Estática dos Sólidos. Prof. Ettore Baldini-Neto

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Estática dos Sólidos

Prof. Ettore Baldini-Neto

baldini@uninove.br

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Plano de Ensino

• Equilíbrio de uma partícula

• Diagrama de Corpo Livre, Sistemas de Forças 2D e 3D

• Resultante de um sistema de forças (2D e 3D)

• Momento: formulações escalar e vetorial

• Princípio dos momentos

• Momento em relação a um eixo, momento binário

• Propriedades Geométricas de Figuras Planas

• Carregamentos

• Equilíbrio de um corpo rígido

• Diagrama de Corpo Livre, equações de equilíbrio

• Análise Estrutural

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Bibliografia

• Hibbeler, R. C. Estática 10

a

_ed/12

a

_{ed, Pearson SP (2005/2011)}

• Schmidt, R. J.; Boresi, A. P., Estática: Thomson, (2003)

• Bedford & Fowler; Engineering mechanics-Statics 3, NJ, Prentice Hall (2002).

• Beer, F. P., Johnston Jr, E. R., Mecânica vetorial para engenheiros: estátcia 5ed. SP

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Avaliação e Faltas

• AV1 Professor

• AV2 integrada

• AV3 projeto integrador

• O curso é presencial! O limite é de 20 faltas! Faltas Reprovam!

• O aluno que perder uma das AV’s tem direito a uma prova

substitutiva

somente mediante a entrega de atestado médico na

secretaria da universidade.

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Mecânica

• É o ramo da física que estuda os movimentos e também o repouso

dos sistemas físicos (corpos, partículas) que estão sujeitos à ação

de forças.

• Mecânica dos corpos rígidos

• Estática (corpos em repouso ou em movimento com

velocidade constante)

• Dinâmica

• Mecânica dos corpos deformáveis

• Mecânica dos Fluidos

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Grandezas Físicas Fundamentais e o Sistema

Internacional de Unidades

• Comprimento

(m)

com

Dimensão [L]

• Tempo

(s)

com

Dimensão [T]

• Massa

(kg)

com Dimensão [M]

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Definições

• Uma

partícula possui massa mas tem suas dimensões desprezadas

• Um

corpo rígido pode ser aproximado como uma combinação de um

grande número de partículas que permanece fixas, umas em relação às

outras tanto antes quanto depois da aplicação de forças. Desprezamos as

deformações que ocorrem em decorrência destas forças.

• Força Concentrada representa o efeito de uma carga que age em um

ponto de um corpo desde que a área de contato sobre a qual ela é

aplicada seja pequena comparada com o tamanho total do corpo. Exemplo:

Força de contato entre uma roda e o solo

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Leis de Newton

• Uma partícula originalmente em repouso ou movendo-se em linha reta

com velocidade constante tende a permanecer neste estado desde que

não seja submetida a uma força externa resultante desbalanceada, ou

seja, diferente de zero.

• Sob a ação de uma força resultante externa não nula, uma partícula sofre

uma aceleração que possui a mesma direção da força aplicada e

intensidade diretamente proporcional à força e inversamente

proporcional `a sua massa.

• As forças mútuas de ação e reação entre duas partículas são iguais,

opostas e colineares.

~

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Popriedades Gerais dos Vetores

• Na natureza, a grosso modo, existem grandezas físicas que podem ser:

• Grandezas Escalares

• São aquelas que são caracterizadas somente por um valor numérico.

• Exemplo: Energia, Trabalho, Temperatura, Corrente Elétrica, etc.

• Grandezas Vetoriais

• São aquelas que para serem caracterizadas necessitam de, além de um

valor numérico, de uma orientação (direção e sentido)!

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Soma e subtração geométricas com vetores

Quando existem mais de dois vetores podemos agrupá-los em qualquer ordem para

efetuar a soma.

~

A

~

B

~

A + ~

B

~

B + ~

A

( ~

A + ~

B) + ~

C

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Subtração (geométrica) de vetores:

Subtrair um vetor de outro é a mesma coisa

que somar o primeiro ao oposto do segundo.

