FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Syllabus
PART I: STATICS
1.
2.
3.
4.
5.
6.
7.
8.
Fundamental Concepts and Definitions
Resultant of Force Systems
Equilibrium of a Rigid Body
Analysis of Structures
Friction
Force Systems in Space
Centroid and Center of Gravity
Moment of Inertia
PART II: DYNAMICS
1. Fundamental Concepts and Definitions
2. Kinematics of a Particle
3. Kinetics of a Particle
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Engineering Mechanics
Mechanics of Deformable
Bodies
Mechanics of Rigid Bodies
Statics
Kinetics
Dynamics
Mechanics of Fluids
1. Strength of Materials
1. Ideal Fluid
2. Theory of Elasticity
2. Viscous Fluid
3. Theory of Plasticity
3. Incompressible Fluid
Kinematics
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Mechanics of Rigid
Bodies
Statics
Force Systems
Dynamics
Applications
Kinematics
Kinetics
Concurrent
Trusses
Translation
Translation
Parallel
Centroids
Rotation
Rotation
Non-Concurrent
Friction
Plane Motion
Plane Motion
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NORIEL M. PURIGAY
Fundamental Concepts and
Definitions
• Engineering Mechanics – The science which
considers the effects of forces on rigid bodies.
• Statics – considers the effects and distribution of
forces on rigid bodies which are and remain at rest
• Dynamics – considers the motion of rigid bodies
caused by the forces acting upon them
• Kinematics – deals with pure motion of rigid bodies
• Kinetics – relates the motion to applied forces
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Fundamental Concepts and Definitions
Basic Quantities
Length – used to locate the position of a point in space
and thereby describe the size of a physical system
Time – is conceived as a succession of events
Mass – is a measure of the quantity of matter that is used
to compare the action of one body with that of another.
Force – a “push or pull” exerted by one body to another
• External Force - changes, or tends to change, the state
of motion of a body. (independent on point of
application)
• Internal Force – produces stress and deformation in
the body. (dependent on point of application)
* Principle of Transmissibility – a force may be moved
anywhere along its line of action without changing its
external effect on a rigid body.
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NORIEL M. PURIGAY
Fundamental Concepts and Definitions
Idealizations
Particle – has a mass, but a size that can be
neglected.
Rigid Body – can be considered as a large
number of particles in which all the particles
remain at a fixed distance from one another,
both before and after applying a load.
Concentrated Force - represents the effect of
a loading which is assumed to act at a point on
a body. We can represent a load by a
concentrated force, provided the area over
which the load is applied is very small
compared to the overall size of the body.
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Fundamental Concepts and Definitions
Newton’s Three Laws of Motion
First Law (Law of Inertia). A particle originally at rest, or
moving in a straight line with constant velocity, tends to
remain in this state provided the particle is not subjected
to an unbalanced force.
Second Law (Law of Acceleration). A particle acted upon
by an unbalanced force experiences an acceleration a that
has the same direction as the force and a magnitude that
is directly proportional to the force.
Third Law. (Law of Action-Reaction). The mutual forces of
action and reaction between two particles are equal,
opposite, and collinear
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Fundamental Concepts and Definitions
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Fundamental Concepts and Definitions
Newton’s Law of Gravitational Attraction
𝑚1 𝑚2
𝐹=𝐺
𝑟2
where
F = force of gravitation between the two particles
G = universal constant of gravitation; according to
experimental evidence, G = 66.73(10-12) m3/(kg · s2)
m1, m2 = mass of each of the two particles
r = distance between the two particles
Weight – force on an object due to gravity
W= 𝐺
𝑚𝑚𝐸
𝑟2
W = mg
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NORIEL M. PURIGAY
Fundamental Concepts and Definitions
Units of Measurement
F = ma ; W=mg
CGS: dyne = (g)(cm/s2)
MKS: N = (kg)(m/s2)
kgf = (kgm)(9.8 m/s2)
kgf = 9.8 N
Under Standard Condition: g = 9.8 m/s2
kgf = kgm
For Non-Standard Condition
𝑔
kgf = kgm( 𝑔𝑙𝑜𝑐 )
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US: lbf = (slug)(ft/s2)
lbf = (lbm)(32.174 ft/s2)
slug = 32.174 lbm
lbf = lbm
𝑔𝑙𝑜𝑐
)
𝑔
lbf = lbm (
NORIEL M. PURIGAY
Fundamental Concepts and Definitions
Force Systems
Force System – any arrangement when
two or more forces act on a body or on a
group of related bodies.
• Coplanar – the lines of action of all the
forces lie in one plane
• Concurrent – the lines of action pass
through a common point
• Parallel – the lines of actions are
parallel
• Non-Concurrent – the lines of action
are neither parallel nor intersect at a
common point
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NORIEL M. PURIGAY
Fundamental Concepts and Definitions
Axioms of Mechanics
1. The Parallelogram Law: The resultant
of two forces is the diagonal of the
parallelogram formed on the vectors
these forces.
2. The forces are in equilibrium only when
equal in magnitude, opposite in
direction, and collinear in action.
3. A set of forces in equilibrium may be
added to any system of forces without
changing the effect of the original
system
4. Action and reaction forces are equal
but oppositely directed.
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NORIEL M. PURIGAY
Fundamental Concepts and Definitions
Parallelogram Law
Triangle Law
Polygon Law
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Fundamental Concepts and Definitions
Scalar and Vector Quantities
Scalars – quantities which posses
magnitude only and can be added
arithmetically.
Vectors – quantities which posses
magnitude and direction and can be
combined only by geometric (vector)
addition.
•
Multiplication or division of a vector by
a scalar will change the magnitude of
the vector. The sense of the vector will
change if the scalar is negative.
• As a special case, if the vectors are
collinear, the resultant is formed by an
algebraic or scalar addition.
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NORIEL M. PURIGAY
Resultant of Force Systems
Resultant – simplest system that can replace the original
system without changing the effect on a rigid body
Components of a Force
𝐹𝑥 = 𝐹𝑐𝑜𝑠𝜃
𝐹𝑦 = 𝐹𝑠𝑖𝑛𝜃
Resultant
𝐹=
𝐹𝑥 2 + 𝐹𝑦 2
Resultant of Three or More Concurrent
Forces
𝑅=
(Σ𝐹𝑥 )2 + (Σ𝐹𝑦 )2
Position of Resultant
𝑡𝑎𝑛θ𝑥 =
Σ𝐹𝑦
Σ𝐹𝑥
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NORIEL M. PURIGAY
Resultant of Concurrent Forces
Resultant of Non-Perpendicular Forces
𝑅=
𝑃1 2 + 𝑃2 2 + 2𝑃1 𝑃2 𝑐𝑜𝑠θ
Position of Resultant
𝑃1 𝑠𝑖𝑛θ
2 +𝑃1 𝑐𝑜𝑠θ
𝑡𝑎𝑛ϕ = 𝑃
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Resultant of Concurrent Forces
Determine the magnitude and direction of
the resultant of the three forces shown.
