SEET 212: APPLIED MECHANICS

STATICS
!.0 Basic Concepts
The study of statics is based on six fundamental principles. These are
_ Newton’s three fundamental laws,
_ parallelogram law for the addition of forces,
_ principle of transmissibility,
_ Newton’s law of gravitation.

A force represents the action of one body on another. This action can be realized
by actual contact or by action at a distance (e.g. gravitational force). A force is
represented by a vector. It is characterized by its magnitude, its point of action, and
its direction. We should distinguish three kinds of vectors, namely a free vector, a
fixed vector, and a vector bound to its line of action. All of these vectors have their
place in mechanics.

1.1 Newton’s laws

First Law
A particle originally at rest or moving in a straight line with constant velocity, will remain in
this state provided the particle is not subjected to an unbalanced force.
Second law
A particle acted upon by an unbalanced force F experiences an acceleration a that has the
same direction as the force and a magnitude that is directly proportional to the force. If the
particle has a mass denoted by m, the law may be expressed by
F= ma 1.1

Third Law
The mutual forces of action and reaction between two particles are equal, opposite and
collinear.

Figure 1.1 Force of A on B and force of B on A

Newton’s law of gravitational attraction

mm
F = G 12 2 1.2
r
where F= force of gravitational attraction between the two particles
G= universal constant of gravity: from experimental results, G=66.73(10-12) m3/(kg.s2)
m1, m2= mass of each of the two particles
r = distance between the two particles
In this regard weight is the force of gravitational attraction between a body and the earth.
mM
i.e W = G 2 e 1.3
r
Letting g= GMe/r2 yields
W= mg 1.4

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1.2 System of forces
• When several forces of different magnitudes and directions act on a body they
constitute a system of forces. They can be classified into coplanar forces and non-
coplanar forces.
• Coplanar forces: Coplanar forces lie on the same plane. This group can be divided
into three (coplanar concurrent forces, coplanar parallel forces and non
concurrent/non parallel forces).

Coplanar concurrent forces Coplanar parallel forces Non-concurrent/non-

Parallel Coplanar forces
Figure 1.2 Types of concurrent forces

Non coplanar forces

• Non coplanar forces are spatial forces since they do not lie in one plane. This group is
classified into non coplanar concurrent and non-concurrent/non-coplanar forces.
Forces on a tripod carrying a camera is an example of the first while forces acting on
a moving vehicle is a good example of the second.

Figure 1.3 Non coplanar forces

Equilibrium, Resultant and Equilibrant

When two or more forces act on a body in such a way that the body remains in a state of rest
or uniform motion, then the system of forces is said to be at equilibrium.
The vector sum of the forces acting on a body is referred to as the resultant force. Thus
resultant force produces the same effect as done by the combined effects of several forces.

The system of forces on a body may be such that the body is not at equilibrium, and the
resultant force can represent all the forces in magnitude and direction. If however a force that
is equal and opposite to the resultant force is applied to the body, such a force will bring the
body to equilibrium. This force that is equal and opposite to the resultant force is known as
the equilibrant of the force system.

Principle of transmissibility
If the point of application of a force that acts on a body is shifted to any other point on the
line of action of the force without changing its direction, the equilibrium state of the body
remains unchanged. This implies that a force acting at any point on a body may also be
considered to act at any other point along its line of action without changing its effect on the
body.

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1.3 Parallelogram law
The addition of two forces acting at the same point of action is governed by
the parallelogram law. This states that two forces may be replaced by a single
force (called the resultant) obtained as the diagonal of the parallelogram the sides
of which are the given forces (see Fig. 1.4).
The following is valid

1.5

Figure 1.4 Parallelogram law

Similarly from the diagram

1.6
1.4 Resultant Force
Consider a force F acting at the origin of the Cartesian (rectangular and right handed)
coordinate system (see Fig. 1.5).
The force F may be resolved into three components
Consider a force F acting at the origin of the Cartesian (rectangular and right handed)
coordinate system (see Fig. 1.5).
The force F may be resolved into three components

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Figure 1.5 Decomposition of a force into three components.

1.7

1.8
Applying the unit vectors i, j i; and k k directed respectively along x, y; and z z
axes, we may express F in the form

1.9
Or in the form

1.10
If the components Fx, Fy and Fz of a force F are given then the magnitude F of the
force is obtained from Equation 1.8 and the direction cosines are

1.11
Given n concurrent forces, we may determine the resultant Fr by summing their
rectangular components:

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 1.12

1.13

1.14
Problem 1: (i) Two forces of equal magnitude P act at an angle ϴ to each other. What will be
their resultant. (ii) Resultant of two equal forces is equal to either of them. Determine the
angle between the forces. (iii) The resultant of two forces (P+Q) and (P-Q) equals √[3P2+Q2].
Show that the forces are inclined to each other at an angle of 60o.
Problem 2: The resultant of two forces P and Q is at right angles to P. Show that the angle
between the forces is Cos-1(-P/Q).
1.5 Theorem of Resolved Parts
The algebraic sum of the resolved parts of two forces in a given direction is equal to the
resolved part of their resultant in the same direction. Consider the forces OA and OB having
a resultant OC as shown in Figure 1.4.