( ~

B + ~

C) + ~

A

( ~

A + ~

B) + ~

C = ( ~

B + ~

C) + ~

A

~

A

B = ~

~

A + ( ~

B)

~

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20 CH A P T E R 2 FO R C E VE C T O R S 2 F_R ! F₁ " F₂ F_R F_R F₁ F1 F1 F₂ _F₂ F₂ (c) (b) (a) v Fig. 2–7

2.3 Vector Addition of Forces

Experimental evidence has shown that a force is a vector quantity since it has a specified magnitude, direction, and sense and it adds according to the parallelogram law. Two common problems in statics involve either finding the resultant force, knowing its components, or resolving a known force into two components. We will now describe how each of these problems is solved using the parallelogram law.

Finding a Resultant Force.

The two component forces F₁ and F₂

acting on the pin in Fig. 2–7a can be added together to form the resultant force F_R = F₁ + F₂, as shown in Fig. 2–7b. From this construction, or using the triangle rule, Fig. 2–7c, we can apply the law of cosines or the law of sines to the triangle in order to obtain the magnitude of the resultant force and its direction.

Finding the Components of a Force.

Sometimes it is

necessary to resolve a force into two components in order to study its pulling or pushing effect in two specific directions. For example, in Fig. 2–8a, F is to be resolved into two components along the two

members, defined by the u and axes. In order to determine the magnitude of each component, a parallelogram is constructed first, by drawing lines starting from the tip of F, one line parallel to u, and the

other line parallel to . These lines then intersect with the and u axes, forming a parallelogram. The force components F_u and F are then

established by simply joining the tail of F to the intersection points on

the u and axes, Fig. 2–8b. This parallelogram can then be reduced to a triangle, which represents the triangle rule, Fig. 2–8c. From this, the law of sines can then be applied to determine the unknown magnitudes of the components. v v v v v F_u u v F_v F F_R F₂ F₁

Using the parallelogram law force F

caused by the vertical member can be resolved into components acting along the suspension cables a and b.

The parallelogram law must be used to determine the resultant of the two forces acting on the hook.

20 CH A P T E R 2 FO R C E VE C T O R S 2 F_R ! F₁ " F₂ F_R F_R F₁ F₁ F₁ F₂ _F₂ F₂ (c) (b) (a) v Fig. 2–7

2.3 Vector Addition of Forces

Finding a Resultant Force.

The two component forces F₁ and F₂

Finding the Components of a Force.

Sometimes it is

necessary to resolve a force into two components in order to study its pulling or pushing effect in two specific directions. For example, in Fig. 2–8a, F is to be resolved into two components along the two

other line parallel to . These lines then intersect with the and u axes, forming a parallelogram. The force components F_u and F are then

20 C

H A P T E R

2 F

O R C E

V

E C T O R S

2

F_R ! F₁ " F₂ F_R F_R F₁ F1 F1 F₂ _F₂ F₂ (c) (b) (a) v

Fig. 2–7

2.3 Vector Addition of Forces

Experimental evidence has shown that a force is a vector quantity since

it has a specified magnitude, direction, and sense and it adds according to

the parallelogram law. Two common problems in statics involve either

finding the resultant force, knowing its components, or resolving a known

force into two components. We will now describe how each of these

problems is solved using the parallelogram law.

Finding a Resultant Force.

The two component forces

F

₁

and

F

₂

acting on the pin in Fig. 2–7a can be added together to form the resultant

force

F

_R

=

F

₁

+

F

₂

, as shown in Fig. 2–7b. From this construction, or using

the triangle rule, Fig. 2–7c, we can apply the law of cosines or the law of

sines to the triangle in order to obtain the magnitude of the resultant

force and its direction.

Finding the Components of a Force.

