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A REVIEW:
FEATI AERO BOARD REFRESHER
Prepared by: Contado, Clovis B.
ENGINEERING MECHANICS
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Mechanics is the branch of physics thatconsiders the action of forces on bodies or fluids
that are both at rest and in motion
Engineering Mechanics is the branch of
engineering that applies the principles of
mechanics to any design that must take into
account the effect of forces
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Outline - Statics
Chapter I. Principles of Statics
Chapter II. Resultants of Force Systems
Chapter III. Equilibrium of Force Systems
Chapter IV. Analysis of Structure Chapter V. Friction
Chapter VI. Force Systems in Space
Chapter VII. Centroids & Centers of Gravity
Chapter VIII. Moments of Inertia
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ENGINERRING
MECHANICS
Statics
Force Systems
Concurrent
Parallel
Non-
Concurrent
Application
Trusses
Centroids
Friction
Dynamics
Kinematics Kinetics
Fig 1.1 Outline of Engineering
Mechanics
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Chapter I
FUNDAMENTAL CONCEPTS &
DEFINITIONS
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Chapter I. Fundamental Concepts &
Definition
Engineering Mechanics the science which considersthe effects of forces on rigid bodies
Statics Consider the effects and distribution of
forces on rigid bodies
Dynamics consider the motion of rigid bodiescaused by the forces acting on them. It deals with
objects or structures with a non-zero acceleration.
Force that which changes , or tend to change the
state of motion of body.
Note: External effects of forces are considered in Engineering
Mechanics; Internal effects, in Strength of Material
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Chapter I. Fundamental Concepts &
Definition
Characteristics of a Force: (1) has magnitude (2)position of its line of action (3) the direction orsense
Classification of Force Systems
1. Coplanar Force System the line of action of all
forces lie on one plane
2. Non-Coplanar (Space Systems of Forces) the
line of action of all forces do not lie on a same
plane
3. Concurrent forces lines of forces pass trough acommon point
4. Non-concurrent
5. Parallel Force System
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Chapter I. Fundamental Concepts &
Definition
Axioms Of Mechanics
1. The Parallelogram Law the resultant of two
forces is the diagonal formed on the vectors of
these forces
2. Two 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 toany system of forces without changing the effect
if the original system
4. Action and reaction forces are equal but
oppositely directed,
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Chapter II.
RESULTANTS OF FORCE
SYSTEMS
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.Systems
The effect of a system of forces on a body isusually expressed in terms of a resultant.
1. Fx = Fcos x
2. Fy = F Sin x
3.
4.
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.Systems
Problems
1. Determine the resultant of
the concurrent forces
shown
2. The resultant of theconcurrent forces shown in
the fig is 300lb pointing up
along the Y-axis. Compute
the values of F and
required to give theresultant.
500
lb240
lb
30
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ap er . esu an s o orceSystems
3. A boat moving at12kph is crossing a river
500m wide in which a
current is flowing at
4kph. In what directionshould the boat head if
it is to reach a point on
the other side of the
river directly opposite its
starting point?
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.Systems
4. Beam AB in the fig below, supports a load whichvaries from an intensity of 50lb/ft at one end to
200lb/ft at the other. Calculate the magnitude &
position of the resultant.
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.Systems
Moment of a Force the moment of a force about a
point or axis measures the tendency of the force tocause the body to rotate around that axis or point.
M= Fd
Where: d perpendicular distance
F Resultant Force
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.Systems
A. Varignons Theorem the moment of the
resultant is equivalent to the moment sum of itscomponent
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.Systems
Problems
1. In a certain non-concurrent
concurrent force system, it is
is found that X = -80 lb, Y
Y = 160lb, M = 480 ft-lb in
a counterclockwise sense.Determine at which the
resultant intersects the x-axis.
2. Two forces P & Q passthrough a point A which is 4ft
to the right of and 3ft above
a moment center O. Force P
is 200lb directed up to the
right at 30 with thehorizontal and force Q is
100lb directed up to the left
at 60 with the horizontal.
Determine the moment of theresultant of these two forces
with respect to O.
Ans: 377ft-lb
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.Systems
3.The 16-ft wing of an
airplane is subjected toa lift which varies from
zero at the tip to 360
lb/ft at the fuselage
according to = 90x^1/2 lb/ft where x is
measured from the tip.
Compute the resultant
and its location from the
wing tip.
Ans: R = 3840 lb at 9.60 ft
4. Determine completely
the resultant of theforces acting on the
step pulley shown in the
fig. Ans: 1250 lb down
to right; h = 44.3750 lb
1250
lb 250 lb
60
1.25
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.Systems
Couples. Sometimes the resultant wil be zero inmagnitude and yet have a resultant moment sum.
The special case in which the resultant has zeromagnitude but does have a moment is said to consista couple.
C = F d
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.Systems
Ex. Transform the couple
shown in the figure into anequivalent couple whose
forces are horizontal abd
act through points C & D.
Solution:
When the forces of thecouple act through points C
& D, the moment arm of
the couple becomes 3in.