B
C

A
D

O B’ x
A’ C’

Figure 1.6
Resolved part of force OC on Ox = OC’ = OA’ +AD
Example1
The resultant of two forces P and Q is 200 N and it makes an angle 30o with the horizontal.
The resultant also makes angle 55o and 40o with P and Q respectively. Determine the
magnitudes of components P and Q.
Solution. Let triangle OAB represent all the forces as shown. The components P and Q lie on
OB and AB respectively.

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 B Q
40o A
P
55o
30o
X
O
Figure Example 1
Angle OBA = 180-55-40 = 85o. Applying sine rule,

OA OB AB
o
= o
=
Sin85 Sin 40 Sin55o
Solving yields P = 129.1 N and Q = 164.4 N from B to A.

1.6 Resultant of Concurrent Coplanar Forces (Analytical Method)

The resultant, R, of a number of concurrent coplanar forces is carried out in a step by step
manner as presented below.
i) Sketch the forces in a free body diagram. ii) Find the components of each force in the
system in two mutually perpendicular X and Y directions. iii) Make algebraic addition of
components in each direction to get ∑Fx and ∑Fy. iv) Combine the last two components to
obtain magnitude as,

R= (∑ Fx2 ⊕ ∑ Fy2 ) 1.15

The inclination, θ of R to the X axis is given by,

tan θ =
∑F y
1.16
∑F x

Quiz 1
Determine the magnitude and direction of the resultant of the following system of forces.
(i) 200 N inclined 30o with East towards North, (ii) 250 N towards the North, (iii) 300 N towards
North-West and (iv) 350 N inclined at 40o with West towards South. (a) Sketch the system of forces.
(b) Determine the magnitude and direction of the resultant force (c) What is the equilibrant of the
force system

1.7 Equilibrium of a Particle

A particle will be in equilibrium when resultant of all forces acting on it is zero.
R = √([∑Fx]2 +[∑Fy]2) 1.17

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By definition of equilibrium R = 0 implies that ∑Fx =0 and ∑Fy = 0 since the squired terms
are always positive. Hence if any number of forces acting on a particle are in equilibrium,
then the algebraic sum of their resolved parts in any two perpendicular directions are
separately zero.

1.7.1 Triangle Law of Forces

If two forces acting simultaneously on a body is represented by two sides of a triangle taken
in order, their resultant is represented by the closing side of the triangle taken in the opposite
order.

a C b

Q R
Q
θ θ
A
P B
P
D C
Q F3 α F2
Q
β ¥ R
O
F1
P B
R
A P
c d
e

Figure 1.7 a & b: triangle law; c to e: three forces at equilibrium

The triangle law of forces is also stated to reflect figure 1.7 c to e, as follows,
If a system of three forces acting on a body can be represented in magnitude and direction by
the three sides of a triangle taken in order then the system will be in equilibrium.
Problem: A uniform ladder weighing 80 N rests against a smooth vertical wall at a height of
12 m above the ground; the foot of ladder being 10 m from the wall. Determine pressure due
to wall.

1.7.2 Polygon Law of Forces

If a number of concurrent forces acting simultaneously on a body are represented in
magnitude and direction by the sides of a polygon taken in order, then the resultant is
represented in magnitude and direction by the closing side of the polygon taken in opposite
order. This law can be shown to be a repeated application of triangle law of forces. It is
presented from equilibrium perspective as follows: If any system of forces acting at a point
can be represented in magnitude and direction by the sides of a polygon taken in order then
the system of forces is in equilibrium.

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1.7.2.1 Graphical Method
Solution involving system of forces can be obtained graphically by applying the following
steps:
a) Construct the position (space) diagram showing the various forces along with their
magnitude and line of action.
b) Name forces in the space diagram in accordance with Bow’s notation: each force is
named by two capital letters placed on its either side.
c) Construct the force diagram starting from a convenient point. Go on adding all the
forces vector ally one by one to some suitable scale.

1.8 Free Body Diagram

Fig 1.8 Space diagram Fig 1.9 Free body diagram

2.1 Moment of a force about a point

The moment of force about a point is defined as the turning tendency of the force about that
point. It is measured by the product of force the perpendicular distance of the line of action of
the force from that point. Let F denote the force acting on a body and l the perpendicular
distance between a point O and the line of action of force F (see figure 2.1)

O
F

Body

Figure 2.1 Moment of Force F about point O.

The moment of force F about point O is given by;
MO = F x l 2.1
The [point O is called the moment centre and the distance l is called the moment arm.