Sometimes it is

necessary to resolve a force into two components in order to study its

pulling or pushing effect in two specific directions. For example, in

Fig. 2–8a,

F is to be resolved into two components along the two

members, defined by the u and

axes. In order to determine the

magnitude of each component, a parallelogram is constructed first, by

drawing lines starting from the tip of

F, one line parallel to u, and the

other line parallel to . These lines then intersect with the and u axes,

forming a parallelogram. The force components

F

_u

and

F are then

established by simply joining the tail of

F to the intersection points on

the u and axes, Fig. 2–8b. This parallelogram can then be reduced to a

triangle, which represents the triangle rule, Fig. 2–8c. From this, the law of

sines can then be applied to determine the unknown magnitudes of the

components.

v

F_u u v F_v F

F

_R

F

₂

F

₁

20 CH A P T E R 2 FO R C E VE C T O R S 2 F_R! F₁ " F₂ F_R F_R F₁ F₁ F₁ F₂ _F₂ F₂ (c) (b) (a) v Fig. 2–7

2.3 Vector Addition of Forces

Finding a Resultant Force. The two component forces F₁ and F₂

Finding the Components of a Force. Sometimes it is necessary to resolve a force into two components in order to study its pulling or pushing effect in two specific directions. For example, in Fig. 2–8a, F is to be resolved into two components along the two

other line parallel to . These lines then intersect with the and u axes, forming a parallelogram. The force components Fu and F are then

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Decomposição de Forças

2.3 VECTOR ADDITION OF FORCES 21

2 F u (b) F F_u F_u (c) F u (a) v _v F_v F_v Fig. 2–8

Addition of Several Forces.

If more than two forces are to be added, successive applications of the parallelogram law can be carried out in order to obtain the resultant force. For example, if three forces F₁,

F₂, F₃ act at a point O, Fig. 2–9, the resultant of any two of the forces is found, say, F₁ + F₂—and then this resultant is added to the third force, yielding the resultant of all three forces; i.e., F_R = (F₁ + F₂)+F₃. Using the parallelogram law to add more than two forces, as shown here, often requires extensive geometric and trigonometric calculation to determine the numerical values for the magnitude and direction of the resultant. Instead, problems of this type are easily solved by using the “rectangular-component method,” which is explained in Sec. 2.4.

F₁ F_R F₁ ! F₂ F₃₃ F₁ F₃ F₂ F₁ F₂ F₁ ! F₂ _F_R F₃ O Fig. 2–9

The resultant force on the hook requires the addition of , then this resultant is added to .F₃F1 + F2

F_R

Como desenhar as componentes de F?

Addition of Several Forces.

If more than two forces are to be added, successive applications of the parallelogram law can be carried out in order to obtain the resultant force. For example, if three forces F₁,

F₂, F₃ act at a point O, Fig. 2–9, the resultant of any two of the forces is found, say, F₁ + F₂—and then this resultant is added to the third force, yielding the resultant of all three forces; i.e., F_R = (F₁ + F₂)+F₃. Using the parallelogram law to add more than two forces, as shown here, often requires extensive geometric and trigonometric calculation to determine the numerical values for the magnitude and direction of the resultant. Instead, problems of this type are easily solved by using the “rectangular-component method,” which is explained in Sec. 2.4.

The resultant force on the hook requires the addition of , then this resultant is added to .F₃F1 + F2

F_R

O paralelogramo é reduzido então a um

triângulo no qual a lei dos senos pode ser

aplicada para se determinar as intensidades

desconhecidas das componentes.

2.3 VECTOR ADDITION OF FORCES

21 Addition of Several Forces.

If more than two forces are to be

added, successive applications of the parallelogram law can be carried

out in order to obtain the resultant force. For example, if three forces

F

₁

,

F

₂

,

F

₃

act at a point O, Fig. 2–9, the resultant of any two of the forces is

found, say,

F

₁

+

F

₂

—and then this resultant is added to the third force,

yielding the resultant of all three forces; i.e.,

F

_R

= (

F

₁

+

F

₂

)+

F

₃

. Using

the parallelogram law to add more than two forces, as shown here, often

requires extensive geometric and trigonometric calculation to determine

the numerical values for the magnitude and direction of the resultant.

