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Materials Engineering Department
Subject: Engineering Mechanics
Class: First
Lecturer: Dr. Emad AL-Hassani
Date : 2009-2010
Lecture # 1
1.1. WHAT IS MECHANICS?
Mechanics can be defined as that science which describes and
predicts the conditions of rest or motion of bodies under the action of
forces. It is divided into three parts: mechanics of rigid bodies, mechanics
of deformable bodies, and mechanics of fluids. The mechanics of rigid
bodies is subdivided into statics and dynamics, the former dealing with
bodies at rest, the latter with bodies in motion. In this part of the study of
mechanics, bodies are assumed to be perfectly rigid. Actual structures and
machines, however, are never absolutely rigid and deform under the loads
to which they are subjected. But these deformations are usually small and
do not appreciably affect the conditions of equilibrium or motion of the
structure under consideration. They are important, though, as far as the
resistance of the structure to failure is concerned and are studied in
mechanics of materials, which is a part of the mechanics of deformable
bodies. The third division of mechanics, the mechanics of fluids, is
subdivided into the study of incompressible fluids and of compressible
fluids. An important subdivision of the study of incompressible fluids is
hydraulics, which deals with problems involving water.
Mechanics is a physical science, since it deals with the study of
physical phenomena. However, some associate mechanics with
mathematics, while many consider it as an engineering subject. Both
these views are justified in part. Mechanics is the foundation of most
engineering sciences and is an indispensable prerequisite to their study.
1
However, it does not have the empiricism found in some engineering
sciences, i.e., it does not rely on experience or observation alone; by its
rigor and the emphasis it places on deductive reasoning it resembles
mathematics. But, again, it is not an abstract or even a pure science;
mechanics is an applied science. The purpose of mechanics is to explain
and predict physical phenomena and thus to lay the foundations for
engineering applications.
1.2. FUNDAMENTAL CONCEPTS AND PRINCIPLES
The basic concepts used in mechanics are space, time, mass, and
force. These concepts cannot be truly defined; they should be accepted on
the basis of our intuition and experience and used as a mental frame of
reference for our study of mechanics. The concept of space is associated
with the notion of the position of a point P. The position of P can be
defined by three lengths measured from a certain reference point, or
origin, in three given directions. These lengths are known as the
coordinates of P.
To define an event, it is not sufficient to indicate its position in space.
The time of the event should also be given.
The concept of mass is used to characterize and compare bodies on
the basis of certain fundamental mechanical experiments. Two bodies of
the same mass, for example, will be attracted by the earth in the same
manner; they will also offer the same resistance to a change in
translational motion.
A force represents the action of one body on another. It can be
exerted by actual contact or at a distance, as in the case of gravitational
forces and magnetic forces. A force is characterized by its point of
2
application, its magnitude, and its direction; a force is represented by a
vector .
In Newtonian mechanics, space, time, and mass are absolute
concepts, independent of each other. (This is not true in relativistic
mechanics, where the time of an event depends upon its position, and
where the mass of a body varies with its velocity.) On the other hand, the
concept of force is not independent of the other three. Indeed, one of the
fundamental principles of Newtonian mechanics listed below indicates
that the resultant force acting on a body is related to the mass of the body
and to the manner in which its velocity varies with time.
You will study the conditions of rest or motion of particles and
rigid bodies in terms of the four basic concepts we have introduced. By
particle we mean a very small amount of matter which may be assumed
to occupy a single point in space. A rigid body is a combination of a
large number of particles occupying fixed positions with respect to each
other. The study of the mechanics of particles is obviously a prerequisite
to that of rigid bodies. Besides, the results obtained for a particle can be
used directly in a large number of problems dealing with the conditions
of rest or motion of actual bodies. The study of elementary mechanics
rests on six fundamental principles based on experimental evidence.
The Parallelogram Law for the Addition of Forces
This states that two forces acting on a particle may be replaced by a
single force, called their resultant, obtained by drawing the diagonal of
the parallelogram which has sides equal to the given forces.
3
The Principle of Transmissibility
This states that the conditions of equilibrium or of motion of a rigid
body will remain unchanged if a force acting at a given point of the rigid
body is replaced by a force of the same magnitude and same direction,
but acting at a different point, provided that the two forces have the same
line of action.
Newton’s Three Fundamental Laws.
Formulated by Sir Isaac Newton in the latter part of the seventeenth
century, these laws can be stated as follows:
FIRST LAW
If the resultant force acting on a particle is zero, the particle will remain
at rest (if originally at rest) or will move with constant speed in a straight
line (if originally in motion) .
SECOND LAW
If the resultant force acting on a particle is not zero, the particle will have
an acceleration proportional to the magnitude of the resultant and in the
direction of this resultant force., this law can be stated as:
F=m a ……. (1.1)
Where F, m, and a represent, respectively, the resultant force acting on
the particle, the mass of the particle, and the acceleration of the particle,
expressed in a consistent system of units.
4
THIRD LAW
The forces of action and reaction between bodies in contact have the
same magnitude, same line of action, and opposite sense .
Newton’s Law of Gravitation
This states that two particles of mass M and m are mutually attracted
with equal and opposite forces F and −F (Fig. 1.1) of magnitude F given
by the formula
F=G (Mm/r2)…….. (1.2)
Where
r = distance between the two particles
G = universal constant called the constant of gravitation
Newton’s law of gravitation introduces the idea of an action exerted at a
distance and extends the range of application of Newton’s third law: the
action F and the reaction −F in Fig. 1.1 are equal and opposite, and they
have the same line of action.
Fig: (1.1)
A particular case of great importance is that of the attraction of the
earth on a particle located on its surface. The force F exerted by the earth
5
on the particle is then defined as the weight W of the particle. Taking M
equal to the mass of the earth, m equal to the mass of the particle, and r
equal to the radius R of the earth, and introducing the constant
g = GM ……… (1.3)
The magnitude W of the weight of a particle of mass m may be
expressed as
W =mg ………. (1.4)
The value of R in formula (1.3) depends upon the elevation of the
point considered; it also depends upon its latitude, since the earth is not
truly spherical. The value of g therefore varies with the position of the
point considered. As long as the point actually remains on the surface of
the earth, it is sufficiently accurate in most engineering computations to
assume that g equals 9.81 m/s2 or 32.2 ft/s2.