Since the moment effect isconstant, the forces acting
at C & D are found from
C = Fd C = 9x4 = F x3
F = 12lb
Therefore C & D each has
magnitude of 12lb.
9lb 9lb
A B
C
D
4
3
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Systems
Replace the system of forces acting on the frame
below by a resultant R at A and a couple actinghorizontally through B & C.
Ans: R = 50lb down; B=
110lb right: C
= 110lb left
20lb
30lb 60lb
3
4
2
1
A
B
C
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Chapter III
EQUILIBRIUM OF FORCE
SYSTEMS
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.Systems
Equilibrium is the term used to designate thecondition where the resultant of a system of forces is
zero. A body is said to be in equilibrium when the
force system acting upon it has zero resultant.
The physical meaning of equilibrium, as applied to abody, is that the body either is at rest or moving in a
straight line with constant velocity
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.Systems
Equilibrium of Concurrent Force System x = 0
y = 0
Equilibrium of Parallel Force System
F = 0
M = 0
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Chapter III. Equilibrium of Force Systems
Equilibrium of Non-Concurrent Force System
x = 0 x = 0
y = 0 or MA = 0
M = 0
MB
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.Systems
Problems1. A 300 lb box is held at rest
on a smooth plane by a
force P inclined at an
angle with the plane as
shown. If = 45,
determine the value of P
and the normal force N,
exerted by the plane.
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Systems
2. A load of 100 lb is hubg
from the middle of arope, which is stretched
between two rigid walls
30ft apart. Due to the
load, the rope sags 4ftin the middle.
Determine the tension
in the rope.
Ans: 194 lbs
3. The 300 lb force and the
400lb force shown in thefig. are to be held in
equilibrium by a third
force F acting at an
unknown angle
with thehorizontal. Determine the
values of F and .
30
F
400 lb
300 lb
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.Systems
4. Determine the load P required to hold bar AB in a
horizontal position on the smooth inclines showninf Fig. Also determine the reactions at A & B.
14 24
4560
P 400 lb
BA
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Chapter IV
ANALYSIS OF STRUCTURES
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Chapter IV. Analysis of Structure
The analysis of a structure is the process by which we
determine how the loads applied are distributed
throughout a structure.
Two types of structures will be studied;
1. Pin-connected trusses the internal force in a bar
is directed along the axis of the frames
2. Pin-connected frames the members are subjected
to bending action
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Chapter IV. Analysis of Structure
A truss is a structure composed of members
fastened together in such a way to resist change
in shape: it is a rigid structure
Trusses are so constructed that all applied loads
act at the ends of the members. Such membersheld in equilibrium by only two forces are called
two-force members.
Members which are stretched are said to be in
tension, while those that are shortened are saidto be in compression.
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Chapter IV. Analysis of Structure
C f S
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Chapter IV. Analysis of Structure
Ch IV A l i f S
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Chapter IV. Analysis of Structure
Ch t IV A l i f St t
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Chapter IV. Analysis of Structure
Ch t IV A l i f St t
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Chapter IV. Analysis of Structure
Ch t IV A l i f St t
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Chapter IV. Analysis of Structure
Analysis of Structure
1. Method of Joints analysing trusses by
applying the principles of equilibrium to the
concurrent force systems
2. Method of Sections the principles of
equilibrium of non-concurrent force systems are
applied.
Ch t IV A l i f St t
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Chapter IV. Analysis of Structure
Ch t IV A l i f St t
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Chapter IV. Analysis of Structure
Problems.
1. Using the method of sections, determine the force
on members BD,CD, & CE of the truss shown on
below figure.
Ch t IV A l i f St t
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Chapter IV. Analysis of Structure
2. For the truss shown in fig, determine the force in BF
by the method of joints and then check this resultusing the method of sections.
1200 lb
F
E
D
2400 lb1200 lb
C
A
B
9
12
9
12
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Chapter V
FRICTION
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Chapter V. Friction
Friction may be defined as the contactresistance exerted by one body upon a second
body when the second body moves or tends to
move past first body.
It is a retarding force always acting opposite tothe motion or the tendency to.
Ch t V F i ti
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Chapter V. Friction
Ch t V F i ti
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Chapter V. Friction
Chapter V Friction
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Chapter V. Friction
Chapter V Friction
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Chapter V. Friction
Chapter V Friction
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Chapter V. Friction
Problems:
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Chapter VI
FORCE SYSTEMS IN SPACE
Ch t VI F S t i S
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Chapter VI. Force Systems in Space
In the preceding chapters on coplanar systems, we
have seen how two fundamental concepts, (1) thatwhich relates a force to its components and (2) the
moment effect of a force, were applied. When we
consider force systems in space, the same basic
concepts are all that are necessary, only they must beextended to include the more general case of space
forces
Ch t VI F S t i S
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Chapter VI. Force Systems in Space
Ch t VI F S t i S
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Chapter VI. Force Systems in Space
Equations:
Ch t VI F S t i S
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Chapter VI. Force Systems in Space
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