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8/10/2019 Statics - Chapter 1-2
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Lecturer: Tran Hoai Nam
StaticsChapter 1 - 2
MINISTRY OF EDUCATION AND TRAINING
DUY TAN UNIVERSITY
INTERNATIONAL SCHOOL
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COURSE OBJECTIVES
1. To introduce the concept of free-body
diagram.2. To show how to solve equilibrium problemsusing the equations of equilibrium.
3. To analyze the forces acting on the membersof trusses and frames.
4. To develop shear force and bending momentdiagram for a simple structure.
5. To develop a method for determining themoment of inertia for an area.
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REFERENCE MATERIALS
I. Engineering Mechanics Statics, Bedford & Fowler University
of Texas at Austin.
II. Vector Mechanics for Engineers
Statics, Beer
Johnston
Mazurek Eisenbeg.
III. Engineering Mechanics - Dao Huy Bich, Pham Huyen.
IV. Mechanic Material - Le Ngoc Hong.
V. Structures Leu Tho Trinh.
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GRADES
Homeworks: (20%)
Exams #1 - #3: (25%)
Final exam: (55%)
Plus/minus grading will be used.
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TENTATIVE COURSE OUTLINE
Overview of the course
Topics covered
Reading chapters
Summary of Course Outline
- Exam #1: 10/29/2013.
- Exam #2: 11/14/2013.
- Exam #3: 12/03/2013.
- Final Exam:.
Mark the exam dates on your calendar
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The architects and engineers are guided by the principles of
statics during each step of the design and construction of a
building. Statics is one of the sciences underlying the art of
structural design.
CHAPTER 1: INTRODUCTION
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CHAPTER 1: INTRODUCTION
Mechanics can be defined as that branch of the
physical sciences concerned with the state of rest
or motion of bodies that are sujected to the actionof forces.
MECHANICS
1.1. Engineering and Mechanics:
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MECHANICS
Statics:Deals with the equilibrium of bodies, that is those that
are either at rest or move with a constant velocity.
Dynamics:
Is concerned with the accelerated motion of bodies.
CHAPTER 1: INTRODUCTION
1.1. Engineering and Mechanics:
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FUNDAMENTAL CONCEPTS
Idealization:
- Particle:
A particle has a mass but size that can be neglected.
- Rigid body:
A rigid body can be considered as a combination of a large
number of particles in which all the particles remain at 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.
CHAPTER 1: INTRODUCTION
1.1. Engineering and Mechanics:
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Newtons Three Laws of Motion:
- First Law:
A particle originally at rest or moving in a straight line with
constant velocity, will remain in this state.
- Second Law:
A particle acted upon by an unbalanced force F experiences an
acceleration a that has the same direction as force and a
magnitude that is directly proportional to the force:
F = m a- Third Law:
The mutual forces of action and
reaction between two particles
are equal, opposite and colliner.
CHAPTER 1: INTRODUCTION
1.1. Engineering and Mechanics:
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Length:
1 inch = 25,4 mm
1 foot = 0,3048 m
1 mile = 5280 feet = 1609,344 m
Force: 1 lb(pound) = 4,448 N
Mass: 1 slug = 14,59 kg
Stress: 1 psi = 6,895 kPa ( 7 kPa)
Angle: 2 radians = 360
0
Units conversion: The U.S Customary Unitand the Internatinonal System of Units:
1.1. Engineering and Mechanics:
CHAPTER 1: INTRODUCTION
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Example:
1. A man is riding a biycle at a speed
of 6 meters per second (m/s). How
fast is he going in kilometers per
hour (km/h).
Solution:
6m/s = 6 m/s . (1km/1000m). (3600s/1h) = 21,6km/h.
1.1. Engineering and Mechanics:
Strategy:
One kilometer is 1000 meter and one hour is 60 minutes x60 second = 3600 seconds. We can use these unit
conversion to determine his speed in km/h.