1.3. SYSTEMS OF UNITS
With the four fundamental concepts introduced in the preceding
section are associated the so-called kinetic units, i.e., the units of length,
time, mass, and force. These units cannot be chosen independently if Eq.
(1.1) is to be satisfied. Three of the units may be defined arbitrarily; they
are then referred to as basic units. The fourth unit, however, must be
chosen in accordance with Eq. (1.1) and is referred to as a derived unit.
Kinetic units selected in this way are said to form a consistent system of
units.
6
International System of Units (SI Units)
In this system, which will be in universal use after the United
States has completed its conversion to SI units, the base units are the units
of length, mass, and time, and they are called, respectively, the meter (m),
the kilogram (kg), and the second (s). All three are arbitrarily defined.
The second, which was originally chosen to represent 1/86 400 of the
mean solar day, is now defined as the duration of 9 192 631 770 cycles of
the radiation corresponding to the transition between two levels of the
fundamental state of the cesium-133 atom. The meter, originally defined
as one ten-millionth of the distance from the equator to either pole, is now
defined as 1 650 763.73 wavelengths of the orange-red light
corresponding to a certain transition in an atom of krypton-86. The
kilogram, which is approximately equal to the mass of 0.001 m3 of water,
is defined as the mass of a platinum-indium standard kept at the
international Bureau of Weights and Measures at Sevres, near Paris,
France. The unit of force is a derived unit. It is called the Newton (N) and
is defined as the force which gives an acceleration of 1 m/s2 to a mass of
1 kg (Fig. 1.2). From Eq. (1.1) we write
1 N = (l kg) (l m/s2) = l kg.m/s2 …….. (1.5)
7
The SI units are said to form an absolute system of units. This
means that the three base units chosen are independent of the location
where measurements are made. The meter, the kilogram, and the second
may be used anywhere on the earth; they may even be used on another
planet. They will always have the same significance.
The weight of a body, or the force of gravity exerted on that body,
should, like any other force, be expressed in newtons. From Eq. (1.4) it
follows that the weight of a body of mass 1 kg (Fig. 1.3) is
W =mg
= (1 kg) (9.81 m/s2)
=9.81N
Multiples and submultiples of the fundamental SI units may be
obtained through the use of the prefixes defined in Table 1.1. The
multiples and submultiples of the units of length, mass, and force most
frequently used in engineering are, respectively, the kilometer (km) and
8
the millimeter (mm); the megagram (Mg) and the gram (g); and the
kilonewton (kN). According to Table 1.1, we have
1 km =1000 m
1 mm = 0.001 m
l Mg= 1000kg
1 g = 0.001 kg
1kN = 1000 N
The conversion of these units into meters, kilograms, and newtons,
respectively, can be effected by simply moving the decimal point three
places to the right or to the left. For example, to convert 3.82 km into
meters, one moves the decimal point three places to the right:
3.82 km = 3820 m
Similarly, 47.2 mm is converted into meters by moving the
decimal point three places to the left:
47.2 mm = 0.0472 m
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Table (1.1): SI Prefixes
Multiplication Factor
Pretixt
Symbol
1000 000 000 000 =1012
tera
T
1 000 000 000 = 109
giga
G
1 000 000 = 106
mega
M
1000 =103
kilo
k
100 = 102
hecto
h
10 = 101
deka
da
0.1= 10−1
deci
d
0.01 = 10−2
centi
c
0.001 = 10−3
milli
m
0.000 001 = 10−6
micro
µ
0.000 000 001 = 10−9
nano
n
0.000 000 000 001=10−12
pico
p
femto
f
atto
a
0.000 000 000 000 001= 10−15
0.000 000 000 000 000 001 = 10−18
Using scientific notation, one may also write
3.82 km = 3.82 x 103 m
47.2 mm = 47.2 x 10−3
The multiples of the unit of time are the minute (min) and the hour (h).
Since 1 min = 60 s and 1 h 60 min = 3600 s, these multiples cannot be
converted as readily as the others.
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By using the appropriate multiple or submultiple of a given unit,
one can avoid writing very large or very small numbers. For example, one
usually writes 427.2 km rather than 427 200 m, and 2.16 mm rather than
0.002 16 m.
Units of Area and Volume
The unit of area is the square meter (m2), which represents the area
of a square of side 1 m; the unit of volume is the cubic meter (m3), equal
to the volume of a cube of side 1 m. In order to avoid exceedingly small
or large numerical values in the computation of areas and volumes, one
uses systems of subunits obtained by respectively squaring and cubing
not only the millimeter but also two intermediate submultiples of the
meter, namely, the decimeter (dm) and the centimeter (cm). Since, by
definition,
1 dm = 0.1 m = 10−1m
1 cm = 0.01 m = 10−2 m
1 mm 0.001 m = 10−2 m
The submultiples of the unit of area are
1dm2 = (1dm) 2= (10−1) 2=10−2 m2
1 cm2 = (1 cm) 2 = (10−2m) 2 =10−4 m2
1 mm2 = (1 mm) 2 = (10−3m) 2 = 10−6m 2
and the submultiples of the unit of volume are
1dm3 = (1 dm) 3 = (10−1m) 3 = 10−3m 3
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1 cm3 = (1 cm) 3 = (10−2m) 3= 10−6m 3
1 mm3 = (1 mm) 3 = (10−3m) 3 = 10−9m 3
It should be noted that when the volume of a liquid is being
measured, the cubic decimeter (dm3) is usually referred to as a liter (L).
Other derived SI units used to measure the moment of a force, the
work of a force, etc., are shown in Table 1.2. While these units will be
introduced in later chapters as they are needed, we should note an
important rule at this time: When a derived unit is obtained by dividing a
base unit by another base unit, a prefix may be used in the numerator of
the derived unit but not in its denominator. For example, the constant k of
a spring which stretches 20 mm under a load of 100 N will be expressed
as
k = 100 N/20 mm = l00N /0.020 m =5000 N/m
Or
k = 5 kN /m
but never as
k = 5 N/mm.