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We shall note the following:
I) The moment of force about a point is a vector which is directed perpendicular to
the plane containing the moment centre and the force.
II) The SI unit of moment is Nm.
III) The action of moment tends to cause a rotational motion to the body.
IV) The general convention is to take the clockwise moments as positive and the anti-
clockwise moments as negative.
V) If a number of moments act on a body (see figure 2.2), then the resultant moment
is the algebraic sum of these moments .

F2

L2

L1 L3 F3

F1
L4

F4
Figure 2.2 Coplanar Forces acting on a body.
The resultant moment M about the point O for the set of coplanar forces shown in Figure
2.2 is
M = -F1L1+ F2L2 -F3L3 + F4L4 2.2
2.2 Graphical Representation of Moment

Let a force F be represented in magnitude and direction by a line AB. Further let O be the
point about which the moment is to be determined (Figure 2.3).

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 O

A B
M F

Figure 2.3
The moment of force F about point O is
M = F x OM
= AB x OM = 2[(1/2)x AB x OM]

= 2(area of triangle OAB) 2.3

Thus the moment of force about any point is geometrically equal to twice the area of the
triangle whose base is the line that represent the force and whose vertex is the point about
which the moment is to be found.

2.3 Law of Moments: Varignon’s Theorem

This law states: “Moment of a resultant of two forces, about a point lying in the plane of
the forces, is equal to the algebraic sum of moments of these two forces about the same
point.”

Consider two concurrent forces P and Q represented in magnitude and direction by AB

and AC respectively. Let O be the point about which moments are to be taken. Through
point O draw a line parallel to the direction of force P and let this line intersect the line of
action of force Q at C. The parallelogram is completed with AB and AC as adjacent sides.
The diagonal of this parallelogram represents the resultant of P and Q in magnitude and
direction. Join O with points A and B as shown in figure 2.4

C D
O
Q

A B
P

Figure 2.4 Law of moments

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 Recall that the moment of a force about a point is equal to twice the area of the triangle
formed. Hence
Moment of force P about O = 2 x area of triangle AOB= 2 x ∆ AOB. Similarly
Moment of force Q about O = 2 x ∆ AOC
Moment of force R about O = 2 x ∆ AOD
The geometrical configuration of figure 2.4 reveals that;
∆ AOD = ∆ AOC+ ∆ ACD = ∆ AOC + ∆ ABD
Observe that the triangles AOB and ABD are on the same base line, AB, and between the
same lines, thereby making them of equal area. Thus
∆ AOD = ∆ AOC + ∆ AOB
Hence the moment of force R about O may be re-written as
= 2 x (∆ AOC + ∆ AOB) 2.4
This agrees with the theorem of Varignon. This principle can be extended to any number
of forces.
2.4 Principle of Moments
A body acted upon by a number of coplanar forces will be in equilibrium, if the algebraic
sum of moments of all the forces about a point lying in the same plane is zero. That is;
∑M=0 2.5
Or : clockwise moments = anticlockwise moments.

Example 1 A force of 200N is acting at point B as shown in figure E1. Determine the
moment of this force about O.

3m P=200N
60 deg
O B

Figure E1

Solution: The moment of force about O is

Mo = P x OM = P x OB Sin 60
= 200x(3x0.866) = 519.6Nm (clockwise)
Parallel Forces
Forces that are parallel but lie in the same plane are called coplanar non-concurrent forces.
There are two types; like parallel forces and unlike parallel forces.
Like parallel forces act in the same direction while unlike parallel forces do not all act in the
same direction.

Resultant of like Parallel forces

Consider two parallel forces P and Q that acts at points A and B respectively as shown.

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 R= P+Q
P
M Q

A
B
C ϴ

Figure 2.5

N
The resultant R = P + Q. Consider that the resultant passes though a point C along AB. The
point of action of the resultant force can be obtained by taking moments about C as shown in
figure 2.5. The moment arms are MC and CN respectively. Thus
P x MC = Q x CN
We know that MC = AC x Cosϴ and CN = CB x Cosϴ
Substituting for MC and CN and rearranging yields
P CN BC cosθ BC
= = =
Q CM AC cos θ AC
Thus the point of application of the resultant of two like parallel forces lies between the two
forces and it divides the distance between the forces internally in the inverse ratio of the
magnitude of forces.
Resultant of Unlike Parallel Forces.
Consider the two unlike parallel forces shown in figure 2.6.

R=(P-Q)
P

C B
ϴ A

M Q

Figure 2.6
N
Taking moments about C as usual
P x AC cos ϴ = Q x CB cos ϴ

P BC
Therefore: =
Q AC

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Hence the point of application of the resultant of two unlike parallel forces lies outside the
two forces nearer to the larger of the two forces. Moreover it divides the distance between
these forces externally in the inverse ratio of the magnitude of forces.

Couple
Two parallel forces that are equal in magnitude but opposite in direction and separated by a
finite distance form a couple. The rotational effect of a couple is measured by its moment
which is defined as the product of either of the forces and the perpendicular distance between
the forces.

