Post on 02-Jul-2015
description
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:
Aplicando las ecuaciones de equilibrio obtenemos:
𝑀! = 0: 36 2,4 − 𝐵 13,5 + 20 9 + 20 4,5 = 0
𝑩 = 36 2,4 + 20 9 + 20 4,5
13,5 = 𝟐𝟔,𝟒 𝒌𝑵 ↑
𝐹! = 0: − 36+ 𝐾! = 0
𝑲𝒙 = 𝟑𝟔 𝒌𝑵 →
𝐹! = 0: 26,4− 20− 20 + 𝐾! = 0
4,5!!!
!!
!!
4,5!!! 4,5!!!
!! !!
!!
B !
20!!"! 20!!"!
2,4!!! !!
36!!"!
COSMOS: Complete Online Solutions Manual Organization System
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
Chapter 4, Solution 19.
Free-Body Diagram:
(a) From free-body diagram of lever BCD
( ) ( )0: 50 mm 200 N 75 mm 0C AB
M TΣ = − =
300AB
T∴ =(b) From free-body diagram of lever BCD
( )0: 200 N 0.6 300 N 0x x
F CΣ = + + =
380 N or 380 Nx x
C∴ = − =C
( )0: 0.8 300 N 0y y
F CΣ = + =
N 240or N 240 =−=∴yy
C C
Then ( ) ( )2 22 2380 240 449.44 N
x yC C C= + = + =
and °=⎟⎠
⎞⎜⎝
⎛
−−=⎟⎟
⎠
⎞⎜⎜⎝
⎛= −−
276.32380
240tantan
11
x
y
C
Cθ
or 449 N=C 32.3°▹
COSMOS: Complete Online Solutions Manual Organization System
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
Chapter 4, Solution 19.
Free-Body Diagram:
(a) From free-body diagram of lever BCD
( ) ( )0: 50 mm 200 N 75 mm 0C AB
M TΣ = − =
300AB
T∴ =(b) From free-body diagram of lever BCD
( )0: 200 N 0.6 300 N 0x x
F CΣ = + + =
380 N or 380 Nx x
C∴ = − =C
( )0: 0.8 300 N 0y y
F CΣ = + =
N 240or N 240 =−=∴yy
C C
Then ( ) ( )2 22 2380 240 449.44 N
x yC C C= + = + =
and °=⎟⎠
⎞⎜⎝
⎛
−−=⎟⎟
⎠
⎞⎜⎜⎝
⎛= −−
276.32380
240tantan
11
x
y
C
Cθ
or 449 N=C 32.3°▹
COSMOS: Complete Online Solutions Manual Organization System
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
Chapter 4, Solution 19.
Free-Body Diagram:
(a) From free-body diagram of lever BCD
( ) ( )0: 50 mm 200 N 75 mm 0C AB
M TΣ = − =
300AB
T∴ =(b) From free-body diagram of lever BCD
( )0: 200 N 0.6 300 N 0x x
F CΣ = + + =
380 N or 380 Nx x
C∴ = − =C
( )0: 0.8 300 N 0y y
F CΣ = + =
N 240or N 240 =−=∴yy
C C
Then ( ) ( )2 22 2380 240 449.44 N
x yC C C= + = + =
and °=⎟⎠
⎞⎜⎝
⎛
−−=⎟⎟
⎠
⎞⎜⎜⎝
⎛= −−
276.32380
240tantan
11
x
y
C
Cθ
or 449 N=C 32.3°▹
𝑲𝒚 = 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:
Aplicando de nuevo las ecuaciones de equilibrio obtenemos:
𝑀! = 0: 36 1,2 − 26,4 2,25 − 𝐹!" 1,2 = 0
𝑭𝑨𝑫 = 36 1,2 − 26,4 2,25
1,2 = −𝟏𝟑,𝟓 𝒌𝑵
𝐹!" = 13,5 𝑘𝑁 𝑪 𝒆𝒍𝒆𝒎𝒆𝒏𝒕𝒐 𝒆𝒏 𝒄𝒐𝒎𝒑𝒓𝒆𝒔𝒊𝒐𝒏
𝑀! = 0: 817 𝐹!" 4,5 = 0
𝐹!" = 0
𝑀! = 0: 1517 𝐹!" 2,4 − 26,4 4,5 = 0
17!
!! !!
36!!"!
1,2!!!
!!"
!!
!! 1,2!!!
4,5!!!
!!"
!!" 26,4!!"!
2,25!!! !!
8!
17! 8!
15! 15!
COSMOS: Complete Online Solutions Manual Organization System
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
Chapter 4, Solution 19.
Free-Body Diagram:
(a) From free-body diagram of lever BCD
( ) ( )0: 50 mm 200 N 75 mm 0C AB
M TΣ = − =
300AB
T∴ =(b) From free-body diagram of lever BCD
( )0: 200 N 0.6 300 N 0x x
F CΣ = + + =
380 N or 380 Nx x
C∴ = − =C
( )0: 0.8 300 N 0y y
F CΣ = + =
N 240or N 240 =−=∴yy
C C
Then ( ) ( )2 22 2380 240 449.44 N
x yC C C= + = + =
and °=⎟⎠
⎞⎜⎝
⎛
−−=⎟⎟
⎠
⎞⎜⎜⎝
⎛= −−
276.32380
240tantan
11
x
y
C
Cθ
or 449 N=C 32.3°▹
COSMOS: Complete Online Solutions Manual Organization System
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
Chapter 4, Solution 19.
Free-Body Diagram:
(a) From free-body diagram of lever BCD
( ) ( )0: 50 mm 200 N 75 mm 0C AB
M TΣ = − =
300AB
T∴ =(b) From free-body diagram of lever BCD
( )0: 200 N 0.6 300 N 0x x
F CΣ = + + =
380 N or 380 Nx x
C∴ = − =C
( )0: 0.8 300 N 0y y
F CΣ = + =
N 240or N 240 =−=∴yy
C C
Then ( ) ( )2 22 2380 240 449.44 N
x yC C C= + = + =
and °=⎟⎠
⎞⎜⎝
⎛
−−=⎟⎟
⎠
⎞⎜⎜⎝
⎛= −−
276.32380
240tantan
11
x
y
C
Cθ
or 449 N=C 32.3°▹
COSMOS: Complete Online Solutions Manual Organization System
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
Chapter 4, Solution 19.
Free-Body Diagram:
(a) From free-body diagram of lever BCD
( ) ( )0: 50 mm 200 N 75 mm 0C AB
M TΣ = − =
300AB
T∴ =(b) From free-body diagram of lever BCD
( )0: 200 N 0.6 300 N 0x x
F CΣ = + + =
380 N or 380 Nx x
C∴ = − =C
( )0: 0.8 300 N 0y y
F CΣ = + =
N 240or N 240 =−=∴yy
C C
Then ( ) ( )2 22 2380 240 449.44 N
x yC C C= + = + =
and °=⎟⎠
⎞⎜⎝
⎛
−−=⎟⎟
⎠
⎞⎜⎜⎝
⎛= −−
276.32380
240tantan
11
x
y
C
Cθ
or 449 N=C 32.3°▹
𝑭𝑪𝑬 = 17 26,4 4,5
15 2,4 = 𝟓𝟔,𝟏 𝒌𝑵
𝑭𝑪𝑬 = 𝟓𝟔,𝟏 𝒌𝑵 𝑻 𝒆𝒍𝒆𝒎𝒆𝒏𝒕𝒐 𝒆𝒏 𝒕𝒓𝒂𝒄𝒄𝒊𝒐𝒏