallel axes theorem to ?1?1 & ?2?2 which is wrong because none of these two is a centroidal axis.
It may also be noted that I??(C.G) will be minimum amongst all parallel axes.

ii) Perpendicular Axis Theorem
Let Ip = Moment of Inertia  of plane lamina lying in xy plane about an axis, perpendicular to xy plane and passing through C.G.
Therefore, By Definition of Moment of Inertia Π
 Ip = ∑ r2dA Ip = ∑ (x2 + y2).dA = ∑ y2.dA + ∑ x2.dA Ip = IXX + IYY
Thus, Moment of Inertia of a plane lamina about an axis perpendicular to the lamina and passing through its centroid is equal to the sum of the Moment of Inertia s of the lamina about two mutually perpendicular axes passing through the Centroid in the plane of the lamina.
Determination of The Moment of Inertia of Some Standard Sections.
1. Rectangular section

ii).Triangular Section
In this problem , it will be easier to find moment inertia about the base of triangle. After this determination , by using parallel axis theorem,I?? can be determined.

iii). Solid Circular Section
Let Ip = Moment of Inertia of plane lamina lying in xy plane about an axis, perpendicular to xy plane passing through C.G.
Consider a small annular ring of radius r and thickness dr, we have by definition of Moment of Inertia.

(iv) Hollow Circular Section

v) Thin Circular Tube of Radius = R and Thickness = t

vi). Semi Circular Section  of radius R.

To Determine  Moment of Inertia of a Composite area about the C.G of the given Composite Area.
Frist , The C.G of the Composite area is determined as per the reference axes discussed earlier .
Let there be “ i  ” elements of different areas in the given composite area section.
From here onwards, there are two methods.
Method one is a regular method.
Method two is an indirect method.
This second method can be used in a tabular form also
Method 1:
Method 2
First, the Moment of Inertia about the reference axis is calculated for all the “ i ” number of elements.
Next, using parallel axis theorem , Moment of Inertia about the C.G of composite section is obtained .
For this equation, a table can be made and the quantities entered in the table.
Above method will be explained by an example of 2 elements( 2 rectangles only) as given below.

Example 1 A tee-Section is made up of 2 rectangles as shown. Determine the Moment of Inertia about horizontal and vertical axis passing through the C.G. of the Section.

 i = 2 elements A1 = 10×2 = 20mm2 A2 = 10×2 = 20mm2 A= A1+A2 = 40mm2 y1 = 5mm , y2 = 11mm
Method 1 Method 2
OR Table Format
 S.No Ixx A y A y2 1 166.67 20 5 500 2 6.67 20 11 2420 ∑(sum) 173.34 40 2920
̤̇ Since Y-axis is passing through C.G. of both the rectangles, it will be easy to find Iyy(C.G.) . Sometimes ,one is confused,about, which dimension should be cubed and which should not to be cubed in the formula Ixx = 1/12 bd3. It is easy to remember that if you wish to calculate Iyy, the dimension parallel to yy axis (say d) should not be cubed but will be allowed to remain same i.e. Iyy = 1/12db3.Similarly in the formula 1/12 bd3, b is a dimension parallel to the axis for which you want M.I. i.e. Ixx = 1/12 bd3
Example 2. Determine Moment of Inertia about x and y axis passing through the C.G of an angle -iron section shown in the figure .
 i= 2 No. of Elements , 1 & 2 A1 = 80 mm2 , y1 = 1, x1 = 20 A2 = 20 mm2 , y2 = 7, x2 = 1
Example 3. Determine Moment of Inertia about x and y axis passing through the C.G of channel section shown in the Figure.

Reference axes X,Y are as shown.x,y are axes passing through the C.G. of the composite area.

 i = 3 number of elements. A1 = 10 X 2 =20mm2 A2 = 30 X 2 =60mm2 A3 = 40 X 2 =80mm2 ∴ A1 + A2 + A3 = 160mm2
Example 4: determine the Moment of Inertia about the x-x & y-y axis shown in the figure.
It may be noted that the section is symmetrical about an x-x axis and the C.G. location has not been asked. Some of the students try to find C.G Location and waste time. This problem is meant to find out whether the students know the correct use of parallel-axis theorem. i=4 type of elements
• Two rectangles identically placed(1,1)
• Two triangles also identically Placed(2,2)
• One Full semicircle (3)
• one hollow semicircle(-4, It means material removed)

Please note that y-y axis is the base for both rectangle & triangle. So base M.I. formula has been used. For semicircle, diameters are parallel to y-y axis.
Therefore M.I.= 0.11R4.