Determine also the horizontal and vertical
component of the resultant.
Solution:
𝑅𝑥 = Σ𝐹𝑥
= 50𝑐𝑜𝑠45 + 75𝑐𝑜𝑠75 − 80𝑐𝑜𝑠60
𝑹𝒙 = 𝟏𝟒. 𝟕𝟕 𝐍
𝑅𝑦 = Σ𝐹𝑦
= 50𝑠𝑖𝑛45 + 75𝑠𝑖𝑛75 + 80𝑠𝑖𝑛60
𝑹𝒚 = 𝟏𝟕𝟕. 𝟎𝟖 𝐍
𝑅=
(Σ𝐹𝑥 )2 + (Σ𝐹𝑦 )2
= 14.772 + 177.082
𝑹 = 𝟏𝟕𝟕. 𝟕𝟎 𝐍
Σ𝐹𝑦
θ = 𝑡𝑎𝑛−1 Σ𝐹 = 𝑡𝑎𝑛−1
𝜽 = 𝟖𝟓. 𝟐𝟑°
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𝑥
177.08
14.77
NORIEL M. PURIGAY
Resultant of Concurrent Forces
Determine the magnitude and direction of R
if P1 and P2 are 100 lb and 150 lb respectively.
P2 lies horizontally while P1 makes 120° with
the horizontal.
Solution 1:
𝑅=
(Σ𝐹𝑥 )2 + (Σ𝐹𝑦 )2
= (150 − 100𝑐𝑜𝑠60)2 + 100𝑠𝑖𝑛602
𝑹 = 𝟏𝟑𝟐. 𝟐𝟗 𝐥𝐛
Σ𝐹𝑦
ϕ = 𝑡𝑎𝑛−1
Σ𝐹𝑥
= 𝑡𝑎𝑛−1
𝝓 = 𝟒𝟎. 𝟖𝟗°
86.60
100
Solution 2:
𝑅=
𝑃1 2 + 𝑃2 2 + 2𝑃1 𝑃2 𝑐𝑜𝑠θ
= 1002 + 1502 + 2(100)(150)𝑐𝑜𝑠120
𝑹 = 𝟏𝟑𝟐. 𝟐𝟗 𝐥𝐛
𝑃1 𝑠𝑖𝑛θ
)
𝑃2 +𝑃1 𝑐𝑜𝑠θ
ϕ = 𝑡𝑎𝑛−1 (
𝝓 = 𝟒𝟎. 𝟖𝟗°
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NORIEL M. PURIGAY
Resultant of Concurrent Forces
Determine the magnitude of R if P1 and P2
are 100 lb and 150 lb respectively. ϕ = 41°
Solution:
Let α be the angle opposite R and β be the
angle opposite P2:
By Sine Law
𝑅
𝑃1
𝑃2
=
=
𝑠𝑖𝑛α 𝑠𝑖𝑛ϕ 𝑠𝑖𝑛β
100
150
=
𝑠𝑖𝑛41 𝑠𝑖𝑛β
β = 80°
α = 180 − 80 + 41 = 59
100
𝑅
=
𝑠𝑖𝑛41 𝑠𝑖𝑛59
𝑹 = 𝟏𝟑𝟎. 𝟔𝟓 𝒍𝒃
By Cosine Law:
𝑅=
=
𝑃1 2 + 𝑃2 2 − 2𝑃1 𝑃2 𝑐𝑜𝑠α
1002 + 1502 − 2(100)(150)𝑐𝑜𝑠59
𝑹 = 𝟏𝟑𝟎. 𝟓𝟕 𝒍𝒃
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Resultant of Concurrent Forces
A boat moving at 12kph is crossing a river 500 m wide
in which a current is flowing at 4 kph. In what
direction should the boat head if it is to reach a point
on the other side of the river directly opposite its
starting point?
Solution:
𝑠𝑖𝑛θ =
4
12
𝜽 = 𝟏𝟗. 𝟒𝟕°, 𝒖𝒑𝒔𝒕𝒓𝒆𝒂𝒎
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NORIEL M. PURIGAY
Resultant of Non-Concurrent Forces
Moment of a Force
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NORIEL M. PURIGAY
Resultant of Non-Concurrent Forces
Moment – is the measure of the ability of a force
to produce turning or twisting about an axis.
𝑀𝑂 = 𝐹𝑑
where d is the moment arm (perpendicular
distance from the axis at point O to the line of
action of the force.
The Principle of Moments (Varignon’s Theorem)
The moment of a force is equal to the sum of the
moments of its components.
𝑀𝑅 = Σ(𝐹𝑑)
𝑀𝑅 = Σ𝑀 = 𝑅𝑑
𝑀𝑅 = 𝐹1 𝑑1 − 𝐹2 𝑑2 + 𝐹3 𝑑3
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NORIEL M. PURIGAY
Resultant of Non-Concurrent Forces
Determine the resultant moment of
the four forces acting on the rod
shown below about point O.
Solution:
𝑀𝑅𝑂 = Σ𝐹𝑑
= −50𝑁 2𝑚 + 60𝑁 0𝑚
+ 20𝑁 3𝑠𝑖𝑛30𝑚
− (40𝑁)(4𝑚 + 3𝑐𝑜𝑠30𝑚)
𝑴𝑹𝑶 = −𝟑𝟑𝟒 𝐍 · 𝒎 = 𝟑𝟑𝟒 𝑵 ↻
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NORIEL M. PURIGAY
Resultant of Non-Concurrent Forces
Couple
Couple – Two parallel, non-collinear forces that
are equal in magnitude and opposite in direction
𝑅 = 0 ; Σ𝑀 ≠ 0
𝐶 = 𝐹𝑑
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NORIEL M. PURIGAY
Resultant of Non-Concurrent Forces
Equivalent Couples
𝐶 = 100 𝑙𝑏 2 𝑓𝑡 = 200 𝑙𝑏 1 𝑓𝑡 = 200 𝑙𝑏 · 𝑓𝑡
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Resultant of Non-Concurrent Forces
Coplanar Force System
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NORIEL M. PURIGAY
Resultant of Non-Concurrent Forces
The force system shown consists of
the couple C and four forces. If the
resultant of this system is a 500-lb·in.
counterclockwise couple, determine
P, Q, and C.
