Bài giảng giáo trình Engineering Mechanics Statics 13th edition Hibbeler (Lecture Notes Slides) (Chương 3456)

Hibbeler and Kai Beng Yap © Pearson Education South Asia Pte Ltd 2013.. Hibbeler and Kai Beng Yap © Pearson Education South Asia Pte Ltd 2013.. Hibbeler and Kai Beng Yap © Pearson Edu

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Mechanics for Engineers: Statics, 13th SI Edition

EQUILIBRIUM OF A PARTICLE, THE FREE-BODY DIAGRAM & COPLANAR FORCE SYSTEMS

Analysis of Spring and Pulleys

Concept Quiz

Group Problem Solving

Attention Quiz

Today’s Objectives:

Students will be able to :

a) Draw a free-body diagram (FBD),

and,

b) Apply equations of equilibrium to

solve a 2-D problem

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READING QUIZ

1) When a particle is in equilibrium, the sum of forces acting

on it equals _ (Choose the most appropriate answer) A) A constant B) A positive number C) Zero D) A negative number E) An integer

2) For a frictionless pulley and cable, tensions in the

cable (T1 and T2) are related as _

The crane is lifting a load To decide if the straps holding the load to the crane hook will fail, you need to know the force in the straps How could you find the forces?

APPLICATIONS

Straps

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For a spool of given weight, how would you find the forces in cables

AB and AC? If designing

a spreader bar like this one, you need to know the forces to make sure the rigging doesn’t fail

APPLICATIONS

(continued)

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COPLANAR FORCE SYSTEMS

This is an example of a 2-D or coplanar force system

If the whole assembly is in equilibrium, then particle A is also in equilibrium

THE WHAT, WHY, AND HOW OF A FREE-BODY DIAGRAM (FBD)

Free-body diagrams are one of the most important things for you to know how to draw and use for statics and other subjects!

What? - It is a drawing that shows all external forces

acting on the particle

Why? - It is key to being able to write the equations of

equilibrium—which are used to solve for the unknowns

(usually forces or angles)

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How?

Active forces: They want to move the particle

Reactive forces: They tend to resist the motion

Note : Cylinder mass = 40 Kg

1 Imagine the particle to be isolated or cut free from its

surroundings

A

3 Identify each force and show all known magnitudes and

directions Show all unknown magnitudes and / or

EQUATIONS OF 2-D EQUILIBRIUM

Or, written in a scalar form,

6 Fx = 0 and 6 Fy = 0

These are two scalar equations of equilibrium (E-of-E)

They can be used to solve for up to two unknowns

Since particle A is in equilibrium, the net force at A is zero

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Write the scalar E-of-E:

+ o 6 Fx = FB cos 30º – FD = 0

+ n 6 Fy = FB sin 30º – 392.4 N = 0

Solving the second equation gives: FB = 785 N ĺ

From the first equation, we get: FD = 680 N ĸ

Note : Cylinder mass = 40 Kg

SIMPLE SPRINGS

Spring Force = spring constant * deformation of spring

or F = k * s

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CABLES AND PULLEYS

With a frictionless pulley and cable

T1 = T2

T 1

T 2

EXAMPLE

Plan:

1 Draw a FBD for point A

2 Apply the E-of-E to solve for the forces in ropes AB and AC

Given: The box weighs 550 N and

geometry is as shown

Find: The forces in the ropes AB

and AC

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A) The weight is too heavy

B) The cables are too thin

C) There are more unknowns than equations

D) There are too few cables for a 1000 N

weight

2) Why?