Table (1.2): Principal SI Units Used in Mechanics
Quantity
Unit
Symbol Formula
Acceleration
Meter per second squared
…….
m/s2
Angle
Radian
rad
*
Angular acceleration Radian per second squared .
rad/s2
Angular velocity
Rad/s
Radian per second
12
.
Area
Square meter
.
m2
Density
Kilogram per cubic meter
.
kg/m3
Energy
Joule
J
N. m
Force
Newton
N
Kg. m/s2
Frequency
Hertz
Hz
s−1
Impulse
Newton-second
..
Kg . m/s
Length
Meter
m
Mass
Kilogram
kg
Moment of a force
Newton-meter
...
N. m
Power
Watt
W
J/s
Pressure
Pascal
Pa
N/m2
Stress
Pascal
Pa
N/ m2
Time
Second
s
Velocity
Meter per second
.
m/s
Solids
Cubic meter
.
m3
Liquids
Liter
L
10−3m 3
Work
Joule
J
N. m
Volume
*Supplementary unit (1 revolution = 2π rad = 360ο).
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U.S. Customary Units
Most practicing American engineers still commonly use a system in
which the base units are the units of length, force, and time. These units
are, respectively, the foot (ft), the pound (lb), and the second (s). The
second is the same as the corresponding SI unit. The foot is defined as
0.3048 m. The pound is defined as the weight of a platinum standard,
called the standard pound, which is kept at the National Institute of
Standards and Technology outside Washington, the mass of which is
0.453 592 43 kg. Since the weight of a body depends upon the earth’s
gravitational attraction, which varies with location, it is specified that the
standard pound should be placed at sea level and at latitude of 45° to
properly define a force of 1 lb. Clearly the U.S. customary units do not
form an absolute system of units. Because of their dependence upon the
gravitational attraction of the earth, they form a gravitational system of
units.
While the standard pound also serves as the unit of mass in
commercial transactions in the United Sates, it cannot be so used in
engineering computations, since such a unit would not be consistent with
the base units defined in the preceding paragraph. Indeed, when acted
upon by a force of 1 lb, that is, when subjected to the force of gravity, the
standard pound receives the acceleration of gravity, g = 32.2 ft/s2 (Fig.
1.4), not the unit acceleration required by Eq. (1.1). The unit of mass
consistent with the foot, the pound, and the second is the mass which
receives an acceleration of 1 ft/s2 when a force of 1 lb is applied to it (Fig.
1.5). This unit, sometimes called a slug, can be derived from the equation
F = ma after substituting 1 lb and 1 ft/s2 for F and a, respectively. We
write
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F=ma
and obtain
1 lb= (1slug) (1ft/s2)
l slug = l lb / l (ft/s2) = 1 lb.s2/ft …….. (1.6)
Comparing Figs. 1.4 and 1.5, we conclude that the slug is a mass
32.2 times larger than the mass of the standard pound.
The fact that in the U.S. customary system of units bodies are
characterized by their weight in pounds rather than by their mass in slugs
will be a convenience in the study of statics, where one constantly deals
with weights and other forces and only seldom with masses. However, in
the study of dynamics, where forces, masses, and accelerations are
involved, the mass m of a body will be expressed in slugs when its weight
W is given in pounds. Recalling Eq. (1.4), we write
m= w/ g ………….. (1.7)
Where g is the acceleration of gravity (g = 32.2 ft/s2).
15
Other U.S. customary units frequently encountered in engineering
problems are the mile (mi), equal to 5280 ft; the inch (in.), equal to (1/12)
ft; and the kilo pound (kip), equal to a force of 1000 lb. The ton is often
used to represent a mass of 2000 lb but, like the pound, must be converted
into slugs in engineering computations.
The conversion into feet, pounds, and seconds of quantities
expressed in other U.S. customary units is generally more involved and
requires greater attention than the corresponding operation in SI units. If,
for example, the magnitude of a velocity is given as v = 30 mi/h, we
convert it to ft/s as follows. First we write
v = 30 mi/h
Since we want to get rid of the unit miles and introduce instead the
unit feet, we should multiply the right-hand member of the equation by an
expression containing miles in the denominator and feet in the numerator.
But, since we do not want to change the value of the right-hand member,
the expression used should have a value equal to unity. The quotient
(5280 ft)/ (1 mi) is such an expression. Operating in a similar way to
transform the unit hour into seconds, we write
v = (30 mi/h) (5280ft/1mi) (1 h/3600s)
Carrying out the numerical computations and canceling out units
which appear in both the numerator and the denominator, we obtain
v = 44 ft /s
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1.4. CONVERSION FROM ONE SYSTEM OF UNITS TO ANOTHER
There are many instances when an engineer wishes to convert into
SI units a numerical result obtained in U.S. customary units or vice versa.
Because the unit of time is the same in both systems, only two kinetic
base units need be converted. Thus, since all other kinetic units can be
derived from these base units, only two conversion factors need be
remembered.
Units of Length
By definition the U.S. customary unit of length is
1 ft =0.3048 m ……. (1.8)
It follows that
1 mi = 5280 ft = 5280(0.3048m) = 1609 m
or
1 mi = 1.609 km ………… (1.9)
also
1 in. = 1/12 ft = 1/12 (0.3048 m) = 0.0254 m
Or
1 in. = 25.4 mm …………. (1.10)
Units of Force
Recalling that the U.S. customary unit of force (pound) is defined
as the weight of the standard pound (of mass 0.4536 kg) at sea level and
at a latitude of 45° (where g = 9.807 m /s2) and using Eq. (1.4), we write
W = mg
1 lb = (0.4536 kg)(9.807 m/s2) = 4.448 kg• m/s2
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or, recalling Eq. (1.5),
l Ib = 4.448N …………… (1.11)
Units of Mass
The U.S. customary unit of mass (slug) is a derived unit. Thus,
using Eqs. (1.6), (1.8), and (1.11), we write
l slug =1lb.s2/ft =1 Ib/ 1 ft/s2 = 4.448N/ 0.3048 m/s2
and, recalling Eq. (1.5),
1 slug = 1 Ib. s2/ft = 14.59 kg ………….. (1.12)
Although it cannot be used as a consistent unit of mass, we recall
that the mass of the standard pound is, by definition,
1 pound mass = 0.4536 kg ……………… (1.13)
This constant may be used to determine the mass in SI units
(kilograms) of a body which has been characterized by its weight in U.S.
customary units (pounds).