P
L

P
Figure 2.7 A Couple

The couple is given by

M=PxL

3.0 PLANE TRUSSES AND FRAMES

In a plane frame all the members lie in one plane and all forces act along the plane of the
frame. Bridge trusses and roof trusses are typical examples of plane frames.
A frame can be said to be statically determinate when the force analysis of the frame can be
made by applying the equation of statics only. On the other hand when the equations of static
equilibrium are not sufficient to determine the forces in the members then the frame is said to
be statically indeterminate. In the latter a consideration of deformation is normally required in
addition to static equilibrium.
A frame is said to be perfect if the number of members is just sufficient to prevent distortion
of shape when subjected to external loads. Denote the number of members of a frame by m
and the number of joints by j. A perfect frame must satisfy the following relationship between
m and j:
m = 2j -3 3.1
When the LHS is less than the RHS the frame is said to be deficient. On the other end is
when the LHS is greater than the RHS the frame is said to be redundant. Typical perfect
frames are shown in figure 3.1 a & b while frame c is deficient.

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 a c
b

Figure 3.1 Typical Frames

Substitution into equation 3.1 will reveal that frame (a) and (b) are perfect frames while (c) is
a deficient frame.

3.1 Reactions at the Support of Frame

Statically determinate frames have two types of supports: hinged and roller supports. At a
hinged support the line of action of the reaction depends on the load system on the frame.
This reaction can be split into horizontal and vertical components; Rv = R sin ϴ and Rh =R
con ϴ. Roller supports are frictionless and provide a reaction at right angles to the roller
base.

3.2 Analysis
The following assumptions are generally made:
a) The truss is perfect and statically determinate.
b) All members are rigid and lie in the same plane.
c) The members are slender and of uniform cross-section
d) The external loads and reactions act at joint only
e) The self weight of the members is negligible compared to the load they carry.
f) The forces are transmitted from one member to another through smooth pines (no
friction).
The stresses induced in the members as a result of externally applied loads are obtained by
analytical method or by graphical method. The analytical method is further classified into
method of joint and method of sections.
3.2.1 Method of Joints
Every joint is treated separately as a free body in equilibrium. The following steps will ensure
a successful solution.
1) All the pin joints are labelled.
2) A free body diagram of the entire frame is drawn and the reactions at the supports are
calculated.
3) Each joint is treated separately as free body. A choice of direction of forces acting on
the joint is made and the forces are worked out by applying condition of equilibrium.
If the magnitude of a force turns positive then the choice of direction is correct but if
negative the direction is simply reversed.
4) Start at a joint where the number of unknown forces is no more than two. The process
is repeated for other joints until all the forces are determined.

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 5) A force in a member is tensile if the member pulls the joint to which it is connected
(force is directed away from the pin) whereas the force in the member is compressive
if the member pushes the joint to which it is connected (force is toward the pin).
Example3.1 Determine the forces in all the members of a truss with the loading and
support system shown in figure 3E1

30kN
C

60deg 30deg B
A

5m

Figure 3E1
Solution

30kN
C

60deg 30deg B
A
D
Ra Rb

Figure 3E1.1 Free body diagram

Roller support (end B) is frictionless and provides a reaction Rb at right angle to the support
base. The reaction at the hinged support (end A) has only the vertical component since the
only external load is vertical. The free-body diagram is as shown in figure 3E1.1. A
consideration of triangle ABC indicate that the angle at C is 90o. The distance AD is the
moment arm of the external force about point A. From the geometry of triangle ABC,
AC = AB cos60 = 5x 0.5 = 2.5m
AD = ACcos60 = 2.5x 0.5 =1.25m
Taking moments about A we obtain

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Rb x 5 = 30x1.25

30 x1.25
Rb = = 7.5kN
5
And Ra = 30-7.5 = 22.5kN
We can start from any of the joints since they all have two unknown forces each. However let
us start from joint A.
Joint A: Consider the free-body diagram of joint A as shown with the direction of forces as
assumed.

F1

60deg 30deg B
A
F2
Ra = 22.5kN

Figure 3E1.2 Free-body diagram of joint A

Considering equilibrium equations in the vertical and horizontal directions yields;
∑Fx = 0 ; F2 – F1cos60 =0 and
∑Fy = 0; F1sin60 –Ra = 0

Ra 22.5
Therefore F1 = = = 25.97 kN (compressive)
sin 60 0.866
And F2 = F1cos60 = 25.97x 0.5 = 12.99kN (tensile)
Joint B: In the free-body diagram shown in figure 3E1.3 force F2 has already been
determined in the previous step.

F3

30deg B
A

F2 = 12.99kN Rb = 7.5kN

Figure 3E1.3

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From equation of equilibrium,
∑Fx = 0; F3cos30 – F2 = 0

F2 12.99
Therefore F3 = = = 15kN (compressive)
cos 30 0.866
The force F3 is acting towards the joint B which means that the member BC is in
compression.