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GROUP PROBLEM SOLVING

1 Draw a FBD for Point D

2 Apply E-of-E at Point D to solve for the unknowns (FCD & FDE)

3 Knowing FCD, repeat this process at point C

Given: The mass of lamp is 20

kg and geometry is as shown

Find: The force in each cable

Plan:

GROUP PROBLEM SOLVING (continued)

Applying the scalar E-of-E at D, we get;

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Applying the scalar E-of-E at C, we get;

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2 Using this FBD of Point C, the sum of

forces in the x-direction (6 FX) is _

Use a sign convention of + o

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THREE-DIMENSIONAL FORCE SYSTEMS

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READING QUIZ

1 Particle P is in equilibrium with five (5) forces acting on it in 3-D space How many scalar equations of equilibrium can be written for point P?

A) 2 B) 3 C) 4

D) 5 E) 6

2 In 3-D, when a particle is in equilibrium, which of the

following equations apply?

A) (6 Fx) i + (6 Fy) j + (6 Fz) k = 0

B) 6 F = 0

C) 6 Fx = 6 Fy = 6 Fz = 0

D) All of the above

E) None of the above

APPLICATIONS

You know the weight of the electromagnet and its load But, you need to know the forces in the chains to see if it is a safe assembly How would you

do this?

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APPLICATIONS

(continued)

This shear leg derrick

is to be designed to lift a maximum of 200

kg of fish

How would you find the effect of different offset distances on the forces in the cable and derrick legs?

Offset distance

THE EQUATIONS OF 3-D EQUILIBRIUM

This vector equation will be satisfied only when

When a particle is in equilibrium, the vector

sum of all the forces acting on it must be

zero (6 F = 0 )

This equation can be written in terms of its

x, y, and z components This form is written

as follows

(6 Fx) i + (6 Fy) j + (6 Fz) k = 0

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EXAMPLE I

1) Draw a FBD of particle O

2) Write the unknown force as

F5 = {Fx i + Fy j + Fz k} N

3) Write F 1, F 2 , F 3 , F 4 , and F 5in Cartesian vector form

4) Apply the three equilibrium equations to solve for the three unknowns Fx, Fy, and Fz

Given: The four forces and

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Equating the respective i, j, k components to zero, we have

6 Fx = 76.8 – 600 + Fx = 0 ; solving gives Fx = 523.2 N

6 Fy = 240 – 102.4 + Fy = 0 ; solving gives Fy = – 137.6 N

6 Fz = 180 – 900 + 153.6 + Fz = 0 ; solving gives Fz = 566.4 N

Thus, F 5 = {523 i – 138 j + 566 k} N

Using this force vector, you can determine the force’s

magnitude and coordinate direction angles as needed

EXAMPLE I (continued)

EXAMPLE II

1) Draw a free-body diagram of Point A Let the unknown

force magnitudes be FB, FC, FD

2) Represent each force in its Cartesian vector form

3) Apply equilibrium equations to solve for the three

unknowns

Given: A 600-N load is

supported by three cords with the geometry as shown

Find: The tension in cords AB,

AC and AD

Plan:

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Solving the three simultaneous equations yields

FC = 646 N (since it is positive, it is as assumed, e.g., in tension)

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CONCEPT QUIZ

1 In 3-D, when you know the direction of a force but not its magnitude, how many unknowns corresponding to that force remain?

A) One B) Two C) Three D) Four

2 If a particle has 3-D forces acting on it and is in static

equilibrium, the components of the resultant force (6 Fx, 6

Fy, and 6 Fz ) _

A) have to sum to zero, e.g., -5 i + 3 j + 2 k

B) have to equal zero, e.g., 0 i + 0 j + 0 k

C) have to be positive, e.g., 5 i + 5 j + 5 k

D) have to be negative, e.g., -5 i - 5 j - 5 k

1) Draw a free-body diagram of Point A Let the unknown force magnitudes be FB, FC, F D

2) Represent each force in the Cartesian vector form

3) Apply equilibrium equations to solve for the three unknowns.

GROUP PROBLEM SOLVING Given: A 17500-N (≈ 1750-kg)

motor and plate, as

shown, are in equilibrium

and supported by three

cables and

d = 1.2 m

Find: Magnitude of the tension

in each of the cables

Plan:

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W = load or weight of unit = 17500 k N

The particle A is in equilibrium, hence

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ATTENTION QUIZ

2 In 3-D, when you don’t know the direction or the

magnitude of a force, how many unknowns do you have corresponding to that force?

