Mecânica Geral - Exercícios Resolvidos - Cap. 05 - Beer e Johnston

x

y

0

w1

L

w2

x

y

=0

w1x + Ly − w 1L − w 2x

=0 Ly + ( w 1 − w 2 ) x − w 1L = 0 Ly = − ( w 1 − w 2 ) x + w 1L ÷ ( L ) w −w ∴ y = −  1 2  ÷ x + w1 L   

L

R

= ∫ ydx 0

Onde Onde :

 w1 − w 2  x + w 1 ÷  L  

y = −

Substituindo Substituindo e Integrand Integrando o:

 w− w  x + w 1 = ∫ ydx ⇔ R = ∫  −  1 2  ÷ L    0 0   L w1 − w 2  x 2  R = − ÷ + w 1x  L   2 0  w − w 2  ×L2 +w L R = − 1 1 ÷  2L    w − w 2  L + w L R = − 1 1 ÷  2   L

L

R

R

=−

R  =

w1L

2 w1L 2

+

+

w 2L

2 w 2L

+ w1L

2  w1 + w 2

∴ R  =   

2

  ÷ L ( I)  

AFONSO CARIOCA – AULAS PARTICULARES E CURSOS LIVRES afonsocarioca!o"#ai$%co# – afonsocarioca!o"#ai$%co#  – CEL& 8'09-(036

 3092-2268

dx

L

x=

∫ xydx

x=

0 L

R  Onde :

∫ ydx 0

Onde:

Qy

L

Qy

= ∫ xydx

L

e

R

0

= ∫ ydx

w 1 + 2w 2 6

2

÷L 

e

R

=  

 w1 + 2w 2  L2  6 ÷ Qy   x= ⇔x= R   w1 + w 2  L  2 ÷    ( w1 + 2w 2 ) L2 2 x= × 6 ( w1 + w 2 ) ( w + 2w 2 ) L ∴x = 1 ( II ) 3 ( w1 + w 2 )

0

Qy

R  Onde:

 w1 − w 2  x + w 1 ÷  L  

y = −

Substituindo e Integrando : L

Qy

=  

w1 + w 2 2

Substituindo :

Assim : x=

Qy

= ∫ xydx

L

0

 w− w  x w = ∫ x  −  1 2  + 1 ÷  dx L      0 L   w − w 2  2  Q y = ∫  −  1 x w x dx + 1  ÷ L      0 L

Qy

L

w1 − w 2  x 3 x2  Qy = −  ÷ 3 + w1 2 L    0 3 2  w − w 2  L + w L Qy = −  1 1 ÷ 2  L   3 w1 − w 2  2 L2  Qy = −  ÷ L + w1 2  3   Qy Qy

=− =

w1L2 3

w1L2 6

+

+

w 2L2 3

w 2L2 2

+

w 1L2 2

w + 2w 2  2 ∴ Q y =  1 ÷L  6  

AFONSO CARIOCA – AULAS PARTICULARES E CURSOS LIVRES afonsocarioca!o"#ai$%co# – CEL& 8'09-(036

 3092-2268



÷

   

L

'entr(ide x : x=

( w1 + 2w 2 ) L 3 ( w1 + w 2 )

Substituindo : x= x=

( 2000 + 2 ×1!00 ) × 2" # m 3 × ( 2000 + 1!00 ) !000 × 2" # 10!00

m⇔x

=

12000 10!00

m

x = 1"1#3 m Res%: x = 1"1#3 m

Dados :

= 2000 N  m w 2 = 1!00 N  m L = 2" # m w1

temos : R

=  

w1 + w 2 

÷  

2

L

e

x

=

( w1 + 2w 2 ) L 3 ( w1 + w 2 )

Resu$ tan te R : R

=  

w1 + w 2 

÷L  

2

Substituindo : R

=  

R

=

2000 + 1!00  2

3!00 × 2"#

2 R = #200 N Re s% : R

÷ × 2" # N  

N

= #" 2 &N AFONSO CARIOCA – AULAS PARTICULARES E CURSOS LIVRES afonsocarioca!o"#ai$%co# – CEL& 8'09-(036