To convert a derived U.S. customary unit into SI units, one simply
multiplies or divides by the appropriate conversion factors. For example,
to convert the moment of a force which was found to be M = 47 lb. in.
into SI units, we use formulas (1.10) and (1.11) and write
M = 47 lb. in. = 47(4.448 N)(25.4 mm)
= 5310 N. mm =5.31 N. m
18
The conversion factors given in this section may also be used to
convert a numerical result obtained in SI units into U.S. customary units.
For example, if the moment of a force was found to be M = 40 N. m, we
write, following the procedure used in the last paragraph of Sec. 1.3,
M= 40N.m = (40N.m)( 1 Ib/4.448 N) ( 1 ft /0.3048 m )
Carrying out the numerical computations and canceling out units
which appear in both the numerator and the denominator, we obtain
M = 29.5 lb.ft
The U.S. customary units most frequently used in mechanics are
listed in Table 1.3 with their SI equivalents.
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20
Materials Engineering Department
Class: First
Date : 2009-2010
Subject: Engineering Mechanics
Lecturer: Dr. Emad AL-Hassani
Lecture # 2
Introduction
• The objective for the current chapter is to investigate the
effects of forces on particles:
replacing multiple forces acting on a particle with a single
equivalent or resultant force,
-
Relations between forces acting on a particle that is in a state of
equilibrium
• The focus on particles does not imply a restriction to miniscule
bodies. Rather, the study is restricted to analyses in which the size
and shape of the bodies is not significant so that all forces may be
assumed to be applied at a single point.
Resultant of Two Forces
• .Force: action of one body on another; characterized by its point of
application, magnitude, line of action, and sense
• Experimental evidence shows that the combined effect of two forces
may be represented by a single resultant force.
• The resultant is equivalent to the diagonal of a parallelogram which
contains the two forces in adjacent legs.
• Force is a vector quantity
1
Vectors
• Vector: parameter possessing magnitude and direction which add
according to the parallelogram law. Examples: displacements,
velocities, accelerations.
• Scalar:
parameter possessing magnitude but not direction.
Examples: mass, volume, temperature
• Vector classifications:
- Fixed or bound vectors have well defined points of
application that cannot be changed without affecting
an analysis.
- Free vectors may be freely moved in space without changing
their effect on an analysis.
- Sliding vectors may be applied anywhere along their line of
action without affecting an analysis.
• Equal vectors have the same magnitude and direction.
•
Negative vector of a given vector has the same magnitude and
the opposite direction
2
Addition of Vectors
• Trapezoid rule for vector addition
•
Triangle rule for vector addition
2
2
2
=
+
− 2 PQ cos B
R
P
Q
• Law of cosines, r
r r
R = P+Q
C
B
C
B
• Law of sines,
sin A sin B sin C
=
=
Q
R
A
r r r r
P+Q =Q+ P
• Vector addition is commutative
• Vector subtraction
• Addition of three or more vectors through repeated application of
the triangle rule
3
• The polygon rule for the addition of three or more vectors.
• Vector addition is associative,
r r r
r r r r r r
P + Q + S = (P + Q ) + S = P + (Q + S )
• Multiplication of a vector by a scalar
Resultant of Several Concurrent Forces
• Concurrent forces: set of forces which all pass through
the same point.
A set of concurrent forces applied to a particle may be
replaced by a single resultant force which is the vector
sum of the applied forces.
• Vector force components: two or more force vectors
which, together, have the same effect as a single force
vector.
4
Sample Problem
The two forces act on a bolt at A. Determine their resultant.
SOLUTION:
• Graphical solution - construct a parallelogram with sides in the
same direction as P and Q and lengths in proportion. Graphically
evaluate the resultant which is equivalent in direction and
proportional in magnitude to the diagonal.
•
Trigonometric solution - use the triangle rule for vector addition in
conjunction with the law of cosines and law of sines to find the
resultant.
• Graphical solution - A parallelogram with sides equal to P and Q is
drawn to scale. The magnitude and direction of the resultant or of
the diagonal to the parallelogram are measured,
R = 98 N = 35°
5
• Graphical solution - A triangle is drawn with P and Q head-to-tail
and to scale. The magnitude and direction of the resultant or of the
third side of the triangle are measured,
R = 98 N α = 35°
• Trigonometric solution - Apply the triangle rule.
From the Law of Cosines,
R 2 = P 2 + Q 2 − 2 PQ cos B
= (40 N ) + (60 N ) − 2(40 N )(60 N ) cos155°
2
2
R = 97.73N
From the Law of Sines,
sin A sin B
=
Q
R
sin A = sin B
Q
R
= sin 155°
60 N
97.73N
A = 15.04°
α = 20° + A
α = 35.04°
6
Rectangular Components of a Force: Unit Vectors
• May resolve a force vector into perpendicular components
r
r so
that the resulting parallelogram is a rectangle. Fx and Fy are
referred to as rectangular vector components and
r r r
F = Fx + Fy
• Define perpendicular unit vectors i and j which are parallel to the
x and y axes.
Vector components may be expressed as products of the unit vectors with
the scalar magnitudes of the vector components.
r
Fx and Fy are referred to as the scalar components of F
7
Addition of Forces by Summing Components
• Wish to find the resultant of 3 or more concurrent forces,
r r r r
R = P+Q+ S
Resolve each force into rectangular components
r
r
r
r
r
r
r
r
Rx i + R y j = Px i + Py j + Qx i + Q y j + S x i + S y j
r
r
= (Px + Qx + S x )i + (Py + Qy + S y ) j
The scalar components of the resultant are equal to the sum of the
corresponding scalar components of the given forces and to find the
resultant magnitude and direction,
Rx = Px + Qx + S x
= ∑ Fx
R = Rx2 + R y2
R y = Py + Q y + S y
= ∑ Fy
θ = tan −1
Ry
Rx
8
Sample Problem
Four forces act on bolt A as shown. Determine the resultant of the force
on the bolt.
