∴ C x = −380 N or C x = 380 N 1) The document analyzes a truss structure using the method of sections to calculate reactions at supports. 2) It draws the free body diagram of member BCD and applies equations of equilibrium to find the reaction forces. 3) It then calculates the inclination of various members and the vertical distances of points using trigonometric functions.

1.
Por
el
método
de
las
secciones:
Dibujaremos
el
diagrama
de
cuerpo
libre
para
hacer
el
calculo
de
las
reacciones
en
los
apoyos:
1!!"!
2!!"!
!!
2!!"!
2!!"!
!!
1!!"!
!!
2,62!!!
!!
!!
0,46!!!
COSMOS: Complete Online Solutions Manual Organization System
COSMOS: Complete Online Solutions Manual Organization System
!!
!!
!!
!!
!!
!!!
Chapter 4, Solution 19. 19.
Chapter 4, Solution
!!!
!!!
Free-Body Diagram:
Free-Body Diagram:
(a) (a)
From free-body diagram of lever BCD
free-body diagram of lever BCD2,4!!!
2,4!!! From 2,4!!!
2,4!!!
ΣM C M C0:= 0:AB T50 (mmmm ) − 200 75(mmmm )0= 0
T ( AB 50 ) − 200 N ( N 75 ) =
Σ=
Aplicando
las
ecuaciones
de
equilibrio
obtenemos
que:
(b) (b) From free-body diagramlever BCD
From free-body diagram of of lever BCD
∴ TABTAB 300
∴ = = 300
ΣFx Σ𝐹 x0:= 0: 𝑅N += + + x0.6 ( 300300 N )0= 0
= = 0: 200 N 0
C + 0.6 ( N ) =
200
Cx
F
!
!
!
∴ C x = x−380380 N or or C x C x 380380 N
= = N
∴C = −N
ΣFy Σ= y0:= 0: y + y + (0.8 ( 300 N )0= 0
C C 0.8 300 N ) =
F
𝐹! = 0:
∴ C y = −240240 N or or C y C y240240 N
= = N
∴ Cy = − N
𝑅!! + Then− 1𝑘𝑁 C = C 2− 2𝑘𝑁 2 (2𝑘𝑁 2− 1𝑘𝑁 )= =2 449.44 N
𝑅!!
− 2𝑘𝑁 + 2 2 = − 380 ) + 2( 240 2 0
C
Then
and and
C = x Cx + C y =
y
⎛ Cy ⎞
C
Cx ⎠ x
⎝ ⎝C ⎠
( 380 )
+ ( 240 ) = 449.44 N
⎛ − 240 ⎞
⎝ − 380380 ⎠
⎝− ⎠
⎞
tan 1 ⎜ ⎜⎟ y ⎟ = 1 ⎜ ⎜ ⎟ = ⎞ = 32. °
θ = θ =−tan −1 ⎛⎟ = tan −tan −1 ⎛ − 240 32.276276°
⎟
⎜ ⎜
⎟
or C = 449449 N 32.3° ▹ ▹
or C = N
32.3°

2.
𝑅!! + 𝑅!! = 8 𝑘𝑁
2,16!!!
2,62!!!
!
0,46!!!
9,6!!!
Con
los
datos
del
dibujo
podemos
sacar:
2,16
𝛼 = tan!!
= 12,68°
9,6

3. 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 tan 𝛼
,
𝑦! = 𝐴𝐸 tan 𝛼
,
𝑦! = 𝐴𝐺 tan 𝛼
𝐴 𝐶 = 2,4 𝑚
,
𝐴 𝐸 = 4,8 𝑚
,
𝐴 𝐺 = 7,2 𝑚
𝐴 𝐵 = 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

4. COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 19.
Free-Body Diagram: cuaciones
de
equilibrio
obtenemos
que:
Aplicando
las
e
(a) From free-body diagram of lever BCD
ΣM C = 0: TAB ( 50 mm ) − 200 N ( 75 mm ) = 0
𝑀! = 0:
∴ TAB = 300
(b) From free-body diagram of lever BCD
−2,4 2 − 4,8 2 − 7,2 Fx = 0: 200 N+ 9,6(𝑅0.6 ( 300 N ) = 0
Σ 2 − 9,6 1 + Cx + !! ) = 0
∴ C x = −380 N
or
C x = 380 N
2,4 2 + 4,8 2ΣFy 7,2 2 C++9,6 (1 N ) = 0
+ = 0: y 0.8 300
𝑅!! =
= 4 𝑘𝑁
9,6 ∴ C y = −240 N
or
C y = 240 N
Luego
tenemos
que:
C =
Then
2
2
Cx + C y =
( 380 )2 + ( 240 )2
= 449.44 N
⎞
𝑅!! = 8𝑘𝑁 − 𝑅!!θ = tan −1 ⎛ −y 4𝑘𝑁 =1 ⎛ − 240 ⎞ = 32.276°
4 𝑘𝑁
⎟
⎟ = tan − ⎜
⎜
and
= 8 𝑘𝑁
C
⎜C ⎟
⎝ x⎠
− 380
⎠
⎝
or C = ual,
Calcularemos
el
valor
de
cada
componente
de
las
reacciones,
por
lo
c449 Nse
32.3° ▹
seccionará
la
armadura
en
los
elementos
CE,
DE
y
DF
!!
2!!"!
!!"
!!
1!!"!
!
!!
!!
!!
!!"
!!
!!"
4!!"!
2,4!!!
2,4!!!
2,4
𝛽 = tan!! Beer, = Russell Johnston, Jr.,
67,38°
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P.1
E.
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.

5. COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 19.
Free-Body Diagram:ecuaciones
de
equilibrio
obtenemos
que:
Aplicando
las
(a) From free-body diagram of lever BCD
ΣM C = 0: TAB ( 50 mm ) − 200 N ( 75 mm ) = 0
𝑀! = 0:
∴ TAB = 300
(b) From free-body diagram of lever BCD
COSMOS: Complete Online Solutions Manual Organization System
1 𝐹!" − 2,4 4𝑘𝑁 =+ 2,4 1𝑘𝑁
ΣFx 0: 200 N +
C= + 0.6 ( 300 N ) = 0
x 0
∴ C x = −380 N
or
C x = 380 N
Chapter 4, Solution 19.
ΣFy = 0: C y + 0.8 ( 300 N ) = 0
2,4 4𝑘𝑁 − 2,4 1𝑘𝑁
𝐹!" =
= 7,2 𝑘𝑁 𝑻 𝑒𝑙 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑜 𝐶𝐸 𝑒𝑠𝑡𝑎 𝑒𝑛 𝑡𝑟𝑎𝑐𝑐𝑖ó𝑛
1
∴ C y = −240 N
or
C y = 240 N
Free-Body Diagram:
2
2
(a)
C = C 2 + C 2 of ( 380 ) +
ThenFrom free-body diagram = lever BCD ( 240 ) = 449.44 N
x
y
ΣM C = 0: TAB ( 50 mm ) − 200 N ( 75 mm ) = 0
𝑀! = C y ⎞
⎛ 0:
⎛ − 240 ⎞
and
θ = tan −1 ⎜ ⎟ = tan −1 ⎜
⎟ = 32.276°
∴ TAB = 300
⎜C ⎟
⎝ − 380 ⎠
⎝ x⎠
(b) From free-body diagram of lever BCD
or C = 449 N
32.3° ▹
−4,8 4𝑘𝑁 + 4,8 1𝑘𝑁 + 2,4 2𝑘𝑁 − 1Fcos 12,68 (𝐹!"Cx− 2,4 (sin 12,680 (𝐹!" ) = 0
Σ x = 0: 200 N + ) + 0.6 300 N ) =
𝐹!"
∴ C = −380 N
or
C x = 380 N
x
4,8 4𝑘𝑁 − 4,8 1𝑘𝑁 − 2,4 2𝑘𝑁
=
= −6,34 𝑘𝑁
Fy 0: C y +
1 cos 12,68 +Σ 2,4= sin 12,680.8 ( 300 N ) = 0
COSMOS: Complete Online Solutions Manual Organization System
∴ C y = −240 N
or
C y = 240 N
Chapter 4, Solution 19.
2
2
2
2
∴ 𝐹!" = 6,34 𝑘𝑁 𝑪 𝑒𝑙 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑜 𝐷𝐹 𝑒𝑠𝑡𝑎 𝑒𝑛 𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑖ó𝑛
449.44 N
C = C x + C y = ( 380 ) + ( 240 ) =
Then
Free-Body Diagram:
⎛ Cy ⎞
⎛ − 240 ⎞
(a)
and From free-body −diagram of lever BCD ⎟ = 32.276°
θ = tan 1 ⎜ ⎟ = tan −1 ⎜
⎟
⎜
⎝ 380 ⎠
ΣM C = 0:⎝ C xAB ( 50 mm )−− 200 N ( 75 mm ) = 0
T ⎠
𝑀! = 0:
or C = 449 N ∴32.3° = 300
TAB ▹
(b) From free-body diagram of lever BCD
2,4 1𝑘𝑁 − 2,4 4𝑘𝑁 + 1 634𝑘𝑁 − 1 N + 67,38 (𝐹300 N ) 0
0
ΣFx = 0: 200 sin Cx + 0.6 ( !" ) = =
∴ C = −380 N
C = 380 N
∴ C y = −240 N
𝐹!"
or
or
C y = 240 N
x
−2,4 1𝑘𝑁 + 2,4 4𝑘𝑁 − 1 x634𝑘𝑁
=
= N) = 0
−0,93 𝑘𝑁
ΣFy =
1 sin 12,68 0: C y + 0.8 ( 300
2
2
2
2
∴ 𝐹!" = 0,93 𝑘𝑁 𝑪 𝑒𝑙 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑜 𝐷𝐸 𝑒𝑠𝑡𝑎 𝑒𝑛 𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑖ó𝑛
449.44 N
C = C x + C y = ( 380 ) + ( 240 ) =
Then
and
⎛ Cy ⎞
− 240 ⎞
⎟ = tan −1 ⎛
⎟ = 32.276°
⎜
⎟
⎝ − 380 ⎠
⎝ Cx ⎠
θ = tan −1 ⎜
⎜
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.
or C = 449 N
32.3° ▹