 3092-2268

temos : R

∑* = 0: x

+A

w1 + w 2 

÷  

2

L

e

x

=

( w1 + 2w 2 ) L 3 ( w1 + w 2 )

=0 Resu$ tan te R :

∑,

=0 2" #-) − 1"1#3 × #" 2 = 0 2"#-) = 1"1#3 × #"2 2" #-) = #" .01 -)

=  

=

A

#".01 2" #

&N ∴ -)

= 2" 0 &N

R

=  

w1 + w 2 

÷L  

2

Substituindo : R

=  

R

=

6"6/ + !"00  2

÷ × 3" 60 &N  

11" 6/ × 3" 6

&N

2 R = 21"01&N

∑* = 0: y

− #" 2 + -) = 0 ⇔ -A − #" 2 + 2" 0 = 0 -A = #" 2 − 2" 0 = 2" 2 &N ∴-A = 2" 2 &N -A

Res%: R

= 21"01&N

'entr(ide x : Re s% : -A

= 2" 2 &N" + A = 0 e -) = 2" 0 &N

x=

( w1 + 2w 2 ) L 3 ( w1 + w 2 )

Substituindo : x= x=

( 6" 6/ + 2 × !" 00 ) × 3" 60 m 3 × ( 6" 6/ + !" 00 ) 16" 6/ × 3" 60 3!" 01

m⇔x

=

60" 01 3!" 01

m

x = 1" /1 m Re s% : x = 1" /1 m

Dados :

= 6" 6/ &N  m w 2 = !" 00 &N  m L = 3"60 m w1

AFONSO CARIOCA – AULAS PARTICULARES E CURSOS LIVRES afonsocarioca!o"#ai$%co# – CEL& 8'09-(036

 3092-2268

∑* = 0: x

+)

=0

∑,

=0 −2"10-) + 1". × 21"01 = 0 −2"10-) = −1". × 21"01 2"10-) = 3"/ -)

=

'

3"/ 2"10

&N ∴ -)

= 1." &N

R

= !# &N

L = 1". m

Substituindo :

∑* = 0:

 w A + w A  ×1". ÷  2   w A + w A !# w + wA = &N  m ⇔ A = 30 &N  m

!# = 

y

-) − 21" 01 + -'

= 0 ⇔ 1."  − 21" 01 + -' = 0 -' = 21" 01 −1."  = 2"11 &N ∴-' = 2"11 &N Re s% : -)

emos :

= 1."  &N" + ) = 0 e -' = 2"11 &N

2

1".

2

Assim : wA

+ w ) = 60 &N  m ∴ w A + w ) = 60 ( I )

AFONSO CARIOCA – AULAS PARTICULARES E CURSOS LIVRES afonsocarioca!o"#ai$%co# – CEL& 8'09-(036

 3092-2268

w A + w ) = 60 w + 2w = 110  A ) Assim :

 60 110 wA =  1 1 

1



2   120 − 110

1



=

2 −1

&N =

10 1

&N

2

Assim : wA

= 10 &N

e w)

Re s% : w A

emos ue :

∑,

= 60 − w A = 60 −10 &N = !0 &N ∴ w ) = !0 &N = 10 &N e w ) = !0 &N

=0 !#x − 1" ! × 30 − 0" 6 × 2# = 0 !#x − #! − 1#" # = 0 ⇔ !#x − !" # = 0 A

Assim :

= !" # ⇔ x =

!#x

!"# !#

m ∴x

= 1"1 m

,as: x=

( w A + 2w ) ) L 3( wA + w ) )

Substituindo : 1"1 =

( w A + 2w ) ) ×1". 3 × 60

⇔ 1"1 =

( w A + 2w ) ) ×1". 1.0

Assim :

( w A + 2w ) ) ×1". = 1.0 ×1"1 Assim : wA

+ 2w ) =

1.0

÷1".

1".

×1"1

÷1".

Assim : wA

+ 2w ) = 100 ×1"1∴ w A + 2w ) =110 ( II )

AFONSO CARIOCA – AULAS PARTICULARES E CURSOS LIVRES afonsocarioca!o"#ai$%co# – CEL& 8'09-(036

 3092-2268