CHAPTER 1: INTRODUCTION
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2. The pressure exerted at point of the
hull of the deep submersible vehicule is
3.106Pa. Determine the pressure in
pound per square foot.
1.1. Engineering and Mechanics:
Solution:
The pressure is:
3.106 N/m2= 3.106N/m2 x (1lb/4,448N) x (0,3m/1ft)2 = 62,700 lb/ft2
Example:
Strategy:
1 pound = 4,448N and 1 foot =
0,3meters.
CHAPTER 1: INTRODUCTION
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1.2. Newtonian Gravitation:
The gravitational force between two
particles of mass m1 and m2 that are
separated by a distance r:
1 22 , 1.1Gm m
F
r
where G is called the universal gravitational
constant.
The weight of an object of mass m
due to the gravitational attraction of the
earth:
2 , 1.2EGm mW
r
where mE is the mass of the earth and ris the distance
from the center of earth to the object.
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1.2. Newtonian Gravitation:
When an objects weight is the only
force acting on it, the resulting
acceleration is called the acceleration due
to gravity (W = ma):
2
. 1.3EGm
ar
The acceleration due to gravity at
sea level is denoted by g. The
acceleration due to gravity at a
distance r from the center of the
earth in terms of the acceleration due
to gravity at sea level:
2
2 . 1.4
ER
a g r
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1.2. Newtonian Gravitation:
The weight of the an object at a
distance rfrom the center of the earth
is:
2
2 . 1.5E
RW ma mg
r
At sea level r = RE, the weight of an
object is given in terms of its mass by
the simple relation:
, 1.6W mg
where m is mass of object, g is the acceleration due to gravity at
sea level (g = 9,81m/s2 in SI units and g = 32,2 ft/s2 in US
Customary units.
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Example:
1. The C- clamp weights 14 oz at sea level (16 oz (ounces) = 1 lb).
And g = 9,81m/s2. what is the mass of the C-lamp in kg?
Strategy:
- Determining the weight of the C clamp in
Newton.
- Using Eq.(1.6) to determine the mass in kg.
1.2. Newtonian Gravitation:
Solution:
14 oz = 14 oz (1lb/16oz).(4,448N/lb)= 3,892N.
m = W/g = 3,982 (kg.m/s2)
/(9,81m/s2) = 0,397kg.
m = 0,397kg
CHAPTER 1: INTRODUCTION
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RESUTS 1.2
CHAPTER 1: INTRODUCTION
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CHAPTER 2 VECTORS
To review vector operations: vector addition, product of
a scalar and a vector, vector subtraction, dot products andcross products.
To express vectors in terms of components.
To present examples of engineering applications of
vectors.
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CHAPTER 2 VECTORS
2.1. Scalars and Vectors2.1.1. Vector
Scalar Vector
- A quantity characterized bya positive or negativenumber.
A quantity that has both amagnitude and direction.
- Mass, volume - Force, velocity
Notations for Vectors:
- For handwritten work, a vector is generally represented by a
letter with an arrow writter over it, such as- The magnitude is designated
In your textbook vectors are symbolized in boldface type; for
example, A is used to designate the vector A. Its magnitude is
symbolized by |A| or simplyA.
A
A or A
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CHAPTER 2 VECTORS
2.1. Scalars and Vectors
Graphical representation of a vector
The arrow represents the vector graphically and used to define
the vector magnitude, direction.
The magnitude is the length of the arrow and the direction is
defined by the angle between a reference axis and the arrows line
of action.
The sense is indicated by the arrowhead.
2.1.1. Vector
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The direction of a vector includes both its inclination and
its sense. You might say the "sense" of a vector is its sign.
The two vectors shown have the same inclination (vertical)
but opposite senses. (One is up, the other, down.)