SOLUTION:
• Resolve each force into rectangular
components.
• Determine the components of the resultant by adding the
corresponding force components.
• Calculate the magnitude and direction of the resultant.
SOLUTION:
• Resolve each force into rectangular components
forc mag
r
F1 150
r
F2
80
r
F3 110
r
F4 100
x − comp
+ 129.9
y − comp
+ 75.0
− 27.4
+ 75.2
0
− 110.0
+ 96.6
− 25.9
Determine the components of the resultant by adding the corresponding
force components.
Calculate the magnitude and direction.
tan α =
Ry
Rx
=
14.3 N
α = 4.1°
199.1 N
9
α = 4.1°
R=
14.3 N
= 199.6 N
sin
Equilibrium of a Particle
• When the resultant of all forces acting on a particle is zero,
the particle is in equilibrium.
• Newton’s First Law: If the resultant force on a particle is zero, the
particle will remain at rest or will continue at constant speed in a
straight line.
• Particle acted upon by two forces:
- equal magnitude
- same line of action
- opposite sense
• Particle acted upon by three or more forces:
- graphical solution yields a closed polygon
- algebraic solution
r
r
R = ∑F = 0
∑ Fx = 0
∑ Fy = 0
10
Free-Body Diagrams
Space Diagram: A sketch showing the
physical conditions of the
problem.
Free-Body Diagram: A
sketch showing only the
forces on the selected
particle
Rectangular Components in Space
The vector F is
contained in the
plane OBAC.
• Resolve
into F
horizontal
and vertical
components.
Fy = F cos θ y
Fh = F sin θ y
11
Resolve Fh into
rectangular
components
Fx = Fh cos φ
= F sin θ y cos φ
Fy = Fh sin φ
= F sin θ y sin φ
With the angles between F and the axes,
r
F
Fx = F cosθ x Fy = F cosθ y Fz = F cosθ z
r
r
r
r
F = Fx i + Fy j + Fz k
r
r
r
= F cosθ x i + cosθ y j + cosθ z k
r
= Fλ
r
r
r
r
λ = cosθ x i + cosθ y j + cosθ z k
(
)
λ is a unit vector along the line of action of F
and cosθx, cosθy, and cosθz are the direction cosines
for F
Direction of the force is defined by the location of two points
M (x1 , y1 , z1 ) and N ( x2 , y2 , z 2 )
r
d = vector joining M and N
r
r
r
= d xi + d y j + d z k
d x = x2 − x1 d y = y2 − y1 d z = z 2 − z1
r
r
F = Fλ
r
r 1 r
r
λ = d xi + d y j + d z k
d
Fd y
Fd x
Fd z
Fx =
Fy =
Fz =
d
d
d
(
)
12
Materials Engineering Department
Class: First
Date : 2009-2010
Subject: Engineering Mechanics
Lecturer: Dr. Emad AL-Hassani
Lecture # 3
Rigid Bodies: Equivalent Systems of Forces
Introduction
• Treatment of a body as a single particle is not always possible. In
general, the size of the body and the specific points of application of
the forces must be considered.
• Most bodies in elementary mechanics are assumed to be rigid, i.e.,
the actual deformations are small and do not affect the conditions of
equilibrium or motion of the body.
• Current chapter describes the effect of forces exerted on a rigid body
and how to replace a given system of forces with a simpler equivalent
system.
- Moment of a force about a point
- Moment of a force about an axis
- Moment due to a couple
• Any system of forces acting on a rigid body can be replaced by an
equivalent system consisting of one force acting at a given point and
one couple.
External and Internal Forces
• Forces acting on rigid bodies are divided into two groups:
- External forces
Internal forces
•
External forces are shown in a free-body diagram.
1
• If unopposed, each external force can impart a motion of translation or
rotation, or both.
Principle of Transmissibility: Equivalent Forces
• Principle of Transmissibility Conditions of equilibrium or motion are not affected by transmitting
a force along its line of action. NOTE: F and F’ are equivalent
forces.
• Moving the point of application of the force F to the rear bumper does
not affect the motion or the other forces acting on the truck.
• Principle of transmissibility may not always apply in determining internal
forces and deformations.
2
Vector Product of Two Vectors
• Concept of the moment of a force about a point is more easily understood
through applications of the vector product or cross product.
• Vector product of two vectors P and Q is defined as the vector V which
satisfies the following conditions:
1. Line of action of V is perpendicular to plane containing P and Q.
2. Magnitude of V is V = PQ sin θ
3. Direction of V is obtained from the righthand rule
• Vector products:
- are not commutative, Q × P = −(P × Q )
- are distributive, P × (Q1 + Q 2 ) = P × Q1 + P × Q 2
-are not associative, (P × Q )× S ≠ P × (Q × S )
Vector Products: Rectangular Components
• Vector products of Cartesian unit vectors,
r r
i ×i = 0
r r r
i× j =k
r r
r
i ×k = − j
r r r r
r r
j × i = −k k × i = j
r r
r r
r
j× j =0
k × j = −i
r r
v r r
j ×k = i
k ×k = 0
3
• Vector products in terms of rectangular coordinates
(
)(
)
r r r r r r r
V=Pxi +Py j +Pzk ×Qxi +Qy j +Qzk
(P
=
y Q
(P
+
z
x
r
i
r
j
r
k
= Px
Qx
Py
Qy
Pz
Qz
− PzQ
Q
y
r
y
)i
− PyQ
+
x
(P z Q
r
)k
x
− PxQ
r
z
)j
Moment of a Force About a Point
• A force vector is defined by its magnitude and direction. Its effect on the
rigid body also depends on it point of application.
