Examen n7

1. Ejercicio Práctico Único: Para la viga y condiciones de carga mostradas en la figura:
a) Diagrama de cuerpo libre de la viga con sus reacciones;
b) Cálculo de las reacciones en los apoyos;
c) Diagramas de fuerza cortante y momento flexionante o flector;
d) Momento de Inercia de la viga;
e) Ubicación del eje neutro de la viga; y
f) determínense los esfuerzos máximos de tensión y de compresión de la viga.
25 mm
25 mm

2. ody Diagram:
r 4, Solution 19.
b)
𝐹! = 0 ,
𝑅! =
12
𝑘𝑁
𝑚
0,9!!!
32.3° ▹
Fuerza Homogénea
1,8!!!
or C = 449 N
⎛ Cy ⎞
− 240 ⎞
⎟ = tan −1 ⎛
⎟ = 32.276°
⎜
⎟
⎝ − 380 ⎠
⎝ Cx ⎠
= 449.44 N
0,3!
θ = tan −1 ⎜
⎜
( 380 )2 + ( 240 )2
C y = 240 N
!!
𝑅! + 𝑅! = 𝟔𝟗, 𝟔 𝒌𝑵
and
2
2
Cx + C y =
or
0,6!!!
C =
𝐷𝑒𝑏𝑖𝑑𝑜 𝑎 𝑙𝑎 𝑠𝑖𝑚𝑒𝑡𝑟𝑖𝑎 𝑑𝑒𝑙 𝑠𝑖𝑠𝑡𝑒𝑚𝑎 𝑹 𝑨 = 𝑹 𝑩
∴ C y = −240 N
C x = 380 N
∴ TAB = 300
!!
Then
or
ΣFy = 0: C y + 0.8 ( 300 N ) = 0
∴ C x = −380 N
ΣFx = 0: 200 N + Cx + 0.6 ( 300 N ) = 0
(b) From free-body diagram of lever BCD
ΣM C = 0: TAB ( 50 mm ) − 200 N ( 75 mm ) = 0
(a) From free-body diagram of lever BCD
: Complete Online Solutions Manual Organization System
a)
1,8 𝑚
= 𝟐𝟏, 𝟔 𝒌𝑵
21,6!!"!
24!!"!
24!!"!
!!
0,9!!!
!!
0,3!
69,6
𝑘𝑁 = 𝟑𝟒, 𝟖 𝒌𝑵
2
0,6!!!
𝑅! + 𝑅! − 24 + 21,6 + 24 𝑘𝑁 = 0