Solution:
𝑅𝑥 = Σ𝐹𝑥
12
4
= − 13 𝑄 + 5 𝑃 + 80 = 0
𝑅𝑦 = Σ𝐹𝑦
5
(1)
3
= − 13 𝑄 + 5 𝑃 − 20 = 0
(2)
Solving Eqns (1) and (2) simultaneously gives
𝑷 = 𝟐𝟎𝟎𝒍𝒃 𝒂𝒏𝒅 𝑸 = 𝟐𝟔𝟎𝒍𝒃
𝐶 𝑅 = 𝑀𝑅 = Σ𝑀𝐴
3
4
500 = −20 3 − 𝐶 + 80 4 + 𝑃 6 + 𝑃(6)
5
5
𝑪 = 𝟏𝟒𝟒𝟎 𝒍𝒃 · 𝒊𝒏
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NORIEL M. PURIGAY
Resultant of Non-Concurrent Forces
Parallel Force System
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NORIEL M. PURIGAY
Resultant of Non-Concurrent Forces
Replace the force and couple moment
system acting on the beam in the figure by
an equivalent resultant force, and find
where its line of action intersects the beam,
measured from point O.
Solution:
𝑅𝑥 = Σ𝐹𝑥
= 8 𝑘𝑁
𝑅𝑦 = Σ𝐹𝑦
3
5
= 4.8 𝑘𝑁
= −4 𝑘𝑁 + 8 𝑘𝑁
4
5
= 2.4 𝑘𝑁
𝑅 = (4. 8𝑘𝑁)2 + (2.4𝑘𝑁)2
𝑹 = 𝟓. 𝟑𝟕 𝒌𝑵
θ = 𝑡𝑎𝑛−1
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2.4
4.8
= 𝟐𝟔. 𝟔°
NORIEL M. PURIGAY
Resultant of Non-Concurrent Forces
Solution:
𝑀𝑅 = Σ𝑀𝑂
2.4𝑘𝑁 𝑑 = −4𝑘𝑁 1.5𝑚 − 15𝑘𝑁 · 𝑚 −
3
8𝑘𝑁 5 0.5𝑚 + 8𝑘𝑁
𝒅 = 𝟐. 𝟐𝟓𝒎
4
5
(4.5𝑚)
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NORIEL M. PURIGAY
Equilibrium of a Rigid Body
Equilibrium – A body is said to be in equilibrium if the
resultant of the force system that acts on the body
vanishes. Equilibrium means that both the resultant force
and the resultant couple are zero.
Free Body Diagram (FBD) - is a sketch of the body
showing all forces that act on it. The term free implies
that all supports have been remove and replaced by the
forces (reactions) that they exert on the body.
Forces that Act on a Body
1. Reactive Forces (Reactions) - forces that are exerted
on a body by the supports to which it is attached.
2. Applied Forces - forces acting on a body that are not
provided by the supports.
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NORIEL M. PURIGAY
Equilibrium of a Rigid Body
Conditions of Equilibrium
1. Graphical Condition: Under this
condition, the forces or vectors
are transformed into a force
polygon. For equilibrium, the
force polygon must close.
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NORIEL M. PURIGAY
Equilibrium of a Rigid Body
2. Directional Condition: If three
or more non-parallel forces or
vectors are in equilibrium, then
they must be concurrent. For a
two-force member, the forces
must be equal and opposite.
3. Analytical Condition: If forces
or vectors are in equilibrium, then
it must satisfy the three static
equations:
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Equilibrium of a Rigid Body
Support Reactions
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NORIEL M. PURIGAY
Equilibrium of a Rigid Body
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Equilibrium of a Rigid Body
Three forces 20 N, 30 N, and 40 N are in
equilibrium. Find the largest angle they
make with each other.
Solution:
402 = 302 + 202 − 2 30 20 𝑐𝑜𝑠θ
𝜽 = 𝟏𝟎𝟒. 𝟒𝟖°
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Equilibrium of a Rigid Body
A load of 100 lb is hung from the middle
of a rope, which is stretch between two
rigid walls 30 ft apart. Due to the load,
the rope sags 4 ft in the middle.
Determine the tension in the rope.
Solution:
15
𝑡𝑎𝑛θ =
4
θ = 75.068°
Σ𝐹𝑦 = 0 :
2𝑇𝑐𝑜𝑠75.068 = 100
𝑻 = 𝟏𝟗𝟒 𝒍𝒃
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Equilibrium of a Rigid Body
A simply supported beam is 5m in length. It Solution:
carries a uniformly distributed load including
its own weight of 300N/m and a concentrated
Σ𝑀𝐴 = 0 :
load of 100 N 2m from the left end. Find the 100𝑁 2𝑚 + 1500𝑁 2.5𝑚 = 𝑅𝐵 5𝑚
reactions if reaction A is at the left end and
𝑹𝑩 = 𝟕𝟗𝟎 𝑵
reaction B is at the right end.
Σ𝐹𝑦 = 0 :
𝑅𝐴 + 𝑅𝐵 = 100𝑁 + 1500𝑁
𝑹𝑨 = 𝟖𝟏𝟎 𝑵
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Equilibrium of a Rigid Body
The homogeneous 60-kg disk supported by
the rope AB rests against a rough vertical wall.
Using the given FBD, determine the force in
the rope and the reaction at the wall.
Solution:
Σ𝑀𝐵 = 0 :
𝑭𝑪 = 𝟎
Σ𝐹𝑦 = 0 :
4
𝑇 = 588.6𝑁
5
𝑻 = 𝟕𝟑𝟓. 𝟖 𝑵
Σ𝐹𝑥 = 0 :
3
𝑇 = 𝑁𝐶
5
𝑵𝑪 = 𝟒𝟒𝟏. 𝟓 𝑵
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Equilibrium of a Rigid Body
The homogeneous, 120-kg wooden beam is
suspended from ropes at A and B. A power
wrench applies the 500-N · m clockwise
couple to tighten a bolt at C. Use the given
FBD to determine the tensions in the ropes.