CHAPTER 2 VECTORS
2.1. Scalars and Vectors2.1.1. Vector
Graphical representation of a vector
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CHAPTER 2 VECTORS
2.1. Scalars and Vectors
U + V = W (2.1)
a) A displacement by thevector U.
b) The displacement Ufollowed by thedisplacement V.
c) The displacement Uand V are equivalentto the dispalement W.
d) The final position of the book doesntdepend on the orderof the displacement.
2.1.2. Vector Addition:
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a) Two vectors Uand V.b) The head of Uplaced at the tail of V.c) The triangle rule for obtaining the sum of Uand V.
d) The sum is independent of the order in which the vertors are added.e) The parallelogram rule for ontaining the sum of Uand V.
The triangle and parallelogram rule:
2.1. Scalars and Vectors2.1.2. Vector Addition:
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Sum of the three vector U, Vand W
The definition of vector addition implies that:
U + V = V + U (2.2) (vector addtion is communitative).
(U + V) + W = U + (V + W) (2.3) (vector addtion is associative).
Three vector U, Vand Wwhosesum is zero.
2.1. Scalars and Vectors2.1.2. Vector Addition:
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Product of a scalar and a vector:
The product of scalar (real number ) a and a vector U is a vector
written asaU:
- Its magnitude is |a|.|U|.
- The direction ofaU is the same as the direction of U when a is
positive and is opposite to the direction ofU when a is negative.
A vector U and some ofits scalar multiples
2.1. Scalars and Vectors2.1.2. Vector Addition:
Notes:
=> (-1)U is written as U and iscalled negative of the vector U.
=>1
( )U
Ua a
CHAPTER 2 VECTORS
O
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The definition of vector addition and the product of a scalar and a
vector imply that:
a(bU) = (ab)U
(a + b)U = aU + bU
a(U + V) = aU + aV
Product of a scalar and a vector:
2.1. Scalars and Vectors2.1.2. Vector Addition:
Vector subtraction:
The difference of two vectors U and V
is obtained by adding U to the vector
(-1)V:
U V = U + (-1) V
a) Two vectorsU and V.
b) The vector
V and (-1)Vc) The sum of
U and (-1)Vis the vectordifference U V.
CHAPTER 2 VECTORS
C O S
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Unit vectors
2.1. Scalars and Vectors2.1.2. Vector Addition:
A unit vector is simply a vector whose magnitude is 1. Any vector
U can be regraded as the product of its magnitude and a unit
vector that has same direction as U:
Ue
U
Since U and ehave the samedirection thevector U equalsthe product of itsmagnitude with e.
CHAPTER 2 VECTORS
VECTORS
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Vector Addition of Forces:
2.1. Scalars and Vectors2.1.2. Vector Addition:
CHAPTER 2 VECTORS
- A force is vector quantity since it has a specified
magnitude and direction.
- Two common problems in statics involve either finding the
resultant force given its components or resolving a known
force into components.
- The law of cosines is often used to find the magnitude,
while the law of sines is used to find the direction.
VECTORS
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1. Part of the roof of a sports stadium is to suppoted by the cableAB andAC. The forces the cables exert on the pylon to which they are attached arerepresented by the vectors FAB and FAC. the magnitudes of the forces are|FAB|=100kN and |FAC|=60kN. Determine the magnitude and direction ofthe sum of forces exerted on the pylon by cables.
Strategy:- Using the parellelogram rule for adding the two forces andmeasuring the magnitude and direction of their sum.
2.1.3. Example:
2.1. Scalars and Vectors
CHAPTER 2 VECTORS
C VECTORS
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Solution:
Using the law of cosine in the
triangle:
Using the law of cosine in the
triangle:
Its direction to be 190 above the
horizonal.
2 2 0
2 | | . | | cos150 155 .AB AB AB ACF F F F kN
2 2 2
0
60 100 150
cos 0.5082.60.100
19 .