Moment of inertia with respect to area is defined as m4 or mm4 units. And is expressed as area multiplied by the distance squared or length “K” squared. Thus we may as well write as
Is called the “Radius of Gyration ” the subscript x to K denote the axis about which M.I. is taken i.e Ixx.
5.Product of inertia
We have already defined the M.I. in which distance square appears and it makes it always positive i.e.
This quantity Ixy is called a product of inertia. Depending upon the position of the area in the axes, product of inertia can be positive, zero or negative.
Let us take a rectangular section b xd with symmetrical axis x-x passing through C.G., dividing the area equally into two part A/2 & A/2
Product of inertia about x-x, y-y axes will be equal to Ixy = (A/2.d/4) +A/2(-d/4) = 0
We conclude that  when we have one or more axes of symmetry, product of inertia Ixy = 0

Product of Inertia and Parallel Axis Theorem
It can be proved that parallel axis theorem will be applicable to product of inertia.
In the figure shown x,y are the axes passing through the C.G. of the section. X,Y are the axes about which product of inertia  is to be determined (i). Rectangular Section Because of Axes of symmetry through C.G of rectangular section bxd.
Ixy(C.G) Is zero.
 Hence or Ixy = 0 + (bd)(b/2)(d/2) Ixy = b2d2/4
(ii). Right angle Triangle
Determine product of inertia about the sides of right triangle from the fundamental definition

Product of Inertia and “Angular Axis Formula ”
So far, we have dealt with parallel axis theorem and perpendicular axis theorem. we have got to calculate what will happen to these quantities(Ixx, Iyy & Ixy) Due to change in angle of the axes(?)
Section is given with known quantities C.G , Area A, Ixx,Iyy & Ixy.
Interesting Conclusion of above 3 equations.
substituting this value of ? in Equation (1) & (2) , we get, & Equations (1),(2) & (3) are known as "Angular Axis-Formally".

Principal Axes and Principal Moment of Inertia
Principal axes are those axes about which product of inertia is zero and therefore the moment of inertia about these axes will be maximum and minimum.
Hence principal moment of inertia are those which are obtained about the centroidal principal axes. On one principal axis it will be maximum and on the other perpendicular axis, it will be minimum. Their magnitude will be as given below :-

Relationship between mass Moment of Inertia and Area Moment of inertia
Mass = density × volume , i.e. m = ?×v = ?t.A= ?t.A i.e. dm = ?txdA
if Ixx =   ∑ y2.dA , Ixx(mass) = ∑ y2.dm = ∑ y2dA.?t
Therefore Ixx(mass) = Ixx(Area). ?t Where t is uniform thickness
General formula will be Imass = ? t.IA = M/A. IA Where M = Total mass

Example 1 Solid Circular disc of radius “R”  and mass “M” about perpendicular axis
 Consider Circular lamina of Area A = ?R2 Ixx = Iyy = ?R4/4 Ixx + Iyy = ?R4/2 Ip (mass) = ?t.IA = M/A. IA = M/?R2. ?R4/2 IP (mass) = MR2/2 Very importannt formula Unit kgm2
Example 2 Solid Circular cylinder of Radius R and Length L.
Example 3. Determine Mass Moment of inertia of a solid cone about its vertical axis passing through the center of base and apex.
Example 4 Determine Mass Moment of inertia of a sphere about any diameter volume of elementary disc of radius “y” is .dx

Summary of the chapter on moment of inertia (i) The center of gravity of a plane figure is called centroid of the area. The units of Moment of Inertia with respect to area are mm4 and that of moment of inertia of mass are Kgm2.
(ii) Moment of inertia of a circular area about an axis perpendicular to the plane of area is ?R4/2 or ?D4/32
(iii) Moment of Inertia of triangular section is 1/36.bh3 about Its central and an axis parallel to the base. I base = 1/12 bh3
(iv) Moment of Inertia of an area is always positive and minimum about the centroidal axis .
Where as product of Inertia  Ixy for an area can be positive, zero or negative. If it has one or more axes of symmetry, product of inertia Ixy=0
(v) The axes for which Ixy = 0 are known as principal axis. And the Moment of Inertia about these axes are known as principal moment of inertia
(vi) About one principal axis, we will get maximum moment of inertia and about another axis will get minimum moment of inertia
The angle for the axes Imax and Imin  is obtained as tan2? = -2Ixy/Ixx-Iyy
(vii) For the plate of thickness t and density  ?, Imass = ?t Iareaor Imass = (M/A).IArea.
(viii) Mass moment of inertia of a circular cylinder of radius R and mass M about is axis is MR2/2 and that of sphere is 2/5 MR2.
(ix) Moment of inertia of area of the circular tube of radius “R” and Thickness “t” about horizontal or vertical axis is Ixx = Iyy = ?R3 t
(x) Mass moment of inertia of a thin spherical shell of mass M and Radius R is 2/3 MR2.