Solution:
Σ𝑀𝐴 = 0 :
𝑇𝐵 (4𝑚) = 1177.2𝑁 3𝑚 + 500𝑁𝑚
𝑻𝑩 = 𝟏𝟎𝟎𝟕. 𝟗 𝑵
Σ𝐹𝑦 = 0 :
𝑇𝐴 + 𝑇𝐵 = 1177.2
𝑻𝑨 = 𝟏𝟔𝟗. 𝟑 𝑵
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NORIEL M. PURIGAY
Equilibrium of a Rigid Body
The structure in Fig. (a) is
loaded by the 240-lb · in.
counterclockwise
couple
applied to member AB.
Neglecting the weights of the
members, determine all
forces acting on member
BCD.
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Equilibrium of a Rigid Body
Solution:
For Fig. (b)
Σ𝑀𝐴 = 0 :
𝑇𝐶 𝑐𝑜𝑠30 8 + 240 = 𝑁𝐷 (12)
𝑁𝐷 = 0.577𝑇𝐶 + 20 (𝟏)
For Fig. (c)
Σ𝑀𝐵 = 0 :
𝑇𝐶 𝑐𝑜𝑠30 4 + 𝑇𝐶 𝑠𝑖𝑛30(3) = 𝑁𝐷 (8)
𝑁𝐷 = 0.620𝑇𝐶
𝟐
Solving Eqs (1) and (2) simultaneously
gives:
𝑵𝑫 = 𝟐𝟖𝟖 𝒍𝒃 ; 𝑻𝑪 = 𝟒𝟔𝟓 𝒍𝒃
Σ𝐹𝑋 = 0 :
𝑁𝐷 + 𝐵𝑋 = 𝑇𝐶 𝑐𝑜𝑠30
𝑩𝑿 = 𝟏𝟏𝟒 𝒍𝒃
Σ𝐹𝑦 = 0 :
𝐵𝑦 = 𝑇𝐶 𝑠𝑖𝑛30
𝑩𝒚 = 𝟐𝟑𝟐 𝒍𝒃
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Analysis of Structures
Simple Trusses
Truss – is a structure composed of slender members
joined together at their end joints.
Planar Trusses - lie in a single plane and are often used
to support roofs and bridges
Simple Trusses - constructed by expanding the basic
triangular truss
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Analysis of Structures
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Analysis of Structures
Truss-Type Fuselage. A Warren Truss
uses mostly diagonal bracing.
Truss Wing Spar
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Analysis of Structures
Assumptions for Design
1. The weights of the members are negligible.
2. The members are joined together by smooth pins.
3. The applied forces act at the joints.
Each member of a truss is a two-force member.
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Analysis of Structures
Method of Joints
When using the method of joints to calculate the forces in
the members of a truss, the equilibrium equations are
applied to individual joints (or pins) of the truss.
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Analysis of Structures
Zero-Force Member – member that does not carry a load
– contributes to the stability of the structure
– can carry loads in the event that variations are introduced in
the normal external loading configuration
Σ𝐹𝑦 = 0 :
𝐺𝐶 = 0
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Analysis of Structures
Stable Structures
for 𝑚 = 2𝑗 − 3 →
𝑚 < 2𝑗 − 3 →
stable
unstable
Determinate Structures
for
r ≤ 3𝑚 → determinate
r > 3𝑚 → indeterminate
where:
m = no. of members
j = no. of joints
r = no. of reactions
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Analysis of Structures
Using the method of joints, determine the force in each member of
the truss shown in the figure. Indicate whether the members are in
tension or compression. (One of the supports is usually designed to
be equivalent to a roller, in order to permit the elongation and
contraction of the truss with temperature changes).
Solution:
Σ𝑀𝐶 = 0 :
𝑁𝐴 6 + 10 6 = 60(3)
𝑁𝐴 = 20 𝑘𝑁
Σ𝐹𝑋 = 0 :
𝐶𝑋 = 10 𝑘𝑁
Σ𝐹𝑦 = 0 :
𝐶𝑦 + 𝑁𝐴 = 60
𝐶𝑦 = 40 𝑘𝑁
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Analysis of Structures
at pin A:
at pin D:
Σ𝐹𝑦 = 0 :
1
𝑁𝐴 + 𝐴𝐵
=0
2
𝐴𝐵 = −28.28 𝑘𝑁
𝑨𝑩 = 𝟐𝟖. 𝟐𝟖 𝒌𝑵 𝑪
Σ𝐹𝑦 = 0 :
2
𝐵𝐷
= 60
5
𝑩𝑫 = 𝟔𝟕. 𝟎𝟖 𝒌𝑵 𝑻
Σ𝐹𝑥 = 0 :
1
𝐴𝐵
+ 𝐴𝐷 = 0
2
𝑨𝑫 = 𝟐𝟎 𝒌𝑵 𝑻
Σ𝐹𝑥 = 0 :
1
𝐵𝐷
+ 𝐶𝐷 = 𝐴𝐷
5
𝐶𝐷 = −10 𝑘𝑁
𝑪𝑫 = 𝟏𝟎 𝒌𝑵 𝑪
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
at pin C:
Σ𝐹𝑦 = 0 :
𝐵𝐶 + 40 = 0
𝐵𝐶 = −40 𝑘𝑁
𝑩𝑪 = 𝟒𝟎 𝒌𝑵 𝑪
NORIEL M. PURIGAY
Analysis of Structures
Method of Sections
• Analyzing the free-body diagram of a part of a truss
that contains two or more joints is called the method of
sections.
• Principle: If the truss is in equilibrium then any
segment of the truss is also in equilibrium.
• It permits us to directly determine the force in almost
any member instead of proceeding to that member by
joint-to-joint analysis.
• The cutting plane must not cut more than three
members whose internal forces are unknown.
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Analysis of Structures
Using the method of sections, determine the forces in members BC, HC, HG, and DF.
Solution:
Σ𝑀𝐴 = 0 :
8000 6 + 3000 12 = 𝑁𝐸 (24)
𝑁𝐸 = 3500 𝑙𝑏
Σ𝐹𝑋 = 0 :
𝐴𝑥 = 0
Σ𝐹𝑦 = 0 :
𝐴𝑦 + 𝑁𝐸 = 8000 + 3000
𝐴𝑦 = 7500 𝑙𝑏
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Analysis of Structures
Solution:
Σ𝐹𝑦 = 0 :
4
7500 + 𝐻𝐶
= 8000
5
𝑯𝑪 = 𝟔𝟐𝟓 𝒍𝒃 𝑻
Σ𝑀𝐻 = 0 :
7500 6 + 𝐵𝐶 8 = 0
𝐵𝐶 = −5625 𝑙𝑏
𝑩𝑪 = 𝟓𝟔𝟐𝟓 𝒍𝒃 𝑪
Σ𝑀𝐶 = 0 :
7500 12 = 8000 6 + 𝐻𝐺(8)
𝑯𝑮 = 𝟓𝟐𝟓𝟎 𝒍𝒃 𝑻
Σ𝐹𝑦 = 0 :
𝑫𝑭 = 𝟑𝟓𝟎𝟎 𝒍𝒃 𝑻
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Friction
Friction – force that resists the movement of two contacting surfaces that slide relative to
one another.