2.1.3. Example:
2.1. Scalars and Vectors
CHAPTER 2 VECTORS
C VECTORS
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RESULTS 2.1
2.1. Scalars and Vectors
CHAPTER 2 VECTORS
CHAPTER 2 VECTORS
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2.2. Components of a vector
(a) Vector U(b) The vector components Uxand Uy(c) The vector components can be expressed in terms of i andj
2.2.1. Components in Two Dimensions
Definition:
U = Ux+ Uy
U = Uxi + Uyj (2.3)
|U| = (2.4)2 2x yU U
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(a) Two pointsA and B and the position vector rAB fromA to B.
(b) The components of rAB can be determined from the
coordinates of pointsA and B.
Position Vectors in Terms of Components
A(xA, yA) and B (xB, yB):
rAB = (xB - xA)i + (yB - yA)j (2.6)
2.2 Components of a vector2.2.1. Components in Two Dimensions
CHAPTER 2 VECTORS
CHAPTER 2 VECTORS
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Solution:
a. W + D + L = 0 => L = -W D = 600j + 200i 100j = 500j + 200i.
b. L + 2D + W = 500j + 200i 400i + 200j 600j = 100j 200i.
Example :
1. The forces acting on the sailplane are
its weight W = -600j (lb), the drag D = -
200i + 100j (lb) and the lift L.
a) If the sum of the forces on the sailplaneis zore, what are the compronents ofL?
b) If the lift L has the component
determined in (a) and the drag
increases by factor of 2, what is the
magnitude of the sum of the forces on
the sailplane?
2.2 Components of a vector2.2.1. Components in Two Dimensions
CHAPTER 2 VECTORS
CHAPTER 2 VECTORS
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2. The cable from point A topoint B exerts a 900N force on
the top of the television
transmission tower that is
represented by the vector F.
Express F in terms of
components using the
coordinate system shown.
Strategy:
Determine the components of the vector F in two ways:
- Determining the angle between F and the y axis and use trigonometry to
determine the components.
- Using the given slope of the cable AB and apply similar triangles to
determine the components ofF.
2.2 Components of a vector2.2.1. Components in Two Dimensions
CHAPTER 2 VECTORS
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2.2.2. Components in Three Dimensions
Definition:
a) A cube viewed with the line of sigt perpendicular to a face
b) An oblique view of the cube.
c) A cartesian coordinate system aligned with the edges of the cube.
d) Three-dimensional representantion of the coordinate system.
2.2 Components of a vector
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Position vector in terms of components:a) The position vector from point A to point B.b) The components of rcan be determined from
the coordinate of points Av B.
A (xA, yA, zA) and B (xB, yB, zB). The components
are obtained by the coodinates of point A from the
coordinate of point B.
rAB = (xBxA)i + (yB yA)j + (zBzA)k (2.12)
2.2.2. Components in Three Dimensions2.2 Components of a vector
CHAPTER 2 VECTORS
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Components of a vector Parallel to a Given Line
a) Two points A and B on a
line parallel to U.
2.2.2. Components in Three Dimensions2.2 Components of a vector
CHAPTER 2 VECTORS
b) The position vector from A to B.
c) The unit vector eAB that points
from A toward B.
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Example:
1. The coordinates of point Cof the truss arex= 4m, y= 0,z= 0
and the coordinates of the point D arex= 2m, y= 3m,z= 1m.
What are the direction cosines of the position vector r from point C
to point D?
Strategy:
- Determining rCD in terms of its
components.
- Calculating the magnitude of r(the distance from Cto D) and use
equation (2.10) to obtain the
direction cosines.
2.2.2. Components in Three Dimensions2.2 Components of a vector
CHAPTER 2 VECTORS
Answer: cosx= - 0.535,cosy= 0.802, cosy= 0.267
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2. The crane exert a 600lb on the caisson. The angle between F andxaxis
is 540, and the angle between F and the yaxis is 400. Thez component of
F is positive. Express F in terms of components.