3.2.2 Method of Sections

In this method the truss is split into two parts by passing an imaginary section. The imaginary
section must be such that it does not cut more than three members in which the forces are to
be determined. The condition of equilibrium is applied to one part of the truss:
∑Fx = 0; ∑Fy = 0 ; and ∑M =0
This method is particularly convenient when the forces in only some of the members of a
frame are required.
Example3.2 Determine the forces in members AC and AB of a truss with the loading and
support system shown in figure 3E1. Use the method of section.
Solution: Let the section 1-1 cut through the members AC and AB as shown in figure 3E2.1

30kN
C
1

F1
60deg 30deg B
A
F2
Ra=22.5kN 1

Figure 3E2.1 Section 1-1

The two reactions had earlier been determined. The unknown force F1 has been assumed
compressive and F2 assumed tensile as shown. Let us take moments of these forces about B;
∑Mb = 0; Ra x AB - F1 x BC = 0
BC = ABco30 = 5x 0.866 = 4.33m

RaxAB 22.5 x5
Therefore F1 = = = 22.98kN (compressive)
BC 4.33
Take moments about C
∑Mc =0; Ra x ACcos60 – F2 x ACsin60 =0.

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 AC cos 60 cos 60 0 .5
Therefore F2 = Rax = Ra × = 22.5 × = 12.99kN (tensile)
AC sin 60 sin 60 0.866
3.3 Graphical Method
The following steps are applied by this method:
1) The truss is drawn accurately to scale. The loads and support reactions are indicated
in direction and magnitude. All the members of the truss are named in accordance to
Bow’s notation. In this notation a force is designated by two letters that are writtenon
either side of the line of action of the force as shown in figure 3.2. The spaces
between the forces have been marked P, Q, R and S. This is followed by the
construction of a vector diagram.
Construction of a Vector Diagram:
i) Take any point p and draw a vertical line pq to represent force PQ = 30kN. If a
scale of 1cm = 5kN is used then pq = 6cm. From point q take qr = 1.5cm to
represent the reaction Rb = 7.5kN acting vertically upwards. Then rp represent the
reaction Ra.
ii) Choose a joint where not more than two unknown forces are acting, (say joint A).
iii) From point p draw a line parallel to PS and from q draw a line parallel to QS.
These two lines meet at point s. Measure ps and sr in the vector diagram.

30kN
C

Q
P
S
60deg 30deg B
A
D
Ra=22.5kN R Rb=7.5kN

Figure3E3.1

s r

Figure3E3.2 Vector diagram

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ps = 5.2cm therefore Force PS (force in member AC) = 5.2x5 = 26kN
sr=2.6cm therefore force SR (force in member AB) = 2.6x5= 13kN
To determine the nature of these forces we move in clockwise direction round the joint A in
the space diagram. First PS, SR and RP:
In the vector diagram p to s means that the arrow must be marked towards the joint Ain the
space diagram. Thus the member AC is subjected to 26kN compressive force. SR implies s to
r in the vector diagram. This implies that the joint A is pulled by the member AB or that
member AB is subjected to 13kN tensile force.
Considering joint B: In the vector diagram sq represent the force in member BC designated
SQ in the space diagram.
sq = 3cm : Therefore force SQ (force in member BC) = 3x5 = 15kN.
Again moving clockwise round joint B, we have RS, SQ and QR. In the vector diagram:
rs : arrow is marked B to A (away from the joint)
sq : arrow is marked C to B (towards the joint)
Thus the member BC is subjected to 15kN compressive force.

4.0 Moment of Inertia: Area and Mass

Recall that the moment of a force was expressed as the product of the force and the
perpendicular distance to a given point about which the moment is being determined. That is:
M = Fx
If the moment of this moment is sought we have Fx2. This is known as the second moment of
force. However if the term force in the above identity is replaced by area or mass of the body,
the resulting parameter is called the moment of inertia. Thus
Moment of inertia of a plane area = Ax2
Mass moment of inertia of a body = mx2
where A and m respectively denote the area and mass of the body.
Inertia is the property of a body by virtue of which the body resists any change in its state of
rest or uniform motion. Thus area moment of inertia is essentially a measure of resistance to
bending and is applied when deflection or deformation is in view. Similarly mass moment of
inertia is a measure of the resistance of a body to change in angular velocity and is used when
dealing with rotation of rigid bodies.