1. Dry Friction - friction force that exists between two unlubricated solid surfaces.
2. Fluid Friction - acts between moving surfaces that are separated by a layer of fluid.
𝑭 = 𝝁𝑵
where:
F = frictional force
μ = coefficient of friction
N = normal force
φ = angle of friction
𝒕𝒂𝒏𝝓 = 𝝁
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Friction
𝑭𝒎𝒂𝒙 = 𝝁𝒔 𝑵
𝐹𝑚𝑎𝑥 always opposes impending
sliding
𝑭𝒌 = 𝝁𝒌 𝑵
𝐹𝑘 always opposes sliding
𝝁𝒔 > 𝝁𝒌 ; 𝑭𝒔 > 𝑭𝒌
For very low velocity:
𝝁𝒔 ≈ 𝝁𝒌 ; 𝑭𝒔 ≈ 𝑭𝒌
where:
𝐹𝑚𝑎𝑥 = maximum static friction
𝐹𝑘 = kinetic friction
𝐹𝑠 = static friction
μ𝑠 = coefficient of static friction
μ𝑘 = coefficient of kinetic friction
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Friction
The 100-lb block in the figure
below is at rest on a rough
horizontal plane before the force
P is applied. Determine the
magnitude of P that would cause
impending sliding to the right
Solution:
𝑃 = 𝐹 = 𝐹𝑚𝑎𝑥 = 𝜇𝑠 𝑁 = 0.5 100𝑙𝑏
𝑷 = 𝟓𝟎 𝒍𝒃
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Friction
A 600N block rests in a surface inclined at
30°. Determine the horizontal force P
required to prevent the block from sliding
down. Angle of friction between the block
and the inclined plane is 15°.
Solution:
Σ𝐹𝑋 = 0 :
𝑃𝑐𝑜𝑠θ + 𝐹 = 𝑊𝑠𝑖𝑛θ
𝑃𝑐𝑜𝑠θ + μ𝑁 = 𝑊𝑠𝑖𝑛θ
𝑃𝑐𝑜𝑠θ + 𝑡𝑎𝑛ϕ𝑁 = 𝑊𝑠𝑖𝑛θ
𝑃𝑐𝑜𝑠30 + 𝑡𝑎𝑛15𝑁 = 600𝑠𝑖𝑛30
𝒂
Σ𝐹𝑦 = 0 :
−𝑃𝑠𝑖𝑛θ + 𝑁 = 𝑊𝑐𝑜𝑠θ
−𝑃𝑠𝑖𝑛30 + 𝑁 = 600𝑐𝑜𝑠30
𝒃
Solving Eqns. (a) and (b) simultaneously
gives:
𝑷 = 𝟏𝟔𝟎. 𝟕𝟕 𝑵
𝑁 = 600 𝑁
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Friction
Solution:
The uniform 100-lb plank in the figure below is
resting on friction surfaces at A and B. The
coefficients of static friction are shown in the figure.
If a 200-lb man starts walking from A toward B,
determine the distance x when the plank will start to
slide.
𝐹𝐴 = 0.2𝑁𝐴
𝐹𝐵 = 0.5𝑁𝐵
Substituting 𝐹𝐴 and 𝐹𝐵 to eqns.
(a), (b), and (c) and solving
simultaneously gives:
𝑁𝐴 = 163.3 𝑙𝑏 ; 𝑁𝐵 = 125.7 𝑙𝑏
𝒙 = 𝟒. 𝟑𝟒 𝒇𝒕
Σ𝐹𝑋 = 0 :
𝐹𝐴 + 𝐹𝐵 𝑐𝑜𝑠40 − 𝑁𝐵 𝑐𝑜𝑠50 = 0
(𝒂)
Σ𝐹𝑦 = 0 :
𝑁𝐴 + 𝑁𝐵 𝑠𝑖𝑛50 + 𝐹𝐵 𝑠𝑖𝑛40 = 300
(𝒃)
Σ𝑀𝐴 = 0 :
𝑁𝐵 𝑠𝑖𝑛50 10 + 𝐹𝐵 𝑠𝑖𝑛40 10 − 200𝑥 = 100 5 (𝒄)
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Force Systems in Space
Six equilibrium equations in three dimensions:
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Force Systems in Space
Assume the three force vectors intersect at a single
point.
𝐹1 = 4𝑖 + 2𝑗 + 5𝑘
𝐹2 = −2𝑖 + 7𝑗 − 3𝑘
𝐹3 = 2𝑖 − 𝑗 + 6𝑘
What is the magnitude of the resultant force vector, R?
Solution:
𝑅 = 𝐹1 + 𝐹2 + 𝐹3
= 4𝑖 + 2𝑗 + 5𝑘 + −2𝑖 + 7𝑗 − 3𝑘
+ 2𝑖 − 𝑗 + 6𝑘
𝑅 = 4𝑖 + 8𝑗 + 8𝑘
𝑅 =
(𝐴𝑖 )2 + (𝐴𝑗 )2 + (𝐴𝑘 )2
= 42 + 82 + 82
𝑹 = 𝟏𝟐 𝒖𝒏𝒊𝒕𝒔
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Force Systems in Space
Find the reactions for the equipment shelf shown in the
sketch. The three applied loads act at the center of the
volume shown. Supports A and B cannot take reactions
in the y direction and support C cannot take a reaction
in the x direction.