• The moment of F about O is defined as M O = r × F
• The moment vector MO is perpendicular to the plane containing O and
the force F.
• Magnitude of MO measures the tendency of the force to cause rotation of
the body about an axis along MO. M O = rF sin θ = Fd
• The sense of the moment may be determined by the right-hand rule.
4
• Any force F’ that has the same magnitude and direction as F, is
equivalent if it also has the same line of action and therefore, produces
the same moment.
• Two-dimensional structures have length and breadth but negligible
depth and are subjected to forces contained in the plane of the structure.
• The plane of the structure contains the point O and the force F. MO, the
moment of the force about O is perpendicular to the plane
• If the force tends to rotate the structure clockwise, the sense of the
moment vector is out of the plane of the structure and the magnitude of
the moment is positive.
• If the force tends to rotate the structure
counterclockwise, the sense of the moment vector is
into the plane of the structure and the magnitude of the
moment is negative
Varignon’s Theorem
• The moment about a give point O of the resultant of several concurrent
forces is equal to the sum of the moments of the various moments about
the same point O.
(
)
r r r
r r r r
r × F1 + F2 + L = r × F1 + r × F2 + L
• Varigon’s Theorem makes it possible to replace the direct determination
of the moment of a force F by the moments of two or more component
forces of F.
5
Rectangular Components of the Moment of a Force
• The moment of F about O,
r
r r
MO = r × F,
r
r r
r
r = xi + yj + zk
r
r
r
r
F = Fx i + Fy j + Fz k
r
r
r
r
M O = M xi + M y j + M z k
r
i
r
j
r
k
= x
Fx
y
Fy
z
Fz
r
r
r
= ( yFz − zFy )i + ( zFx − xFz ) j + (xFy − yFx )k
• The moment of F about B,
r
r
r
M B = rA / B × F
r
r r
rA / B = rA − rB
r
r
r
= ( x A − xB ) i + ( y A − y B ) j + ( z A − z B ) k
r
r
r
r
F = Fx i + Fy j + Fz k
r
i
r
M B = ( x A − xB )
Fx
r
j
( y A − yB )
Fy
r
k
(z A − z B )
Fz
6
• For two-dimensional structures
r
r
M O = (xFy − yFz )k
MO = MZ
= xFy − yFz
r
r
M O = ( x A − x B )F y − ( y A − y B )Fz k
[
]
MO = MZ
= ( x A − x B )F y − ( y A − y B )Fz
Sample Problem
A 100-lb vertical force is applied to the end of a lever which is attached to a
shaft at O.
Determine:
a) moment about O,
b) horizontal force at A which creates the same moment,
c) smallest force at A which produces the same moment,
d) location for a 240-lb vertical force to produce the same moment,
e) whether any of the forces from b, c, and d is equivalent to the original
force
7
a) Moment about O is equal to the product of the force and the
perpendicular distance between the line of action of the force and O. Since
the force tends to rotate the lever clockwise, the moment vector is into the
plane of the paper
M O = Fd
d = (24 in.) cos 60° = 12 in.
M O = (100 lb )(12 in.)
M O = 1200 lb ⋅ in
b) Horizontal force at A that produces the same moment,
d = (24 in.)sin 60° = 20.8 in.
M O = Fd
1200 lb ⋅ in. = F (20.8 in.)
1200 lb ⋅ in.
F=
20.8 in.
F = 57.7 lb
c) The smallest force A to produce the same moment occurs when the
perpendicular distance is a maximum or when F is perpendicular to OA.
M O = Fd
1200 lb ⋅ in. = F (24 in.)
1200 lb ⋅ in.
F=
24 in.
F = 50 lb
d) To determine the point of application of a 240 lb force to produce the
same moment,
M O = Fd
1200 lb ⋅ in. = (240 lb )d
1200 lb ⋅ in.
= 5 in.
240 lb
OB cos60° = 5 in.
d=
OB = 10 in.
8
e)Although each of the forces in parts b), c), and d) produces the same
moment as the 100 lb force, none are of the same magnitude and sense, or on
the same line of action. None of the forces is equivalent to the 100 lb force.
Sample Problem
The rectangular plate is supported by the brackets at A and B and by a wire
CD. Knowing that the tension in the wire is 200 N, determine the moment
about A of the force exerted by the wire at C.
SOLUTION:
The moment MA of the force F exerted by the wire is obtained by evaluating
the vector product,
r
r
r
r r
rC A = rC − rA = (0.3 m )i + (0.08 m ) j
r
r
r
M A = rC A × F
9
r
r
r
rC D
F = Fλ = (200 N )
rC D
r
r
r
− (0.3 m )i + (0.24 m ) j − (0.32 m )k
= (200 N )
0.5 m
r
r
r
= −(120 N ) i + (96 N ) j − (128 N )k
r
v
r
r
M A = −(7.68 N ⋅ m ) i + (28.8 N ⋅ m ) j + (28.8 N ⋅ m )k
Scalar Product of Two Vectors
• The scalar product or dot product between two vectors P and Q is
defined as
r r
P • Q = PQ cos θ (scalar result )
•
Scalar products:
r r r r
Q•P
-are commutative,r P •r Q =
r
r r
(
)
r r
•
+
=
•
+
• Q2
P
Q
Q
P
Q
P
1
2
1
-are distributive,
(
)
r r r
P
-are not associative • Q • S = undefined
10
• Scalar products with Cartesian unit components
(
)(
r
r
r r
r
r
r
r
P • Q = Px i + Py j + Pz k • Qx i + Q y j + Qz k
r r
r r
r r
r r
i •i =1 j • j =1 k •k =1 i • j = 0
r r
P • Q = Px Qx + Py Q y + Pz Qz
r r
P • P = Px2 + Py2 + Pz2 = P 2
11
)
r r
v r
j •k = 0 k •i = 0
Materials Engineering Department
Class: First
Date : 2009-2010
Subject: Engineering Mechanics
Lecturer: Dr. Emad AL-Hassani
Lecture # 4
Equilibrium of Rigid Bodies
Introduction
• For a rigid body in static equilibrium, the external forces and moments
are balanced and will impart no translational or rotational motion to
the body.