Moment of Inertia and Radius of Gyration

Moment of inertia of a lamina is the second moment of all elemental areas dA comprising the
lamina. Consider Figure 4.1. The moment of inertia about x-axis is given by,
Ixx = ∑(ydA)y

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where ydA is the first moment of area dA about x-axis and (ydA)y is the moment of first
moment (called second moment) of area dA about the x-axis.

x
Area dA

Lamina of area A
x

Figure 4.1
It follows that Ixx = ∑y2dA
Similarly the moment of inertia about the y-axis Iyy is
Iyy= ∑x2dA
The unit of I is m4 or mm4
4.1.1Parallel Axes Theorem :
Statement: The moment of inertia of a plane lamina about any axis is equal to the sum of its
moment of inertia about an axis passing through its centre of gravity G and parallel to the given
axis and the product of area (mass) of the body with the square of the distance between the two
axes.
Proof: Consider the lamina shown in figure 4.2. The lamina consists of infinite number of

B y

j
Area dA

y
x x
G h

Lamina of area A
A A
y
B

Figure 4.2

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small elemental components parallel to the x-axis. One such element of area dA is located at
distance y from x-axis as shown. The distance of this element from the axis AA is (y + h).
Moment of inertia of this elemental component about axis AA will be
= dA (h + y)2
Thus moment of inertia of the entire lamina about axis AA;
=∑dA(h + y)2 = ∑dAh2 +∑dAy2 + ∑dA(2hy)
=h2∑dA + ∑dAy2 + 2h∑ dAy
Observe that h2∑dA = Ah2 (∑dA= A)
∑dAy2 =moment of inertia of the lamina about axis x-x
∑dAy =0 because x-x is a centroidal axis
Hence: IAA= Ixx + Ah2
Also IBB = Iyy + Aj2

Perpendicular Axis Theorem

The moment of inertia of a plane lamina about an axis perpendicular to the plane of the
lamina is equal to the sum of the moments of inertia of the lamina about the two axes at right
angles to each other and intersecting each other at the point where the perpendicular axis
passes through it.

x
Area dA
z

y
r
Lamina of area A
x
o

Figure 4.3
Moment of the elemental component about the axis oz is;
= dAr2 = dA (x2 +y2)
Moment of the lamina about the axis oz is
Izz = ∑dA(x2 + y2) = ∑dAx2 + ∑dAy2

21
Thus Izz = Ixx + Iyy
Radius of Gyration: If the entire area (or mass ) of a lamina is considered to be concentrated
at a point such that there is no change in the moment of inertia about a given axis, then
distance of that point from the given axis is called radius of gyration. Mathematically;
I = Ak2; k = √(I/A)

Moment of Inertia of Lamina of Different Shapes

Rectangular Lamina: Consider a rectangular lamina ABCD of with b and depth d. Let xx and
yy be the axis which pass through the centroid of the area and are parallel to the sides of the
lamina. Consider a small strip of thickness dy located at distance y from the axis xx. Thearea
of the elemental strip,
dA = bdy
Moment of inertia of the elemental component about the axis xx,
dAx y2 = bdy x y2 = by2 dy
Moment of inertia of the entire lamina about axis xx,
d /2
d /2
y3  d 3 d 3  bd 3
Ixx = ∫ by dy = b
2
= b +  =
−d / 2
3 −d / 2  24 24  12

y
D C

dy
y

x
x

A B

Figure 4.4
Similarly the moment of inertia of the lamina about the axis yy is,

db 3
Iyy =
12

Let IAB be the moment of the lamina about its bottom face AB. Then from the theorem of
parallel axis

22
 IAB = Ixx + Ah2 = bd3/12 + bd(d/2)2 = bd3/3
The polar moment of inertia from the theorem of perpendicular axis.
Ip = Ixx + Iyy = bd3/12 + db3/12
For a rectangular (BxD) with a rectangular hole (bxd) made centrally the moment of inertia
about any centroidal axis is
MOI of bigger rectangle – MOI of smaller rectangle
Thus Ixx = BD3/12 – bd3/12
Triangular Lamina
Let ABC be the triangle of base width b and height h.
Consider the elemental strip of length l and height dy

y-axis Area of strip = ldy

dy

l
h
y

x-axis
Figure 4.5
MOI of the strip about the base;
= y2dA = y2ldy
Let us express l in term of y: Note that triangle ADE and ABC are similar
Therefore l/b = (h-y)/h
Thus: l = b(1-y/h)
MOI of the triangle about the base
Ibase = ∫h0y2b(1-y/h)dy = b[y3/3-y4/4h]0h = bh3/12
For a triangle the centroidal axis is at a distance of yc = h/3 from the base. Applying the
theorem of parallel axis:
Ibase = Ixx +Ayc 2
Therefore Ixx = Ibase –Ayc2 = bh3/36

23
Circular Lamina
Let us consider an element of sides rdθ and dr within a circular lamina of radius R. Moment
of this element about a diametrical axis x-x,

= y 2 dA = (r sin θ ) 2 × rdθdr = r 3 sin 2 θdθdr

Therefore MOI of the entire circular lamina

2π 2π
R R 1 − cos 2θ R r3 sin 2θ 2π
I xx = ∫ ∫ r sin θdθdr = ∫ ∫r dθdr = ∫ [θ −
3 2 3
]0 dr
0
0
0
0
2 0 2 2

R
R r3 r4  π
= ∫ (2π )dr = 2π   = R 4
0 2  8 0 4
If d is the diameter of the circular lamina, then;