3. (b) From free-body diagram of lever BCD
ΣM C = 0: TAB ( 50 mm ) TAB = 300
∴ − 200 N ( 75
ΣFx = 0: 200 N + Cx + 0.6 ( 300 N ) = 0
ΣFx = 0: 200 N (b) xFrom free-body= 0
+ C + 0.6 ( 300 N ) diagram of lever BCD
∴ C x = −380 N
or
C x = 380 N
∴ C x = −380 N
or ΣF = x0:= 380 (b) +From free-body diagram of lever BCD
C
200 N Cx + 0.6 ( 300 N ) = 0
N
x
ΣFy = 0: C y + 0.8 ( 300 N ) = 0
ΣF = 0: C y + 0.8 ( 300 N ) = 0
(a) From free-body diagram of lever BCD
ΣFx = 0: x 200 N + Cx + 0.6 ( 300 N )
∴ C x = −380 N
or
C = 380 N
∴ C y = −240 N
or y
C y = 240 N
ΣM C = 0: TAB ( 50 mm ) − 200 N ( 75 mm ) = 0
∴ C y = −240 N
or ΣFy C y 0: 240 N 0.8 ( 300 N ) = 0 ∴ C x = −380 N
or
= = Cy +
2
2
2
2
C = C x + C y = ∴(T )= + ( 240 ) = 449.44 N
380 300
Then
AB
2
2
ΣF
2
2
or y = 0: y C y + 0.8 ( 300 N ) = 0
C = 240 N
C = C x + C y = ( 380 ) + ( 240 ) = ∴ C y =N 240 N
449.44 −
Then
(b) From free-body diagram of lever BCD
⎛ Cy ⎞
⎛ − 240 ⎞
or
2
2 ∴ C y = −240 N
and
θ = tan −1 ⎜ ⎟ = tan −1 ⎜
2
2
⎟ = 32.276°
C = Cx
Then
ΣFx = 0: 200 N + Cx + 0.6 ( 300 N ) = ⎜ C x ⎟ and ⎝ − 380 ⎠= tan −1 ⎛ C y ⎞ = tan −1 ⎛ − 240 ⎞ = 32.276° + C y = ( 380 ) + ( 240 ) = 449.44 N
0 ⎠
⎝
⎟
⎜
θ
⎟
⎜
⎜C ⎟
2
2
2
2
− 380 ⎠
C
⎠
⎝
or C =x 449 N ⎝32.3° ▹
⎛ C y ⎞ Then −1 ⎛ − 240 ⎞ = C x + C y = ( 380 ) + ( 240 )
∴ C x = −380 N
or
C x = 380 N
and
θ or tan −=⎜ 449⎟N tan32.3° ▹ ⎟ = 32.276°
= C 1⎜
⎜
⎟=
⎝ − 380 ⎠
⎝ Cx ⎠
⎛ Cy ⎞
⎛ − 240 ⎞
ΣFy = 0: C y + 0.8 ( 300 N ) = 0
and
θ = tan −1 ⎜ ⎟ = tan −1 ⎜
⎟
⎟
⎜ C 449 N ⎝32.3° ▹= 32.2
− 380 ⎠
or C =x ⎠
⎝
∴ C y = −240 N
or
C y = 240 N
or
2
2
2
2
C = C x + C y = ( 380 ) + ( 240 ) = 449.44 N
Then
Free-Body Diagram:
(a) From free-body diagram of lever BCD
Chapter 4, Solution 19.
(a) From 200 N
ΣM C = 0: Free-Body Diagram: ( 75 mm ) = of lever BCD
TAB ( 50 mm ) −free-body diagram 0
ΣM C = 0: TAB ( 50 mm ) − 200 N ( 75 mm ) = 0
(a) Diagram:
TAB free-body diagram of lever BCD
Free-Body∴From = 300
TAB 300
ΣM = 0: T ( 50∴From =200 N ( 75 mm ) = of lever BCD
(b) From free-body diagram of lever BCD
(a) mm ) −free-body diagram 0
𝑹 𝑨 = 𝑹 𝑩 = 𝟑𝟒, 𝟖 𝒌𝑵
: Complete Online Solutions Manual Organization System
c) Se consideran 5 secciones en la viga:
COSMOS: Complete Online Solutions Manual Organization System
er 4, Solution 19.
AB
Body Diagram:
1.
𝐹! = 0 , 34,8 𝑘𝑁 − 𝑉! = 0 ⇒ 𝑉! = 𝟑𝟒, 𝟖 𝒌𝑵
(a) From free-body diagram of lever BCD
ΣM C = 0: TAB ( 50 mm ) − 200 N ( 75 mm ) = 0
𝑀! = 0 , 34,8 𝑘𝑁 0 𝑚 + 𝑀! = 0 ⇒ 𝑀! = 𝟎
ion 19.
m:
C
ne Solutions Manual Organization System
∴ TAB = 300
(b) From free-body diagram of lever BCD
ΣFx = 0: 200 N + Cx + 0.6 ( 300 N ) = 0
2.
∴ C = −380 N
or
C = 380 N
x
𝐹! = 0 , x 34,8 − 24 𝑘𝑁 − 𝑉! = 0 ⇒ 𝑉! = 𝟏𝟎, 𝟖 𝒌𝑵
ΣFy = 0: C y + 0.8 ( 300 N ) = 0
(a) From free-body diagram of lever BCD
∴ ) y 200 N 75
ΣM = 0: T ( 50 mmC − = −240(N mmor = 0 C y = 240 N
)
C
AB
𝑀! = 0 , 34,8 0,6 𝑘𝑁 ∙ 𝑚 + 𝑀! = 0 ⇒ 𝑀! = −𝟐𝟎, 𝟖𝟖 𝒌𝑵 ∙ 𝒎
2
2 ∴ T
2
2
AB = 300
C = C x + C y = ( 380 ) + ( 240 ) = 449.44 N
Then
(b) From free-body diagram of lever BCD
⎛ Cy ⎞
⎛ − 240 ⎞
andFx = 0: 200 tan+1Cx + ⎟0.6tan −1 ⎜N ) = 0⎟ = 32.276°
θ = N − ⎜ ⎟ = ( 300
Σ
⎜C
⎝ − 380 ⎠
⎝ x⎠
3.
∴ C x = −380 N
or
C x = 380 N
or C = 449 N
32.3° ▹
ΣFy = 𝐹 =C y + 0.834,8 N ) 240 21,6 𝑘𝑁 − 𝑉! = 0 ⇒ 𝑉! = −𝟏𝟎, 𝟖 𝒌𝑵
0: 0 , ( 300 − = −
!
32.3° ▹
(a) From free-body diagramC y lever BCD
∴ of = −240 N
C y = 240 N
or
ΣM C = 0: TAB ( 50 2mm ) − 2002N ( 75 mm ) = 0
0, x + C y = 1,8 − 24240 )2 = 𝑘𝑁 ∙ 𝑚N 𝑀! = 0 ⇒ 𝑀! = −𝟑𝟑, 𝟖𝟒 𝒌𝑵 ∙ 𝒎
C 2 34,8
( 380 ) + ( 1,2 449.44 +
∴ TAB = 300
⎛ Cof⎞lever BCD 240 ⎞
−
(b) and free-body=diagram y ⎟ = tan −1 ⎛
From
θ tan −1 ⎜ ⎟
⎟ = 32.276°
⎜
⎜C
⎝−
x
⎝ N ⎠ C + 0.6 380 ⎠N = 0
4. Fx = 0: 200 + x
Σ
( 300 )
or C = 449 N
32.3° ▹
∴ C x = −380 N
or
C x = 380 N
𝐹! = 0 , 34,8 − 24 − 21,6 − 24 𝑘𝑁 − 𝑉! = 0 ⇒ 𝑉! = −𝟑𝟒, 𝟖 𝒌𝑵
ΣFy = 0: C y + 0.8 ( 300 N ) = 0
(a) From free-body diagram of lever BCD
∴ C = ( 50 N
ΣM = 0: y T −240 mm ) −or N ( 75 = 240 = 0
200 C y mm ) N
C
AB
𝑀! = 0, 34,8 3 − 24 2,4 − 21,6 1,2 𝑘𝑁 ∙ 𝑚 + 𝑀! = 0
2
2
C = C x + C y = ( 380 ) + ( 240 ) = 449.44 N ∴ TAB = 300
Then
⇒ 𝑀! = −𝟐𝟎, 𝟖𝟖 𝒌𝑵 ∙ 𝒎
(b) From free-body diagram of lever BCD
⎛ Cy ⎞
⎛ − 240 ⎞
and
tan =⎜
θ =ΣFx −1 ⎜ 0: ⎟200 N−1+ Cx + 0.6 (32.276) = 0
⎟ = tan ⎜ − 380 ⎟ = 300 N °
⎠
⎝
⎝ Cx ⎠
5.
∴ C x = −380 N
or
C x = 380 N
or C = 449 N
32.3° ▹
ΣFy = 0: C y + 0.8 ( 300 N ) = 0 𝐹 = 0 ⇒ 𝑉! = 𝟎
!
lutions Manual Organization System
2
∴ C y = −240 N
Then
C =
and
θ = tan −1 ⎜
⎜
19.
Free-Body Diagram:
⎛ Cy ⎞
− 240 ⎞
⎟ = tan −1 ⎛
⎟ = 32.276°
⎜
⎟
⎝ − 380 ⎠
⎝ Cx ⎠
m:
2
2
Cx + C y =
2
or
( 380 )2 + ( 240 )2
C y = 240 N
= 449.44 N
⎛ Cy ⎞
− 240 ⎞
⎟ = tan −1 ⎛
⎟ = 32.276°
⎜
Cx ⎟
⎝ − 380 ⎠
⎠
⎝
θ = tan −1 ⎜
⎜
n 19.
and
Chapter 4, Solution 19.
𝑀! =
Then
Solutions Manual Organization System C =
or C = 449 N
19.
Chapter 4, Solution 19.
utions Manual Organization System