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Force Systems in Space
Solution:
Σ𝑀𝑥 = 0 :
8𝑅3 + 16𝑅2 + 200 5 − 400(8) = 0
𝑹𝟐 = 𝟔𝟖. 𝟕𝟓 𝒍𝒃
Σ𝐹𝑧 = 0 :
𝑅1 + 𝑅2 + 𝑅3 − 400 = 0
𝑹𝟏 = 𝟏𝟗𝟑. 𝟕𝟓 𝒍𝒃
Σ𝑀𝑦 = 0 :
300 5 − 400 12 + 24𝑅3 = 0
𝑹𝟑 = 𝟏𝟑𝟕. 𝟓 𝒍𝒃
Σ𝑀𝑧 = 0 :
24𝑅4 − 16𝑅6 − 200 12 + 300 8 = 0
𝑹𝟔 = 𝟑𝟎𝟎 𝒍𝒃
Σ𝐹𝑦 = 0 :
𝑹𝟒 = 𝟐𝟎𝟎 𝒍𝒃
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
Σ𝐹𝑥 = 0 :
𝑅5 + 𝑅6 − 300 = 0
𝑹𝟓 = 𝟎
NORIEL M. PURIGAY
Centroid and Center of Gravity
CG and CM
CG or Center of Weight
• It is the point at which the resultant of the gravitational forces (weight) act on a
body.
• It is a property of the distribution of weight within the body.
Center of Mass
• It is the point through which the resultant inertia force acts on a body.
• It is a property of the distribution of mass within the body.
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Centroid and Center of Gravity
CG by Tabular Summation
CM by Tabular Summation
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Centroid and Center of Gravity
Centroid
• It is the point at which area (or volume or line) can be concentrated.
• It is the point at which the static moment is zero.
• The centroid represents the geometric center of a body. This point
coincides with the center of mass or the center of gravity only if the
material composing the body is uniform or homogeneous.
• Formulas used to locate the center of gravity or the centroid simply
represent a balance between the sum of moments of all the parts of
the system and the moment of the “resultant” for the system.
• In some cases the centroid is located at a point that is not on the
object, as in the case of a ring, where the centroid is at its center.
Also, this point will lie on any axis of symmetry for the body.
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Centroid and Center of Gravity
Centroid by Intergration
Centroid of a Volume
Centroid of an Area
Centroid of a Line
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Centroid and Center of Gravity
Find the location of the centroidal axis
that is parallel to the base of the triangle
in sketch.
Solution:
𝐴𝑦 =
𝑦𝑑𝑎 =
𝑦𝑢𝑑𝑦
By similar triangles:
𝑢 𝑕−𝑦
𝑏
=
→
𝑢 = 𝑕−𝑦
𝑏
𝑕
𝑕
Then
𝑏𝑕
𝐴𝑦 =
𝑦=
2
=
=
𝑏𝑕2
𝑦=
6
𝒉
𝒚=
𝟑
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
ℎ
𝑏
𝑕
𝑕 − 𝑦 𝑦𝑑𝑦
0
𝑏 ℎ
2 𝑑𝑦)
(𝑕𝑦𝑑𝑦
−
𝑦
0
ℎ
𝑏 𝑕𝑦 2 𝑦 3 𝑕
−
𝑕 2
3 0
𝑏 𝑕3 𝑕3
𝑏𝑕2
=
−
=
𝑕 2
3
6
2
𝑏𝑕
NORIEL M. PURIGAY
Centroid and Center of Gravity
Solution:
The 16-ft wing of an airplane is subjected
to a lift which varies from zero at the tip to
360 lb/ft at the fuselage according to
𝑅 = 𝑎𝑟𝑒𝑎 𝑢𝑛𝑑𝑒𝑟 𝑡𝑕𝑒 𝑐𝑢𝑟𝑣𝑒
𝑤 = 90𝑥 lb/ft where x is measured from
the tip. Compute the resultant and its
location from the wingtip.
𝑹 = 𝟑𝟖𝟒𝟎 𝒍𝒃
1
2
16
𝑅=
90 𝑥𝑑𝑥
0
𝑥=
𝑥𝑑𝑎
=
𝐴
𝒙 = 𝟗. 𝟔𝟎 𝐟𝐭
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
16
𝑥
0
90 𝑥𝑑𝑥
3480
NORIEL M. PURIGAY
Centroid and Center of Gravity
Centroid of Common Geometric Shapes
Area and Centroid
Area and Centroid
𝐴 = 𝑏𝑑
1
𝑥= 𝑏
2
1
𝑦= 𝑑
2
𝐴 = π𝑟 2
𝑥=0
𝑦=0
1
𝐴 = 𝑏𝑕
2
1
𝑦= 𝑕
3
1 2
𝐴 = π𝑟
2
𝑥=0
4𝑟
𝑦=
3π
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Centroid and Center of Gravity
Centroid of Common Geometric Shapes
Area and Centroid
1 2
𝐴 = π𝑟
4
4𝑟
𝑥=
3π
4𝑟
𝑦=
3π
𝐴 = π𝑎𝑏
𝑥=0
𝑦=0
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
Area and Centroid
1
𝐴 = π𝑎𝑏
2
𝑥=0
4𝑏
𝑦=
3π
1
𝐴 = π𝑎𝑏
4
4𝑎
𝑥=
3π
4𝑏
𝑦=
3π
NORIEL M. PURIGAY
Centroid and Center of Gravity
Centroid by Tabular Summation
Centroid of a Composite Area
Centroid of a Composite Curve
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Centroid and Center of Gravity
Find the centroidal axes of the section
in the sketch.
Solution:
Element
A
y
x
Ay
Ax
1
.60
2.85
1.0
1.71
.60
2
.60
.15
1.0
.09
.60
3
.48
1.50
.10
.72
.048
ΣAy=2.52
ΣAx=1.248
ΣA=1.68
Σ(𝐴𝑥) 1.248
=
=. 𝟕𝟒𝟑 𝒊𝒏
Σ𝐴
1.68
Σ(𝐴𝑦) 2.52
𝑦=
=
= 𝟏. 𝟓𝟎 𝒊𝒏
Σ𝐴
1.68
𝑥=
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Moment of Inertia
Moment of Inertia
• Also called the second moment of area
• For structural cross-sections, the moment of inertia of interest are
those about the centroidal axes.
