1) The document describes calculating forces in the reinforcement bars of a small structure using a method of sections. 2) Free body diagrams are drawn and equilibrium equations are applied to solve for the forces in each section. 3) The forces are calculated in sections AD, CD, and CE and the reinforcement bars are designed accordingly.

1. Usaremos el metodo de las secciones debido a que calcularemos las fuerzas de
un pequeño numero de barras de la armadura
Primero dibujamos el diagrama de cuerpo libre:
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20!!"!
20!!"!
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36!!"!
2,4!!!
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omplete Online Solutions Manual Organization System
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B
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4, Solution 19.
4,5!!!
4,5!!!
4,5!!!
COSMOS: Complete Online Solutions Manual Organization System
dy Diagram:
Ahora aplicando las ecuaciones de equilibrio tenemos:
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
2,4
9 50 N −
ΣFx = 0:36 ΣM C +=C20 TAB (+ 20 )4,50 N ( 75 mm ) = 0
200 N +0: + 0.6 300mm )= 200
x
𝐵 =
= 26,4 𝑘𝑁 ↑
(a) From free-body diagram 13,5 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
ΣFy = 𝐾! = y 36 𝑘𝑁 → N ) = 0
0: C + 0.8 ( 300
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! = 40 − 26,4 = 13,6 𝑘𝑁 ↑
Como ya se calculo el valor en cada componente de las reacciones,
seccionaremos la armadura en los elementos AD, CD y CE:
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36!!"! !!
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17
15!
1,2!!!
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15!
17
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2,25!!!
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26,4!!"!
ter 4, Solution 19.
4,5!!!
Ahora aplicando nuevamente las ecuaciones de equilibrio tenemos:
-Body Diagram:
(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+ Cx 26,4 ( 300 N ) = 0
200 N − + 0.6 2,25
𝐹!" =
Chapter 4, Solution 19.
Free-Body Diagram:
∴ C x = −3801,2
N
= −13,5 𝑘𝑁
or
C x = 380 N
ΣFy = 0: 𝒆𝒍 𝒆𝒍𝒆𝒎𝒆𝒏𝒕𝒐 𝒔𝒆 𝒆𝒏𝒄𝒖𝒆𝒏𝒕𝒓𝒂 𝒆𝒏 𝒄𝒐𝒎𝒑𝒓𝒆𝒔𝒊𝒐𝒏
𝐹!" = 13,5 𝑘𝑁 𝑪 C y + 0.8 ( 300 N ) = 0
Then
and
(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
θ = tan −1 ⎜ ⎟ = tan −1 ⎜𝐹!" = ⎟ = 32.276°
⎜C ⎟
⎝N+C
⎠
⎝
ΣFx x = 0: 200 − 380 ⎠x + 0.6 ( 300 N ) = 0
32.3°
∴ C x = −380 N or C = 449 N = 380 N ▹
or
Cx
ΣFy = 0: C y + 0.8 ( 300 N ) = 0
∴ C y = −240 N
Then
C =
2
2
Cx + C y =
or
( 380 )2 + ( 240 )2
C y = 240 N
= 449.44 N

3. ter 4, Solution 19.
-Body Diagram:
(a) From free-body diagram of lever BCD
15
ΣM C = 0: TAB ( 50 mm ) − 200 N ( 75 mm ) = 0
𝑀! = 0:
17
𝐹!"
2,4 − 26,4 4,5 = 0
∴ TAB = 300
(b) From free-body diagram of lever BCD
ΣFx = 0: 200 17 Cx + 0.64,5 N ) = 0
N + 26,4 ( 300
𝐹!" =
15
∴ C x = −380 N 2,4 or
= 56,1 𝑘𝑁
C x = 380 N
ΣFy = 0: C y + 0.8 ( 300 N ) = 0
𝐹!" = 56,1 𝑘𝑁 𝑻 𝒆𝒍 𝒆𝒍𝒆𝒎𝒆𝒏𝒕𝒐 𝒔𝒆 𝒆𝒏𝒄𝒖𝒆𝒏𝒕𝒓𝒂 𝒆𝒏 𝒕𝒓𝒂𝒄𝒄𝒊𝒐𝒏
∴ C = −240 N
or
C = 240 N
y
Then
C =
2
2
Cx + C y =
and
y
( 380 )2 + ( 240 )2
= 449.44 N
θ = tan −1 ⎜
⎜
⎛ Cy ⎞
− 240 ⎞
⎟ = tan −1 ⎛
⎟ = 32.276°
⎜
⎟
⎝ − 380 ⎠
⎝ Cx ⎠
Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
The McGraw-Hill Companies.
or C = 449 N
32.3° ▹