A) One B) Two C) Three D) Four

1 Four forces act at point A and

point A is in equilibrium Select the

correct force vector P

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MOMENT OF A FORCE (SCALAR FORMULATION), CROSS PRODUCT, MOMENT OF A FORCE (VECTOR FORMULATION), & PRINCIPLE OF MOMENTS

• Attention Quiz

Today’s Objectives :

Students will be able to:

a) understand and define moment, and

b) determine moments of a force in 2-D

and 3-D cases

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APPLICATIONS

Beams are often used to bridge gaps in walls

We have to know what the effect of the force

on the beam will have on the supports of the

beam

What do you think is happening at points A and B?

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APPLICATIONS (continued)

Carpenters often use a hammer in this way to pull a

stubborn nail Through what sort of action does the force

FH at the handle pull the nail? How can you mathematically model the effect of force FH at point O?

MOMENT OF A FORCE – SCALAR FORMULATION

(Section 4.1)

The moment of a force about a point provides a measure of the tendency for rotation (sometimes called a torque)

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MOMENT OF A FORCE - SCALAR FORMULATION

(continued)

As shown, d is the perpendicular distance from point O to the line of action of the force

In 2-D, the direction of MO is either clockwise (CW) or

counter-clockwise (CCW), depending on the tendency for

rotation

In a 2-D case, the magnitude of the moment is Mo = F d

MOMENT OF A FORCE - SCALAR FORMULATION

(continued)

Often it is easier to determine

MO by using the components

of F as shown

Then MO = (FY a) – (FX b) Note the different signs on the

terms! The typical sign convention for a moment in 2-D is that counter-clockwise is considered positive We can determine the direction of rotation by imagining the body pinned at O and deciding which way the body would rotate because of the

force

For example, MO = F d and the direction is counter-clockwise

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VECTOR CROSS PRODUCT (Section 4.2)

While finding the moment of a force in 2-D is straightforward when you know the perpendicular distance d, finding the perpendicular distances can be hard—especially when you are working with forces in three dimensions

So a more general approach to finding the moment of a

force exists This more general approach is usually used when dealing with three dimensional forces but can be used

in the two dimensional case as well

This more general method of finding the moment of a force uses a vector operation called the cross product of two

vectors

CROSS PRODUCT (Section 4.2)

In general, the cross product of two vectors A and B results

in another vector, C , i.e., C = A uu B The magnitude and direction of the resulting vector can be written as

C = A u B = A B sin T u C

As shown, u C is the unit vector perpendicular to both A and

B vectors (or to the plane containing the A and B vectors)

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CROSS PRODUCT (continued)

The right-hand rule is a useful tool for determining the

direction of the vector resulting from a cross product

For example: i uu j = k

Note that a vector crossed into itself is zero, e.g., i u i = 0

CROSS PRODUCT (continued)

Also, the cross product can be written as a determinant

Each component can be determined using 2 u 2 determinants

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MOMENT OF A FORCE – VECTOR FORMULATION (Section 4.3)

Moments in 3-D can be calculated using scalar (2-D) approach, but it can be difficult and time consuming Thus, it is often

easier to use a mathematical approach called the vector cross product

Using the vector cross product, M O = r uu F

Here r is the position vector from point O to any point on the line of action of F

MOMENT OF A FORCE – VECTOR FORMULATION

(continued)

So, using the cross product, a

moment can be expressed as

By expanding the above equation using 2 u 2 determinants (see Section 4.2), we get (sample units are N - m or lb - ft)

M O = (ry FZ - rZ Fy) (rx Fz - rz Fx ) j + (rx Fy - ry Fx ) k

The physical meaning of the above equation becomes evident

by considering the force components separately and using a

2-D formulation

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1) Resolve the 100 N force along x and y-axes

