Statics - Chapter 1-2

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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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.




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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.




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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.




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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:


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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.


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.



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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.


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.



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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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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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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.


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



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RESUTS 1.2



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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


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:

CHAPTER 2 VECTORS


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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.


CHAPTER 2 VECTORS


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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


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:


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


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:


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:


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:


CHAPTER 2 VECTORS

C VECTORS


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RESULTS 2.1


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

CHAPTER 2 VECTORS


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CHAPTER 2 VECTORS


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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?


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.


CHAPTER 2 VECTORS

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

CHAPTER 2 VECTORS


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CHAPTER 2 VECTORS


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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

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.


CHAPTER 2 VECTORS

b) The position vector from A to B.

c) The unit vector eAB that points

from A toward B.

CHAPTER 2 VECTORS


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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.


CHAPTER 2 VECTORS

Answer: cosx= - 0.535,cosy= 0.802, cosy= 0.267

CHAPTER 2 VECTORS


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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:


CHAPTER 2 VECTORS

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

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

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


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


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


CHAPTER 2 VECTORS


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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

CHAPTER 2 VECTORS


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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


CHAPTER 2 VECTORS


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RESULTS 2.3


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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Statics - Chapter 1-2