• The necessary and sufficient condition for the static equilibrium of a
body are that the resultant force and couple from all external forces
form a system equivalent to zero,
r
F
∑ =0
(
)
r
r r
M
=
r
∑ O ∑ ×F =0
• Resolving each force and moment into its rectangular components
leads to 6 scalar equations which also express the conditions for static
equilibrium,
∑ Fx = 0 ∑ Fy = 0 ∑ Fz = 0
∑Mx = 0 ∑M y = 0 ∑Mz = 0
Free-Body Diagram
• First step in the static equilibrium analysis of a rigid body is
identification of all forces acting on the body with a free-body
diagram.
• Select the extent of the free-body and detach it from the ground and all
other bodies
1
• Indicate point of application, magnitude, and direction of external
forces, including the rigid body weight
• Indicate point of application and assumed direction of unknown
applied forces. These usually consist of reactions through which the
ground and other bodies oppose the possible motion of the rigid body
• Include the dimensions necessary to compute the moments of the
forces
Reactions at Supports and Connections for a Two-Dimensional
Structure
• Reactions equivalent to a force with known line of action.
2
•
Reactions equivalent to a force of unknown direction and magnitude.
• Reactions equivalent to a force of unknown direction and magnitude
and a couple.of unknown magnitude
Equilibrium of a Rigid Body in Two Dimensions
• For all forces and moments acting on a two-dimensional structure
Fz = 0 M x = M y = 0 M z = M O
3
• Equations of equilibrium become
∑F
x
=0
∑F
y
=0
∑M
A
=0
where A is any point in the plane of the structure
• The 3 equations can be solved for no more than 3 unknowns
• The 3 equations can not be augmented with additional equations, but
they can be replaced
∑F
x
=0
∑M
A
∑M
=0
B
=0
Statically Indeterminate Reactions
More unknowns than
equations
Fewer unknowns than
equations, partially
constrained
4
Equal number
unknowns and
equations but
improperly
constrained
Sample Problem
A fixed crane has a mass of 1000 kg and is used to lift a 2400 kg crate. It
is held in place by a pin at A and a rocker at B. The center of gravity of
the crane is located at G.
Determine the components of the reactions at A and B.
SOLUTION:
• Create a free-body diagram for the crane
• Determine B by solving the equation for the sum of the moments of
all forces about A. Note there will be no contribution from the
unknown reactions at A.
• Determine the reactions at A by solving the equations for the sum
of all horizontal force components and all vertical force
components
• Check the values obtained for the reactions by verifying that the
sum of the moments about B of all forces is zero.
• Create the free-body diagram
•
Determine B by solving the equation for the sum of the moments of
all forces about A.
∑M
A
= 0 : + B(1.5m ) − 9.81 kN(2m )
− 23.5 kN(6m ) = 0
B = +107.1 kN
5
• Determine the reactions at A by solving the equations for the sum
of all horizontal forces and all vertical forces
∑ Fx = 0 : Ax + B = 0
Ax = −107.1 kN
∑F
y
= 0:
A y − 9 . 81 kN − 23 . 5 kN = 0
A y = + 33 .3 kN
• Check the values obtained
Equilibrium of a Rigid Body in Three Dimensions
• Six scalar equations are required to express the conditions for the
equilibrium of a rigid body in the general three dimensional case.
∑ Fx = 0 ∑ Fy = 0 ∑ Fz = 0
∑Mx = 0 ∑My = 0 ∑Mz = 0
• These equations can be solved for no more than 6 unknowns which
generally represent reactions at supports or connections
• The scalar equations are conveniently obtained by applying the vector
forms of the conditions for equilibrium
r
r
r r
∑ F = 0 ∑ M O = ∑ (r × F ) = 0
6
Reactions at Supports and Connections for a Three-Dimensional
Structure
7
Sample Problem
A sign of uniform density weighs 270 lb and is supported by a ball-andsocket joint at A and by two cables.
Determine the tension in each cable and the reaction at A.
SOLUTION:
• Create a free-body diagram for the sign
• Apply the conditions for static equilibrium to develop equations for
the unknown reactions
• Create a free-body diagram for the sign.
Since there are only 5 unknowns, the sign is partially constrain. It is free
to rotate about the x axis. It is, however, in equilibrium for the given
loading.
r r
r
rD − rB
TBD = TBD r r
rD − rB
r
r
r
− 8i + 4 j − 8k
= TBD
12
r
r 2r
2
1
= TBD − 3 i + 3 j − 3 k
r r
r
rC − rE
TEC = TEC r r
rC − rE
r
r
r
− 6i + 3 j + 2 k
= TEC
7
r
r 2r
6
3
= TEC − 7 i + 7 j + 7 k
(
)
(
)
8
•
Apply the conditions for static equilibrium to develop equations for
the unknown reactions.
r
r r r
r
(
)
F
=
A
+
T
+
T
−
270
lb
j =0
∑r
BD
EC
i : Ax − 23 TBD − 76 TEC = 0
r
j : Ay + 13 TBD + 73 TEC − 270 lb = 0
r
k : Az − 23 TBD + 72 TEC = 0
r
r
r
r r
r r
∑ rM A = rB × TBD + rE × TEC + (4 ft )i × (− 270 lb) j = 0
j : 5.333TBD − 1.714 TEC = 0
r
k : 2.667 TBD + 2.571TEC − 1080 lb = 0
Solve the 5 equations for the 5 unknowns,
TBD = 101.3 lb TEC = 315 lb
r
r
r
v
A = (338 lb )i + (101.2 lb ) j − (22.5 lb )k
9
Materials Engineering Department
Class: First
Subject: Engineering Mechanics
Lecturer: Dr. Emad AL-Hassani
Centroids and Centers of Gravity
Introduction:
• The earth exerts a gravitational force on each of the particles forming
a body. These forces can be replace by a single equivalent force equal
to the weight of the body and applied at the center of gravity for the
body
• The centroid of an area is analogous to the center of gravity of a body.
The concept of the first moment of an area is used to locate the
centroid.