4. hapter 4, Solution 19.
Free-Body Diagram:
(a) From free-body diagram of lever BCD
ΣM C = 0: TAB ( 50 mm ) − 200 N ( 75 mm ) = 0
𝑀! = 0 , 𝑃𝑜𝑟 𝑠𝑖𝑚𝑒𝑡𝑟𝑖𝑎 ⇒ 𝑀! = 𝟎
∴ TAB = 300
(b) From free-body diagram of lever BCD
ΣFx = 0: 200 N + Cx + 0.6 ( 300 N ) = 0
∴ C x = −380 N
1
2
34,8!!"!
C x = 380 N
or
ΣFy = 0: C y + 0.8 ( 300 N ) = 0
∴ C y = −240 N
3
Then
C =
2
2
Cx + C y =
and
5
4
C y = 240 N
or
( 380 )2 + ( 240 )2
= 449.44 N
θ = tan −1 ⎜
⎜
⎛ Cy ⎞
− 240 ⎞
⎟ = tan −1 ⎛
⎟ = 32.276°
⎜
⎟
⎝ − 380 ⎠
⎝ Cx ⎠
or C = 449 N
32.3° ▹
−34,8!!"!
0,6!!! 0,3!
0,9!!!
0,9!!! 0,3! 0,6!!!
ctor Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
iot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
2007 The McGraw-Hill Companies.
−33,84!!" ∙ !!