• Used in determining the stiffness and bending stresses in beams
and the buckling loads of columns
• For beams, the moment of area of interest is the one about the
bending axis, for columns, it is the minimum moment of inertia
First Moment of Area (Static Moment of Area)
𝑄=
𝑦𝑑𝐴 = Σ𝑎𝑦
• Used to find the shear stress distribution over a cross-section of a
shear carrying member
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Moment of Inertia
Moment of Inertia by Integration
Polar Moment of Inertia
Moment of Inertia by
Tabular Summation
𝐼𝑥 = Σ𝐼𝑥 + ΣA𝑦 2
𝑰𝒙 = 𝑰𝒙 − 𝐀𝒚𝟐
𝐼𝑦 = Σ𝐼𝑦 + ΣA𝑥 2
𝑰𝒚 = 𝑰𝒚 − 𝐀𝒙𝟐
Parallel –Axis Theorem (Transfer Formula)
where:
𝐼𝑎 = moment of inertia about
an arbitrary axis
𝐼𝑎 = moment of inertia about
the parallel axis that passes
through the centroid
𝐴 = area
𝑑 = distance between the
axes (transfer distance)
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Moment of Inertia
Centroidal Moment of Inertia (with respect to an axis passing through the centroid):
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Moment of Inertia
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Moment of Inertia
Moment of Inertia with respect to an axis passing through the base:
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Moment of Inertia
A rectangle has a base of 3 cm and a height
of 6 cm. What is its second moment of area
(in cm4) about an axis through the center of
gravity and parallel to the base.
Solution:
𝐼𝑥0
𝑏𝑕3 3(63 )
= 𝐼𝑥 =
=
12
12
𝟒
𝑰𝒙𝟎 = 𝟓𝟒 𝒄𝒎
𝑏𝑕3 3(63 )
𝐼𝑥 =
=
3
3
𝑰𝒙 = 𝟐𝟏𝟔 𝒄𝒎𝟒
Or by Transfer Formula:
𝐼𝑥 = 𝐼𝑥 + A𝑑 2 = 54 + 3𝑥6 32
𝑰𝒙 = 𝟐𝟏𝟔 𝒄𝒎𝟒
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Moment of Inertia
Find the moment of inertia of the section
about the centroidal axes parallel to axes x
and y
Solution:
From previous example
𝑥 = .743 𝑎𝑛𝑑 𝑦 = 1.50
𝐈𝐱
𝐈𝐲
Element
A
y
x
Ay2
Ax2
1
.60
2.85
1.00
4.8735
.60
.0045 .200
2
.60
.15
1.00
.0135
.60
.0045 .200
3
.48
1.50
.10
1.0800
.0048
.2304 .002
5.967
1.2048 .2394 .402
1.68
𝐼𝑥 = Σ𝐼𝑥 + ΣA𝑦 2 = .2394 + 5.967 = 6.2064 𝐼𝑦 = Σ𝐼𝑦 + ΣA𝑥 2 = .402 + 1.2048 = 1.6068
𝐼𝑥 = 𝐼𝑥 − 𝐴𝑦 2 = 6.2064 − 1.68 1.502
𝑰𝒙 = 𝟐. 𝟒𝟐𝟔 𝒊𝒏𝟒
𝐼𝑦 = 𝐼𝑦 − 𝐴𝑥 2 = 1.6068 − 1.68 .7432
𝑰𝒚 = 𝟎. 𝟔𝟕𝟗𝟓 𝒊𝒏𝟒
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Mechanics of Rigid
Bodies
Statics
Force Systems
Dynamics
Applications
Kinematics
Kinetics
Concurrent
Trusses
Translation
Translation
Parallel
Centroids
Rotation
Rotation
Non-Concurrent
Friction
Plane Motion
Plane Motion
FEATI
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UNIVERSITY
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NORIEL M. PURIGAY
Fundamental Concepts and
Definitions
• Motion – Change of position of an object with
respect to time and reference point.
• Translation – Motion involving change in
displacement over a period of time.
• Rotation – Motion involving change in angle over a
period of time.
• Rectilinear Motion – Straight line motion
• Curvilinear Motion – Motion along a curved path
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Fundamental Concepts and Definition
• Displacement, s – Shortest distance
between initial and final position of a
particle.
• Velocity, V – Instantaneous rate of change
of displacement with respect to time.
𝑑𝑠
𝑉=
𝑑𝑡
• Speed – Refers to the magnitude of velocity.
• Acceleration, a - Instantaneous rate of
change of velocity with respect to time.
𝑑𝑉
𝑎=
𝑑𝑡
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Kinematics of a Particle
Rectilinear Translation
Equations of Motion with Constant Acceleration:
1
• 𝑠 = 𝑉0 𝑡 + 2 𝑎𝑡 2
• 𝑉 = 𝑉0 + 𝑎𝑡
2
• 𝑉 2 = 𝑉0 + 2𝑎𝑠
For a free falling body, 𝑉0 = 0
1
• 𝑦 = 2 𝑔𝑡 2
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Kinematics of a Particle
A ball is thrown vertically upward at a speed of 20
m/s.
a. How high is it after 3s?
1
𝑦 = 𝑉0 𝑡 + 𝑔𝑡 2
2
𝑚
𝑚
𝑦 = 20
3𝑠 + .5 −9.8 2 (3𝑠)2
𝑠
𝑠
𝒚 = 𝟏𝟓. 𝟗 𝒎
b. How high does it rise?
2
𝑉 2 = 𝑉0 + 2𝑔𝑦
𝑚 2
𝑚
0 = 20
+ 2 −9.8 2 𝑦𝑚𝑎𝑥
𝑠
𝑠
𝒚𝒎𝒂𝒙 = 𝟐𝟎. 𝟒𝟏 𝒎
d. How long does it take for the
ball to reach the ground?
𝒕𝒖𝒑 = 𝒕𝒅𝒐𝒘𝒏 = 𝟐. 𝟎𝟒 𝒔
e. What is its velocity when it
returns to the level from which
it started?
𝑽𝟎 = 𝑽𝒇 = 𝟐𝟎 𝒎/𝒔
c. How long does it take to reach the highest point?
𝑉 = 𝑉0 + 𝑔𝑡
𝑚
𝑚
0 = 20
+ −9.8 2 𝑡
𝑠
𝑠
𝒕 = 𝟐. 𝟎𝟒 𝒔
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Kinematics of a Particle
If a particle’s position is given by the expression
𝑥 𝑡 = 3.4𝑡 3 − 5.4𝑡 meters.
a. What is its velocity after t = 3 seconds?
𝑑𝑥
𝑑
𝑉=
= (3.4𝑡 3 − 5.4𝑡)
𝑑𝑡 𝑑𝑡
𝑡=3
𝑽 = 𝟖𝟔. 𝟒 𝒎/𝒔
b. What is the acceleration of the particle after t = 5
seconds?