π d  π
4

I xx =   = d4
4 2 64

dθ

Figure 4.6

π
Similarly I yy = d4
64
If zz is the axis through the centroid and normal to the plane of the lamina, then,

π π π
I zz = I xx + I yy = d4 + d4 = d4
64 64 32

24
The axis zz is called the polar axis and Izz is referred to as the polar moment of inertia.
When a circular lamina has a central circular hole of diameter d, the relevant moments of
inertia are;

π
I xx = I yy = [D 4 − d 4 ]
64
π
I zz = I p = [D 4 − d 4 ]
32
Semi-Circular Lamina
From the previous section the MOI of a circular lamina about x-x axix was found to be

π
I xx = d4
64
For the semi-circular lamina with x-x or AB as the base, the MOI about the base is

1 π 4 π 4
I AB = x( d ) = d
2 64 128

Mass Moment of Inertia

The mass moment of inertia of a body about a particular axis is the product of the mass and
the square of the distance between the mass centre of the body and the axis. It measure the
resistance of the body to change in angular velocity. Consider a body made up of elemental
masses m1, m2, ….mn at distance r1, r2, ….rn. Then;
I = m1r12 + m2r22 + ……mnrn2
= ∑mr2
From the fore going
I = ∫r2dm
The radius of gyration k of the body with respect to the prescribed axis is defined by;
I = k2m; Implies that k = k = √(I/m)
where k gives a measure of the distance at which the entire mass of the body is assumed to be
concentrated.

MOI of a Thin Uniform Rod

y
x dx
Y

l/2 l/2
y
Y

25
Let mass per unit length of the rod be m;
dm for the small element of length dx is mdx.
MOI of the element about axis y-y is
= dmx2 = mx2dx
Thus Iyy = m∫-l/2l/2(x2dx = m[x3/3]l/2-l/2 =(ml3)/12 = Ml2/12
where M = ml.
IYY = Iyy + M(l/2)2 = (Ml2)/3

Moment of Inertia of a rectangular plate

Consider a rectangular plate of dimensions b x t x d and density ρ;
Mass M = ρbtd
Consider an element that is parallel to b and t and of depth dy;
dm = ρbtdy
MMOI of element about xx through the centre of gravity;
= dmy2 = ρbty2dy

d/2

∫ (y
2
Ixx = ρbt )dy = ρbt[y3/3]-d/2d/2 = ρbtd(d2)/12
-d/2
= Md2/12
Siomilarly
Iyy = (Mb2)/12
Izz = Ixx + Iyy = (1/12)M[d2 + b2]

Circular Lamina
Consider a thin circular plate of radius R and uniform thickness t. If the density of the plate is ρ, then mass
M of the plate =
Density x volume = ρπR2t

For an elemental ring of radius r and width dr, mass dm = ρ[π(r+dr)2-πr2]t.

= 2πtρrdr

Mass moment of inertia of this elementary ring about the polar axis zz is,

= dmr2 = 2πtρr3dr

MOI of circular plate about zz is,

R R 2 MR 2
2πtρ ∫ r 3 dr = ρπR 2 t × =
0 2 2
where M is the mass of the circular lamina.

1
Ixx =Iyy = Izz/2 = MR 2
4
It can be shown that for a solid sphere the polar MOI is (4/5)MR2, while Ixx = Iyy = (2/5)MR2.

5.0 Friction

If a body rests on an incline plane, the friction force exerted on it by the surface
prevents it from sliding down the incline. The question is, what is the steepest incline
on which the body can rest?
26
A body is placed on a horizontal surface. The body is pushed with a small horizontal
force F. If the force F is sufficiently small, the body does not move.

Figure 5.1Free-body diagram of the body

Figure 5.1 shows the free-body diagram of the body, where the force W is the
weight force of the body, and N is the normal force exerted by the surface on the
body. The force F is the horizontal force, and Ff is the friction force exerted by
the surface. Friction force arises in part from the interactions of the roughness, or
asperities, of the contacting surfaces. The body is in equilibrium and Ff = F.
The force F is slowly increased. As long as the body remains in equilibrium,
the friction force Ff must increase correspondingly, since it equals the force F. The
body slips on the surface. The friction force, after reaching the maximum value,
cannot maintain the body in equilibrium. The force applied to keep the body moving
on the surface is smaller than the force required to cause it to slip. Why more force
is required to start the body sliding on a surface than to keep it sliding is explained in part by
the necessity to break the asperities of the contacting surfaces before sliding
can begin.
The theory of dry friction, or Coulomb friction, predicts:
.the maximum friction forces that can be exerted by dry, contacting surfaces that
are stationary relative to each other;
. the friction forces exerted by the surfaces when they are in relative motion, or
sliding.
Static Coefficient of Friction
The magnitude of the maximum friction force, Ff , that can be exerted between
two plane dry surfaces in contact is
Ff = μsN
5.1
where μs is a constant, the static coefficient of friction, and N is the normal component
of the contact force between the surfaces. The value of the static coefficient of
friction, μs, depends on:
_ the materials of the contacting surfaces;
_ the conditions of the contacting surfaces namely smoothness and degree of contamination.
Typical values of μs for various materials are shown in Table 5.1.