Strategy:
- Using the Eq(2.17) to determine the third angle.
- Using the Eq (2.16) to determine the components of F.
Example:
2.2.2. Components in Three Dimensions2.2 Components of a vector
CHAPTER 2 VECTORS
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RESULTS 2.2:
CHAPTER 2 VECTORS
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RESULTS 2.2:
CHAPTER 2 VECTORS
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RESULTS 2.2
CHAPTER 2 VECTORS
CHAPTER 2 VECTORS
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RESULTS 2.2
CHAPTER 2 VECTORS
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2.3. Dot Products and Cross Products:2.3.1. Dot Products
The dot product of U and V:
U.V = |U|.|V|cos (2.13)
The dot product has the properties:
U.V = V.U (The dot product is commutative).
a(U.V) = aU.V = U.(aV) (The dot product is
associate with respect scalar multiplication).
U.(V + W) = U.V + U.W (The dot product
is associate with respect to vector addtion).
Definition:
CHAPTER 2 VECTORS
CHAPTER 2 VECTORS
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Dot products in terms of component:
Vectors U and V: U = Ux.i + Uy.j+ Uz.k ; V = Vxi + Vy.j + Vz .k
Vector components paralleland normal to a line:
The parallel component:
|Up| = |U|.cosUp = (eL .U)eL (2.16)
(e is unit vector parallel to L)
The normal component:
Un = U Up (2.17)
. 2.14
.cos 2.15
x x y y z z
x x y y z z
U V U V U V U V
U V U V U V U V
U V U V
2.3. Dot Products and Cross Products:2.3.1. Dot Products
CHAPTER 2 VECTORS
CHAPTER 2 VECTORS
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Example1. The components of two vector U and V are U = 6i 5j 3k and V = 4i+ 2j + 3k.a) What is the value of U.Vb) What is the angle between U and V when they are placed tail to tail?
Strategy:
- Using Eq (2.14) to determine the value of U.V- Using Eq (2.15) to calculate the angle between the vectors
2.3. Dot Products and Cross Products:2.3.1. Dot Products
Solution:
a) b)
CHAPTER 2 VECTORS
CHAPTER 2 VECTORS
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2. What is the angle between the linesAB andAC?
Strategy:- Determine the components of thevector rAB and vector rAC.- Using Eq (2.15) to determine .
Example
2.3. Dot Products and Cross Products:2.3.1. Dot Products
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CHAPTER 2 VECTORS
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Cross product in term components:
2.3. Dot Products and Cross Products:2.3.2. Cross Products
Vectors U and V: U = Ux.i + Uy.j+ Uz.k ; V = Vxi + Vy.j + Vz .k
2.19x y z
x y z
i j k
U V U U U
V V V
Mixed Triple Product:
In chapter 4, when we discuss the moment of a force about a
line, we will use an operation called the mixed triple product:
.( ) 2.20
x y z
x y z
x y z
U U U
U V W V V V
W W W
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Example:
1. The components of two vectors U and V are U = 6i 5j k and V = 4i
+ 2j + 2k.
a) Determine the cross product U x V.
b) Use the dot product to prove that U x V is perpendicular to V.
Strategy:
- Use Eq (2.19) to determine U x V
- Due to prove U x V is perpendicular to U by showing that
(U x V).U = 0 (Use Eq (2.14)) .
2.3. Dot Products and Cross Products:2.3.2. Cross Products
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CHAPTER 2 VECTORS
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RESULTS 2.3
2.3. Dot Products and Cross Products:
CHAPTER 2 VECTORS
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RESULTS 2.3
2.3. Dot Products and Cross Products:
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RESULTS 2.3
2.3. Dot Products and Cross Products:
MINISTRY OF EDUCATION AND TRAINING
DUY TAN UNIVERSITY
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Tran Hoai Nam
T: 0918.230.728
DUY TAN UNIVERSITY
INTERNATIONAL SCHOOL
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