5. d)
𝑆𝑖 𝑟 = 𝟐𝟓 𝒎𝒎
𝑏 = 25 𝑥 2 = 𝟓𝟎 𝒎𝒎
!! !
25!!!!
!!
1
!! !
25!!!!
2
Áreas
𝜋𝑟 !
𝜋 25 𝑚𝑚
𝐴! =
=
2
2
!
= 𝟗𝟖𝟏, 𝟕𝟒 𝒎𝒎 𝟐
𝐴! = 𝑏ℎ = 50 𝑚𝑚 𝑥 25 𝑚𝑚 = 𝟏𝟐𝟓𝟎 𝒎𝒎 𝟐
Centroide áreas
𝑦! =
4𝑟
4 𝑥 25 𝑚𝑚
=
= 𝟏𝟎, 𝟔𝟏 𝒎𝒎
3𝜋
3𝜋
𝑦! = −
ℎ 25 𝑚𝑚
=
= −𝟏𝟐, 𝟓 𝒎𝒎
2
2

6. Centroide de la figura
𝑦=
𝐴! 𝑦! + 𝐴! 𝑦!
=
𝐴! + 𝐴!
981,74 10,61 + 1250 −125 𝑚𝑚!
= −𝟐, 𝟑𝟑𝟒 𝒎𝒎
981,74 + 1250 𝑚𝑚!
Así enocntramos responder e) que es la ubicación del eje neutro, y este se
encuentra en el centroide de la figura.
𝐼! = 𝐼𝑥! − 𝐴! 𝑦!
!
𝜋𝑟 !
𝜋 25 𝑚𝑚
=
− 𝐴! 𝑦! ! =
8
8
!
− 981,74 𝑚𝑚! 10,61 𝑚𝑚
!
𝑰 𝟏 = 𝟒𝟐𝟖𝟖𝟏, 𝟓𝟒 𝒎𝒎 𝟒
𝑑! = 𝑦! − 𝑦 = 10,61 𝑚𝑚 + 2,334 𝑚𝑚 = 𝟏𝟐, 𝟗𝟒𝟒 𝒎𝒎
𝑰 𝟏 = 𝐼! + 𝐴! 𝑑! ! = 42881,54 𝑚𝑚! + 981,74 𝑚𝑚! 12,944
𝐼! =
𝑏ℎ!
50 𝑚𝑚 25 𝑚𝑚
=
12
12
!
!
= 𝟐𝟎𝟕𝟑𝟔𝟗, 𝟐𝟔 𝒎𝒎 𝟒
= 𝟔𝟓𝟏𝟎𝟒, 𝟏𝟔 𝒎𝒎 𝟒
𝑑! = 𝑦! − 𝑦 = −12,5 𝑚𝑚 + 2,334 𝑚𝑚 = 10,166 𝑚𝑚
𝑰 𝟐 = 𝐼! + 𝐴! 𝑑! ! = 65104,16 𝑚𝑚! + 1250 𝑚𝑚! 10,166
!
= 𝟏𝟗𝟒𝟐𝟖𝟖, 𝟔 𝒎𝒎 𝟒
El momento de Inercia de la pregunta d) seria:
𝐼 = 𝑰 𝟏 + 𝑰 𝟐 = 207369,26 𝑚𝑚! + 194288,6 𝑚𝑚! = 𝟒𝟎𝟏𝟔𝟓𝟕, 𝟖𝟔 𝒎𝒎 𝟒
𝑰 = 𝟒𝟎𝟏, 𝟔𝟓 𝒙 𝟏𝟎!𝟗 𝒎𝒎 𝟒
f) Esfuerzo máximo de tensión y compresión
𝑦! = 25 𝑚𝑚 + 2,334 𝑚𝑚 = 27,334 𝑚𝑚 = 𝟎, 𝟎𝟐𝟕𝟑𝟑𝟒 𝒎
𝑦! = −25 𝑚𝑚 + 2,334 𝑚𝑚 = −22,666 𝑚𝑚 = 𝟎, 𝟎𝟐𝟐𝟔𝟔𝟔 𝒎
𝜎! = −
𝑀𝑦!
−33,84 𝑘𝑁 ∙ 𝑚 0,027334 𝑚
=−
= 𝟐, 𝟑 𝑮𝑷𝒂
𝐼
401,65 𝑥 10!! 𝑚!

7. 𝜎! = −
𝑀𝑦!
−33,84 𝑘𝑁 ∙ 𝑚 −0,022666 𝑚
=−
= −𝟏, 𝟗 𝑮𝑷𝒂
𝐼
401,65 𝑥 10!! 𝑚!