27
Table 5.1

Equation (5.1) gives the maximum friction force that the two surfaces can exert
without causing it to slip. If the static coefficient of friction ms between the body
and the surface is known, the largest value of F one can apply to the body without
causing it to slip is F = Ff = μsN. Equation (5.1) determines the magnitude of the
maximum friction force but not its direction. The friction force resists the impending
motion.
Kinetic coefficient of friction
The magnitude of the friction force between two plane dry contacting surfaces
,l,m
that are in motion relative to each other is
Ff = μkN
5.2
where μk is the kinetic coefficient of friction and N is the normal force between the
surfaces. The value of the kinetic coefficient of friction is generally smaller than the
value of the static coefficient of friction, ms.
To keep the body in Fig. 5.1 in uniform motion (sliding on the surface) the force
exerted must be F = Ff = μkN. The friction force resists the relative motion, when
two surfaces are sliding relative to each other.
The body RB shown in Fig. 5.2(a) is moving on the fixed surface 0.

28
Figure 5.2 Directions of the friction forces
The direction of motion of RB is the positive axis x. The friction force on the
body RB acts in the direction opposite to its motion, and the friction force on the
fixed surface is in the opposite direction as shown in Fig. 5.2(b).

Angles of Friction
The angle of friction, ϴ, is the angle between the friction force, Ff =│Ff │, and
the normal force to the surface N = │N│, as shown in Fig. 5.3.
The magnitudes of the normal force and friction force, and ϴ are related by
Ff = Rsin ϴ,
N = Rcos ϴ,

29
 Figure 5.3 Angle of friction

where R = │R│ = │N+Ff │.

The value of the angle of friction when slip is impending is called the static angle
of friction, ϴs,
tanϴs = μs:
The value of the angle of friction when the surfaces are sliding relative to each other
is called the the kinetic angle of friction, ϴk,
tanϴk = μk:

Motion on a Plane
Consider the pending motion of a body of weight W under a force P that is applied at an
angle β to the direction of impending motion as shown Figure 5.4a. The friction force will
oppose motion and therefore acts in direction opposite to that of impending motion. The free-
body diagram is shown in Figure 5.4b. The coefficient of friction and the friction force can
be determined by considering the equilibrium of forces in the vertical and horizontal
directions.

RN
RN
P
P
β β
F

motion
W W

a b
Figure 5.4

considering equilibrium in x-axis: ∑Fx = 0 gives

Pcos β – F = 0 5.3
That is F = Pcos β.

30
However we know that F = μRN
where μ is the coefficient of friction. Thus μW = Pcos β.

P cos β
Therefore, µ = 5.4
RN

Considering equilibrium along y-axis:

∑Fy = 0; Implies that RN + Psin β = W
Therefore RN = Psin β + W 5.5
We have two equations and two unknowns.

Motion Up an Inclined Plane

Consider the motion of a body of weight W up a plane inclined at angle α under the action of
a force P which acts up the plane. Let P be parallel to the plane. The weight W has
components parallel and perpendicular to the plane.
. Parallel to the plane = Wsin α
. Perpendicular to the plane = Wcos α
Applying equilibrium condition to the direction parallel to the plane yields;
P = Wsinα + μWcos α 5.6

P
P

F
Wcosα
F
α
Wsinα
ααα
Figure 5.5 Motion of a body up an inclined plane

Motion Down an Inclined Plane

Let us now consider the motion of the body in figure 5.6 down the inclined plane under the
influence of force P acting parallel to the plane. Since the impending motion is down the
plane the direction of friction force will be up the plane. For equilibrium of forces;
P = Wsinα - μWcosα 5.7
If Wsinα < μWcosα (i e α < ϴ) the direction of P is reversed and must be applied to push the
body down the plane with uniform speed.

31
 P
P

F
F Wcosα

α α Wsinα
ααα
Figure 5.6: Motion of body down an inclined plane.
Example: An effort of 1500N is required to just move a certain body up an inclined plane of
angle 12o, force acting parallel to the plane. If the angle of inclination is increased to 15o,
then the effort required is 1720N. Find the weight of the body and the coefficient of friction.
Solution: Let W = weight of the body and μ = the coefficient of friction.
Applying Equation 5.6
P = 1500 = W(sinα + μcosα) = W( sin12 + μcos12) ( a)
Similarly P = 1720 = W(sin15 + μcos15) (b)
Dividing equation (b) by equation (a) yields;

1720 W (sin15 + µ cos15)

= (c)
1500 W (sin 12 + µ cos12)

Solving equation (c) yields μ = 0.131

Substituting the value of μ into equation (a)
1500 = W(sin12 +0.131cos12)
Therefore W = 1500/(sin12 +0.131cos12) = 4464N

32