2) Determine MO using a scalar analysis for the

two force components and then add those two

moments together

EXAMPLE I

Given: A 100 N force is

applied to the frame

Find: The moment of the

force at point O

Plan:

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EXAMPLE II

1) Find F = F 1 + F 2 and r OA.2) Determine M O = r OA uu F

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CONCEPT QUIZ

1 If a force of magnitude F can be applied in four different 2-D configurations (P,Q,R, & S), select the cases resulting in the maximum and minimum torque values on the nut (Max, Min) A) (Q, P) B) (R, S)

GROUP PROBLEM SOLVING I

Since this is a 2-D problem: 1) Resolve the 100 N force along the handle’s x and y axes

2) Determine MA using a scalar analysis

Given: A 100 N force is

applied to the hammer

Find: The moment of the

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GROUP PROBLEM SOLVING I (continued)

Solution:

+ n Fy = 100 sin 30° N + o Fx = 100 cos 30° N

x y

+ MA = {–(100 cos 30°)N (450 mm) – (100 sin 20°)N (125 mm)}

= – 43.2464 N·mm = 43.2 N·m (clockwise or CW)

GROUP PROBLEM SOLVING II

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GROUP PROBLEM SOLVING II (continued)

Find the moment by using the cross product

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MOMENT ABOUT AN AXIS

Students will be able to determine the moment of a force

about an axis using

a) scalar analysis, and

b) vector analysis

2 The triple scalar product u • ( r uu F ) results in

A) a scalar quantity ( + or - ) B) a vector quantity

C) zero D) a unit vector

E) an imaginary number

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APPLICATIONS

With the force P, a person is creating a moment M A using this flex-handle socket wrench Does all of M A act to turn the socket? How would you calculate an answer to this question?

APPLICATIONS

Sleeve A of this bracket can provide a maximum resisting moment of 125 N·m about the x-axis How would you determine the maximum magnitude of F before turning about the x-axis occurs?

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SCALAR ANALYSIS

Recall that the moment of a scalar force about any point O

is MO= F dO where dO is the perpendicular (or shortest) distance from the point to the force’s line of action This concept can be extended to find the moment of a force about an axis

Finding the moment of a force about an axis can help

answer the types of questions we just considered

SCALAR ANALYSIS

In the figure above, the moment about the y-axis would

be My= Fz (dx) = F (r cos θ) However, unless the force can easily be broken into components and the “dx” found quickly, such calculations are not always trivial and vector analysis may be much easier (and less likely to produce errors)

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VECTOR ANALYSIS

First compute the moment of F

about any arbitrary point O that lies on the a’- a axis using the cross product

M O = r uu F

Now, find the component of M O along the a-axis using the dot product

Ma’-a= u a • M O

Our goal is to find the moment of

F (the tendency to rotate the body) about the a-axis

VECTOR ANALYSIS (continued)

In the this equation,

u a represents the unit vector along the a-axis,

r is the position vector from any point on the a-axis to

any point A on the line of action of the force, and

F is the force vector

M a can also be obtained as

The above equation is also called the triple scalar product

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EXAMPLE

1) Use Mz = u • (r uu F)

2) First, find F in Cartesian vector form

3) Note that u = 1 i in this case

4) The vector r is the position vector from O to A

A

B

Given: A force is applied to

the tool as shown

Find: The magnitude of

the moment of this force about the x axis

of the value

Plan:

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CONCEPT QUIZ (continued)

2 The force F is acting

along DC Using the

triple scalar product to

determine the moment

of F about the bar BA,

you could use any of

the following position

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GROUP PROBLEM SOLVING

1) Use My = u • (r OA uu F)

2) Find u AB fromr AB

3) Find F in Cartesian vector form using u AB

Given: The hood of the

automobile is supported

by the strut AB, which

exerts a force F = 120 N

Find: The moment of F about

the hinged axis y

Plan:

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