𝑑𝑥
𝑑
𝑉=
=
3.4𝑡 3 − 5.4𝑡 = 10.2𝑡 2 − 5.4
𝑑𝑡 𝑑𝑡
𝑑𝑉
𝑑
𝑎=
= (10.2𝑡 2 − 5.4)
𝑑𝑡 𝑑𝑡
𝑡=5
𝒂 = 𝟏𝟎𝟐 𝒎/𝒔𝟐
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Curvilinear Translation
General Equation of Projectile:
𝑦 = 𝑥𝑡𝑎𝑛θ −
𝑔𝑥 2
2𝑉0 2 𝑐𝑜𝑠 2 θ
• 𝑉𝑦 deceases as it goes up, zero at maximum height, and increases as
it goes down.
• 𝑉𝑥 is constant.
𝑉0𝑥 = 𝑉𝑜 𝑐𝑜𝑠θ
𝑉0𝑦 = 𝑉𝑜 𝑠𝑖𝑛θ
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Kinematics of a Particle
A shot is fired at an angle of 45° with the horizontal
and a velocity of 300 fps. Calculate the range of the
projectile.
Solution:
𝑉0 2 𝑠𝑖𝑛2θ
𝑅=
𝑔
(300 𝑓𝑡/𝑠)2 sin(2𝑥45)
=
32.2 𝑓𝑡/𝑠 2
𝑹 = 𝟐𝟕𝟗𝟓 𝒇𝒕
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Kinematics of a Particle
A projectile leaves with a velocity of 50 m/s at an
angle of 30° with the horizontal. Find the maximum
height that it could reach.
Solution:
𝑉0 2 𝑠𝑖𝑛2 θ
𝐻=
2𝑔
(50 𝑚/𝑠)2 (sin30)2
=
𝑚
2 9.81 2
𝑠
𝑯 = 𝟑𝟏. 𝟖𝟔 𝒎
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Rotation
• 𝑆 = 𝑟θ
• 𝑉 = 𝑟ω
• 𝑎 = 𝑟α
Where:
• r = radius
• θ, ω, and α are angular displacement,
angular velocity, and angular acceleration,
respectively.
• S, V, and a are linear dimensions.
• Linear velocity acts tangent to the point.
• Linear acceleration has tangential and
normal components.
• 𝑎 = 𝑎𝑛 2 + 𝑎𝑡 2
•
•
𝑉2
𝑎𝑛 =
𝑟
𝑑𝑉
𝑎𝑡 = 𝑑𝑡
Also,
1
• θ = ω0 𝑡 + 2 α𝑡 2
• ω = ω0 + α𝑡
2
• ω2 = ω0 + 2α𝑠
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Kinematics of a Particle
A turbine started from rest to 180 rpm in 6 minutes
at a constant acceleration. Find the number of
revolutions that it makes within the elapsed time.
Solution:
ω = ω0 + α𝑡
180 𝑟𝑝𝑚 = 0 + α 6 𝑚𝑖𝑛𝑢𝑡𝑒𝑠
𝑟𝑒𝑣
α = 30
𝑚𝑖𝑛2
2
ω2 = ω0 + 2α𝑠
(180 𝑟𝑝𝑚)2 = 0 + 2 30
𝒔 = 𝟓𝟒𝟎 𝒓𝒆𝒗
𝑟𝑒𝑣
(𝑠)
𝑚𝑖𝑛2
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Kinematics of a Particle
A flywheel is 15 cm in diameter accelerates
uniformly from rest to 500 rpm in 20 seconds.
What is its angular acceleration?
Solution:
𝑟𝑒𝑣 2π𝑟𝑎𝑑 𝑚𝑖𝑛
500
𝑥
𝑥
= 52.36 𝑟𝑎𝑑/𝑠
𝑚𝑖𝑛
𝑟𝑒𝑣
60 𝑠
ω = ω0 + α𝑡
𝑟𝑎𝑑
52.36
= 0 + α 20 𝑠
𝑠
𝜶 = 𝟐. 𝟔𝟐 𝒓𝒂𝒅/𝒔𝟐
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Kinetics of a Particle
D’ Alembert’s Principle
When a body is subjected to an acceleration, there exists a force
opposite the direction of motion and equal to the product of
mass and acceleration . This force is known as the reverse
effective force, REF or inertial force, ma.
𝜮𝑭 = 𝒎𝒂 =
𝒘
𝒂
𝒈
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Kinetics of a Particle
What force is necessary to accelerate a 30,000 lb railway
electric car at the rate of 1.25 ft/sec2, if the force
required to overcome frictional resistance is 400 lb?
Solution:
Σ𝐹ℎ = 𝑚𝑎
𝑊
𝑎
𝑔
30,000 𝑙𝑏
𝑃 − 400 =
32.2 𝑓𝑝𝑠 2
𝑷 = 𝟏𝟓𝟔𝟒. 𝟔 𝒍𝒃
𝑃−𝐹 =
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
1.25𝑓𝑝𝑠 2
NORIEL M. PURIGAY
Kinetics of a Particle
An elevator weighing 2,000 lb attains an upward velocity
of 16 fps in 4 seconds with uniform acceleration. What is
the tension in the supporting cables?
Solution:
𝑉 = 𝑉0 + 𝑎𝑡
𝑓𝑡
16 = 0 + 𝑎 4𝑠
𝑠
𝑎 = 4 𝑓𝑡/𝑠 2
Σ𝐹𝑣 = 𝑚𝑎
𝑊
𝑇−𝑊 =
𝑎
𝑔
2000 𝑙𝑏
𝑇 − 2000 =
4 𝑓𝑝𝑠 2
2
32.2 𝑓𝑝𝑠
𝑷 = 𝟐𝟐𝟒𝟖. 𝟒𝟓 𝒍𝒃
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Kinetics of a Particle
Centripetal and Centrifugal Force
𝑉2
𝐶𝐹 = 𝑚𝑎𝑛 = 𝑚
𝑟
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
Kinetics of a Particle
A cyclist on a circular track of radius
800 ft is traveling at 27 fps. His
speed in the tangential direction
increases at the rate of 3 fps2. What
is the cyclist’s total acceleration?
Solution:
𝑉 2 (27 𝑓𝑝𝑠)2
𝑎𝑛 =
=
𝑟
800 𝑓𝑡
= 0.91 𝑟𝑎𝑑/𝑠 2
𝑎 = 𝑎𝑡 2 + 𝑎𝑛 2
𝑎 = 32 + 0.912
𝒂 = 𝟑. 𝟏𝟒 𝒇𝒑𝒔𝟐
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY
FEATI UNIVERSITY – PROFESSIONAL SCHOOLS / AERONAUTICAL ENGINEERING REVIEW CENTER
NORIEL M. PURIGAY