Eligheor2
5
Por el método de las secciones: Dibujaremos el diagrama de cuerpo libre para hacer el calculo de las reacciones en los apoyos: Aplicando las ecuaciones de equilibrio obtenemos que: ! = 0: ! ! = 0 ! = 0: ! ! + ! ! − 1 − 2 − 2 − 2 − 1 = 0 1 !" 2 !" 2 !" 2 !" 1 !" ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! 2 ,4 ! 2 ,4 ! 2 ,4 ! 2 ,4 ! 2 ,62 ! 0 ,46 !
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Eligheor2
Transcript of Eligheor2
Por el método de las secciones: Dibujaremos el diagrama de cuerpo libre para hacer el calculo de las reacciones en los apoyos:
Aplicando las ecuaciones de equilibrio obtenemos que:
𝐹! = 0: 𝑅!! = 0
𝐹! = 0:
𝑅!! + 𝑅!! − 1𝑘𝑁 − 2𝑘𝑁 − 2𝑘𝑁 − 2𝑘𝑁 − 1𝑘𝑁 = 0
1!!"!
2!!"!
2!!"! 2!!"!
1!!"!
!!! !!!
!!!
!!
!! !! !! !! !!
!!
!! !!
!!
2,4!!! 2,4!!! 2,4!!! 2,4!!!
2,62!!!
0,46!!! 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°▹
𝑅!! + 𝑅!! = 8 𝑘𝑁
Con los datos del dibujo podemos sacar:
𝛼 = tan!!2,169,6 = 12,68°
2,16!!!
0,46!!!
9,6!!!
2,62!!! !
Si se conoce la inclinación de la armadura, podemos encontrar la medida que poseen verticalmente los elementos DC, FE y HG
𝐷𝐶 = 𝑦! + 𝐴𝐵
𝐹𝐸 = 𝑦! + 𝐴𝐵
𝐻𝐺 = 𝑦! + 𝐴𝐵
Donde tenemos: 𝑦! = 𝐴𝐶 tan𝛼 , 𝑦! = 𝐴𝐸 tan𝛼 , 𝑦! = 𝐴𝐺 tan𝛼 𝐴𝐶 = 2,4 𝑚 , 𝐴𝐸 = 4,8 𝑚 , 𝐴𝐺 = 7,2 𝑚 𝐴𝐵 = 0,46 𝑚 , 𝛼 = 12,68° Por lo tanto
𝐷𝐶 = 𝑦! + 𝐴𝐵 = 2,4 tan 12,68 + 0,46 = 1 𝑚
𝐹𝐸 = 𝑦! + 𝐴𝐵 = 4,8 tan 12,68 + 0,46 = 1,54 𝑚
𝐻𝐺 = 𝑦! + 𝐴𝐵 = 7,2 tan 12,68 + 0,46 = 2,08 𝑚
!! !! !! !! !!
!!
!! !!
!! !!
!!
!!
!!
Aplicando las ecuaciones de equilibrio obtenemos que:
𝑀! = 0:
−2,4 2 − 4,8 2 − 7,2 2 − 9,6 1 + 9,6(𝑅!!) = 0
𝑅!! = 2,4 2 + 4,8 2 + 7,2 2 + 9,6 1
9,6 = 4 𝑘𝑁
Luego tenemos que:
𝑅!! = 8𝑘𝑁 − 𝑅!! = 8 𝑘𝑁 − 4𝑘𝑁 = 4 𝑘𝑁 Calcularemos el valor de cada componente de las reacciones, por lo cual, se seccionará la armadura en los elementos CE, DE y DF
𝛽 = tan!!2,41 = 67,38°
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°▹
!! !! !!
!!
!!
!!
4!!"!
2!!"!
1!!"!
!
2,4!!! 2,4!!!
!!"
!!"
!!"
Aplicando las ecuaciones de equilibrio obtenemos que:
𝑀! = 0:
1 𝐹!" − 2,4 4𝑘𝑁 + 2,4 1𝑘𝑁 = 0
𝐹!" = 2,4 4𝑘𝑁 − 2,4 1𝑘𝑁
1 = 7,2 𝑘𝑁 𝑻 𝑒𝑙 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑜 𝐶𝐸 𝑒𝑠𝑡𝑎 𝑒𝑛 𝑡𝑟𝑎𝑐𝑐𝑖ó𝑛
𝑀! = 0: −4,8 4𝑘𝑁 + 4,8 1𝑘𝑁 + 2,4 2𝑘𝑁 − 1 cos 12,68 (𝐹!")− 2,4 sin 12,68 (𝐹!") = 0
𝐹!" = 4,8 4𝑘𝑁 − 4,8 1𝑘𝑁 − 2,4 2𝑘𝑁1 cos 12,68 + 2,4 sin 12,68 = −6,34 𝑘𝑁
∴ 𝐹!" = 6,34 𝑘𝑁 𝑪 𝑒𝑙 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑜 𝐷𝐹 𝑒𝑠𝑡𝑎 𝑒𝑛 𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑖ó𝑛
𝑀! = 0:
2,4 1𝑘𝑁 − 2,4 4𝑘𝑁 + 1 634𝑘𝑁 − 1 sin 67,38 (𝐹!") = 0
𝐹!" = −2,4 1𝑘𝑁 + 2,4 4𝑘𝑁 − 1 634𝑘𝑁
1 sin 12,68 = −0,93 𝑘𝑁
∴ 𝐹!" = 0,93 𝑘𝑁 𝑪 𝑒𝑙 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑜 𝐷𝐸 𝑒𝑠𝑡𝑎 𝑒𝑛 𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑖ó𝑛
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°▹