Instead, problems of this type are easily solved by using the

“rectangular-component method,” which is explained in Sec. 2.4.

The resultant force on the hook

requires the addition of , then this

resultant is added to .F₃F1 + F2 F_R

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Alguma relações importantes

22 C

H A P T E R

2 F

O R C E

V

E C T O R S

2 A

C

B

b

(c)

c

a

Sine law:

sin a

A

!

sin b

B

! C

sin c

Cosine law:

C ! A

2 "

B

2 #

2AB cos c

F

_R

F

₁

F

₂

F

_u

u

(b)

(a)

v

F

_v

Fig. 2–10

Procedure for Analysis

Problems that involve the addition of two forces can be solved as

follows:

Parallelogram Law.

• Two “component” forces

F

₁

and

F

₂

in Fig. 2–10a add according to

the parallelogram law, yielding a resultant force

F

_R

that forms the

diagonal of the parallelogram.

• If a force

F is to be resolved into components along two axes u

and , Fig. 2–10b, then start at the head of force

F and construct

lines parallel to the axes, thereby forming the parallelogram. The

sides of the parallelogram represent the components,

F and F .

• Label all the known and unknown force magnitudes and the

angles on the sketch and identify the two unknowns as the

magnitude and direction of

F

_R

, or the magnitudes of its

components.

Trigonometry.

• Redraw a half portion of the parallelogram to illustrate the

triangular head-to-tail addition of the components.

• From this triangle, the magnitude of the resultant force can be

determined using the law of cosines, and its direction is

determined from the law of sines. The magnitudes of two force

components are determined from the law of sines. The formulas

are given in Fig. 2–10c.

v

u

v

Important Points

• A scalar is a positive or negative number.

• A vector is a quantity that has a magnitude, direction, and sense.

• Multiplication or division of a vector by a scalar will change the

magnitude of the vector.The sense of the vector will change if the

scalar is negative.

• As a special case, if the vectors are collinear, the resultant is

formed by an algebraic or scalar addition.

A

sen(a)

=

B

sen(b)

=

C

sen(c)

Lei dos Senos

Lei dos Cosenos

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Exemplo 2.1 (Hibbeler, 12a Ed.)

O gancho na figura abaixo está sujeito a duas forças

F

1

e F

2

. Determine a

intensidade e a direção da força resultante.

The screw eye in Fig. 2–11a is subjected to two forces,

F

₁

and

F

₂

.

Determine the magnitude and direction of the resultant force.

SOLUTION

Parallelogram Law.

The parallelogram is formed by drawing a line

from the head of

F

₁

that is parallel to

F

₂

, and another line from the

head of

F

₂

that is parallel to

F

₁

.The resultant force

F

_R

extends to where

these lines intersect at point A, Fig. 2–11b. The two unknowns are the

magnitude of

F

_R

and the angle (theta).

Trigonometry.

From the parallelogram, the vector triangle is

constructed, Fig. 2–11c. Using the law of cosines

Ans.

Applying the law of sines to determine ,

Thus, the direction (phi) of

F

_R

, measured from the horizontal, is

Ans.

NOTE:

The results seem reasonable, since Fig. 2–11b shows F

_R

to have

a magnitude larger than its components and a direction that is

between them.

f

= 39.8° + 15.0° = 54.8°

f

u

= 39.8°

sin u

₌

150 N

212.6 N

(sin 115º)

150 N

sin u =

212.6 N

sin 115°

u

= 213 N

= 210 000 + 22 500 - 30 000(-0.4226) = 212.6 N

F

_R

= 2(100 N)

2

+ (150 N)

2

- 2(100 N)(150 N) cos 115°

u

EXAMPLE 2.1

2.3 VECTOR ADDITION OF FORCES

23

F₁ ! 100 N F₂ ! 150 N 10" 15" (a) Fig. 2–11 (c) F_R _{150 N} 100 N 15" 115" u f 2 F_R 90" # 25" ! 65" 10" 15" 100 N A 65" 115" 150 N (b) ! 115" 360 # 2(65") 2 u

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Solução:

Desenhando o paralelogramo e as forças que o geram,

determinamos

os respectivos ângulos (geometria)

The screw eye in Fig. 2–11a is subjected to two forces, F₁ and F₂.