• Determination of the area of a surface of revolution and the volume of
a body of revolution are accomplished with the Theorems of PappusGuldinus
Center of Gravity of a 2D Body
• Center of gravity of a plate
∑ M y x W = ∑ x ∆W
= ∫ x dW
∑ M y yW = ∑ y∆W
= ∫ y dW
• Center of gravity of a wire
1
Centroids and First Moments of Areas and Lines
• Centroid of an area
x W = ∫ x dW
x (γAt ) = ∫ x (γt )dA
x A = ∫ x dA = Q y
= first moment with respect to y
yA = ∫ y dA = Qx
•
= first moment with respect to x
• Centroid of a line
x W = ∫ x dW
x (γ La ) = ∫ x (γ a )dL
x L = ∫ x dL
yL = ∫ y dL
First Moments of Areas and Lines
•
An area is symmetric with respect to an axis BB’ if for every point P
there exists a point P’ such that PP’ is perpendicular to BB’ and is
divided into two equal parts by BB’.
2
•
The first moment of an area with respect to a line of symmetry is zero.
• If an area possesses a line of symmetry, its centroid lies on that axis
• If an area possesses two lines of symmetry, its centroid lies at their
intersection
• An area is symmetric with respect to a center O if for every element
dA at (x,y) there exists an area dA’ of equal area at (-x,-y).
• The centroid of the area coincides with the center of symmetry
3
Centroids of Common Shapes of Areas
Centroids of Common Shapes of Lines
4
Composite Plates and Areas
• Composite plates
X ∑W = ∑ x W
Y ∑W = ∑ y W
• Composite area
X ∑ A = ∑ xA
Y ∑ A = ∑ yA
Sample Problem
For the plane area shown, determine the first moments with respect to the
x and y axes and the location of the centroid.
SOLUTION:
• Divide the area into a triangle, rectangle, and semicircle with a
circular cutout.
• Calculate the first moments of each area with respect to the axes
• Find the total area and first moments of the triangle, rectangle, and
semicircle. Subtract the area and first moment of the circular
cutout
5
• Compute the coordinates of the area centroid by dividing the first
moments by the total area.
•
Find the total area and first moments of the triangle, rectangle, and
semicircle. Subtract the area and first moment of the circular
cutout.
Qx = +506.2 ×103 mm3
Q y = +757.7 × 103 mm3
• Compute the coordinates of the area centroid by dividing the first
moments by the total area
∑ x A = + 757.7 ×10 mm
X=
∑ A 13.828 ×10 mm
3
3
3
2
X = 54.8 mm
∑ y A = + 506.2 ×10 mm
Y =
∑ A 13.828 ×10 mm
3
3
Y = 36.6 mm
6
3
2
Materials Engineering Department
Subject: Engineering Mechanics
Class: First
Lecturer: Dr. Emad AL-Hassani
Friction
Introduction
• In preceding chapters, it was assumed that surfaces in contact were
either frictionless (surfaces could move freely with respect to each
other) or rough (tangential forces prevent relative motion between
surfaces).
• Actually, no perfectly frictionless surface exists. For two surfaces in
contact, tangential forces, called friction forces, will develop if one
attempts to move one relative to the other.
• However, the friction forces are limited in magnitude and will not
prevent motion if sufficiently large forces are applied.
• The distinction between frictionless and rough is, therefore, a matter
of degree.
• There are two types of friction: dry or Coulomb friction and fluid
friction. Fluid friction applies to lubricated mechanisms. The present
discussion is limited to dry friction between nonlubricated surfaces.
The Laws of Dry Friction. Coefficients of Friction
• Block of weight W placed on horizontal surface. Forces acting on
block are its weight and reaction of surface N.
• Small horizontal force P applied to block. For block to remain
stationary, in equilibrium, a horizontal component F of the surface
reaction is required. F is a static-friction force.
1
• As P increases, the static-friction force F increases as well until it
reaches a maximum value Fm.
Fm = µ s N
• Further increase in P causes the block to begin to move as F drops to a
smaller kinetic-friction force Fk.
Fk = µ k N
• Maximum static-friction force:
Fm = µ s N
• Kinetic-friction force:
Fk = µ k N
µ k ≅ 0.75µ s
2
•
Four situations can occur when a rigid body is in contact with a
horizontal surface:
No friction,
(Px = 0)
No motion
,(Px < Fm)
• Motion impending,
(Px = Fm)
Motion, (Px > Fm)
3
Angles of Friction
• It is sometimes convenient to replace normal force N and friction force
F by their resultant R:
No friction
No motion •
tan φ s =
Fm µ s N
=
N
N
tan φ s = µ s
• Motion impending
Motion •
tan φ k =
Fk µ k N
=
N
N
tan φ k = µ k
4
• Consider block of weight W resting on board with variable inclination
angle θ.
• No friction
No motion
• Motion impending
Motion
5
Problems Involving Dry Friction
• All applied forces known
• Coefficient of static friction is known
• Determine whether body will remain at rest or slide
• All applied forces known
• Motion is impending
• Determine value of coefficient of static friction.
• Coefficient of static friction is known
• Motion is impending
• Determine magnitude or direction of one of the applied forces
6
Sample Problem
A 450 N force acts as shown on a 1350 N block placed on an inclined
plane. The coefficients of friction between the block and plane are ms =
0.25 and mk = 0.20. Determine whether the block is in equilibrium and
find the value of the friction force.
SOLUTION:
• Determine values of friction force and normal reaction force from
plane required to maintain equilibrium.
• Calculate maximum friction force and compare with friction force
required for equilibrium. If it is greater, block will not slide.
• If maximum friction force is less than friction force required for
equilibrium, block will slide. Calculate kinetic-friction force.
∑F
x
= 0:
450 N − 53 (1350 N ) − F = 0
F = −360 N
∑F
y
= 0:
N − 54 (1350 N ) = 0
N = 1080 N
Fm = µ s N
Fm = 0.25(1080 N ) = 270 N
The block will slide down the plane
Factual = Fk = µ k N
= 0.20(1080 N )
Factual = 216 N
7