1. Se usará el método de secciones, ya que se desea calcular las fuerzas en un
número reducido de barras en la armadura.
Dibujamos el diagrama de cuerpo libre:
!!
20!!"!
20!!"!
!!
36!!"!
2,4!!!
!!
!!
!!
omplete Online Solutions Manual Organization System
!!
B
!
4, Solution 19.
4,5!!!
4,5!!!
4,5!!!
COSMOS: Complete Online Solutions Manual Organization System
dy Diagram:
Aplicando las ecuaciones de equilibrio obtenemos:
OSMOS: Complete Online Solutions (a) From free-body diagram
Manual Organization System
Chapter 4, Solution 19.
of lever BCD
ΣM C = 0: TAB ( 50 mm ) − 200 N ( 75 mm ) = 0
𝑀! = 0: 36 2,4 − 𝐵 13,5 + 20 9 + 20 4,5 = 0
hapter 4, Solution 19.
Free-Body Diagram:
(b) From free-body diagram of lever BCD of lever BCD
(a) From free-body diagram
Free-Body Diagram:
Then
and
∴ TAB = 300
36 2,4 + 0: + 9 ( 300mm4,5 200
ΣFx = 0: 200M C +=C20 TAB + 20 ) )= 0 N ( 75 mm ) = 0
0.6 50 N −
Σ N
x
𝑩 =
= 𝟐𝟔, 𝟒 𝒌𝑵 ↑
(a) From free-body diagram13,5 BCD
of lever
∴ C x = −380 N
or
C x = 380 N
∴ TAB = 300
ΣM C = 0: TAB ( 50 mm ) − 200 N ( 75 mm ) = 0
Σ(b) =From C y + 0.8 ( 300 N ) = of lever BCD
Fy 0: free-body diagram 0
∴ TAB = 300
ΣFx = 0: 200 N C36 + 300 0
x
∴ C y = −240 N 0: +−C y+=0.6 (𝐾! =N ) = 0
240 N
𝐹! = or
(b) From free-body diagram of lever BCD
∴ C x = −380 N
or
C x = 380 N
2
2
2
2
C = CΣFx C y 0: 200 N)+ + x 240 ) ( 300 N ) = N
C ( + 0.6 = 449.44 0
x + = = ( 380
𝑲 C + 0.8 ( 300
ΣFy = 0:𝒙 = y 𝟑𝟔 𝒌𝑵 →N ) = 0
or
C x = 380 N
C y ⎞ ∴ C x−1= − 380 ⎞
⎛
⎛ −240 N
θ = tan −1 ⎜ ⎟ = tan ⎜∴ C y = =240276° or
C y = 240 N
⎟ − 32. N
Σ⎜ C x=⎟ 0: C y ⎝ − 380 ⎠ N ) = 0
F
+ 0.8 ( 300
⎝y ⎠
𝐹! = 0: 26,4 − 20 −2 20 + 𝐾! = 0
2
2
2
C =C C x −+ C yN= or or = 449240240= 449.44 N
( 380 ) +C( y N ) 32.3° ▹
C
∴ y = 240
=
N
Then
2
2
⎛ − 240 =
2
2 ⎛C ⎞
C x = tany−1 = y( 380tan+1 ( 240 ) ⎞ =449.44 N
θ + C ⎜ ⎟= ) − ⎜
⎟ 32.276°
⎜C ⎟
⎝ − 380 ⎠
⎝ x⎠
⎛ Cy ⎞
⎛ − 240 ⎞
θ = tan −1 ⎜ ⎟ = tan −1 ⎜
⎟ = 32.276° or C = 449 N
⎜C ⎟
− 380
Then and C =
and
32.3° ▹
2. 𝑲 𝒚 = 40 − 26,4 = 𝟏𝟑, 𝟔 𝒌𝑵 ↑
Al calcular el valor en cada componente de las reacciones, seccionara la
armadura en los elementos AD, CD y CE:
!!"
!!"
36!!"! !!
!!
!
17
1,2!!!
8!
15!
15!
!!
1,2!!!
17
!
8!
!!
!!
OS: Complete Online Solutions Manual Organization System
2,25!!!
!!"
26,4!!"!
ter 4, Solution 19.
-Body Diagram:
4,5!!!
Aplicando de nuevo las ecuaciones de equilibrio obtenemos:
(a) From free-body diagram of lever BCD
ΣM C = 0: TAB ( 50 mm ) − 200 N ( 75 mm ) = 0
𝑀! =
COSMOS: Complete Online Solutions Manual Organization System
0: 36 1,2 − 26,4 2,25 − 𝐹!" 1,2 = 0
∴ TAB = 300
(b) From free-body diagram of lever BCD
ΣFx = 0: 36 1,2+ −x26,4 ( 300 N ) = 0
200 N C + 0.6 2,25
𝑭 𝑨𝑫 =
Chapter 4, Solution 19.
∴ C x = −3801,2
N
OS: Complete Online Solutions Manual Organization System F
Σ y
Free-Body Diagram:
𝐹!" =
ter 4, Solution 19.
-Body Diagram:
Then
= −𝟏𝟑, 𝟓 𝒌𝑵
or
C x = 380 N
= 0: C y + 0.8 𝒆𝒍𝒆𝒎𝒆𝒏𝒕𝒐 𝒆𝒏 𝒄𝒐𝒎𝒑𝒓𝒆𝒔𝒊𝒐𝒏
13,5 𝑘𝑁 𝑪 ( 300 N ) = 0
(a) From free-body diagram of lever BCD
∴ C y = −240 N
or
C y = 240 N
ΣM C = 0: TAB ( 50 mm8 − 200 N ( 75 mm ) = 0
) 2
𝑀!2 = ( 380 )2
2
C = C x + C y = 0: + ( 240 ) 𝐹!" 449.44 N 0
= 4,5 =
17
∴ TAB = 300
⎛ Cy ⎞
(b) From free-body diagram− 240 ⎞ 0 BCD
⎛ of lever
and
θ = tan −1 ⎜ ⎟ = tan −1 ⎜𝐹!" = ⎟ = 32.276°
⎜C ⎟
⎝ +C
(a) From free-body diagram of leverN380 ⎠x + 0.6 ( 300 N ) = 0
⎠
⎝
ΣFx x = 0: 200 −BCD
N
32.3°
ΣM C = 0: TAB (∴ C 15 380 N ( 75C = ) = C = 380 N ▹
50 mm ) − 200 N or or 4490
−
mm
x =
x
𝑀! = 0:
𝐹!"
2,4 − 26,4 4,5 = 0
17
∴ TAB = 300
ΣFy = 0: C y + 0.8 ( 300 N ) = 0
(b) From free-body diagram of lever BCD
∴ C y = −240 N
or
C y = 240 N
ΣFx = 0: 200 N + Cx + 0.6 ( 300 N ) = 0
2
2
2
2
C = C x + C y = ( 380 ) + ( 240 ) = 449.44 N
Then
∴ C x = −380 N
or
C x = 380 N
⎛ C y ⎞ N ⎛ − 240 ⎞
300
and ΣFy = 0: = C y −1 ⎜0.8 (⎟ = tan)−1= 0
θ tan +
⎟ = 32.276°
⎜