Determine the magnitude and direction of the resultant force.

SOLUTION

Parallelogram Law. The parallelogram is formed by drawing a line

from the head of F₁ that is parallel to F₂, and another line from the

head of F₂ that is parallel to F₁.The resultant force F_R extends to where these lines intersect at point A, Fig. 2–11b. The two unknowns are the

magnitude of F_R and the angle (theta).

Trigonometry. From the parallelogram, the vector triangle is constructed, Fig. 2–11c. Using the law of cosines

Ans.

Applying the law of sines to determine ,

Thus, the direction (phi) of F_R, measured from the horizontal, is

Ans. NOTE: The results seem reasonable, since Fig. 2–11b shows F_R to have

a magnitude larger than its components and a direction that is between them. f = 39.8° + 15.0° = 54.8° f u = 39.8° sin u ₌ 150 N 212.6 N (sin 115º) 150 N sin u = 212.6 N sin 115° u = 213 N = 210 000 + 22 500 - 30 000(-0.4226) = 212.6 N F_R _{= 2(100 N)}2 _{+ (150 N)}2 _{- 2(100 N)(150 N) cos 115°} u

EXAMPLE 2.1

The screw eye in Fig. 2–11a is subjected to two forces, F₁ and F₂. Determine the magnitude and direction of the resultant force.

SOLUTION

Parallelogram Law. The parallelogram is formed by drawing a line from the head of F₁ that is parallel to F₂, and another line from the head of F₂ that is parallel to F₁.The resultant force F_R extends to where these lines intersect at point A, Fig. 2–11b. The two unknowns are the magnitude of F_R and the angle (theta).

Ans.

NOTE: The results seem reasonable, since Fig. 2–11b shows F_R to have

a magnitude larger than its components and a direction that is between them. f = 39.8° + 15.0° = 54.8° f u = 39.8° sin u ₌ 150 N 212.6 N (sin 115º) 150 N sin u = 212.6 N sin 115° u = 213 N = 210 000 + 22 500 - 30 000(-0.4226) = 212.6 N F_R = 2(100 N)2 + (150 N)2 - 2(100 N)(150 N) cos 115° u

EXAMPLE 2.1

F

_R

=

q

F

₁2

+ F

₂2

2F

₁

.F

₂

.cos(115)

=

p

(100)

2

+ (150)

2

2.100.150.cos(115)

=

213N

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Falta determinar a direção desta força resultante, ou seja, o ângulo ø da figura.

Para isto, vamos utilizar a

lei dos senos para o triângulo formado pela força

resultante e suas componentes.

The screw eye in Fig. 2–11a is subjected to two forces, F₁ and F₂. Determine the magnitude and direction of the resultant force.

SOLUTION

Parallelogram Law. The parallelogram is formed by drawing a line from the head of F₁ that is parallel to F₂, and another line from the head of F₂ that is parallel to F₁.The resultant force F_R extends to where these lines intersect at point A, Fig. 2–11b. The two unknowns are the magnitude of F_R and the angle (theta).

Ans.

NOTE: The results seem reasonable, since Fig. 2–11b shows F_R to have

a magnitude larger than its components and a direction that is between them. f = 39.8° + 15.0° = 54.8° f u = 39.8° sin u ₌ 150 N 212.6 N (sin 115º) 150 N sin u = 212.6 N sin 115° u = 213 N = 210 000 + 22 500 - 30 000(-0.4226) = 212.6 N F_R _{= 2(100 N)}2 _{+ (150 N)}2 _{- 2(100 N)(150 N) cos 115°} u