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Ejerccio 9 enrico
- 1. Mecánica Estática
Ejercicios
Fuerza Cortante y Momento Flector
1.- Determine: a) La Fuerza Cortante b) Momento Flector para una viga simplemente
apoyada c) Grafique Diagrama Fuerza Cortante d) Grafique Momento Flector
Respuesta: Diagrama de Cuerpo Libre
∑ 𝐹𝑥 = 0 𝑨 𝒙 = 𝟎
∑ 𝐹𝑦 = 0
𝑨 𝒚 + 𝑩 𝒀 − 𝟓𝑲𝒏 = 𝟎 (𝑰)
∑ 𝑀𝐴 = 0
(−5𝐾𝑛)(20𝑚) + 𝐵 𝑦(25𝑚) = 0
𝐵 𝑦 =
5𝐾𝑛 ∙ 20𝑚
25𝑚
= 𝟒𝑲𝒏 (𝑰𝑰)
𝑺𝒖𝒔𝒕𝒊𝒕𝒖𝒚𝒆𝒏𝒅𝒐 ( 𝑰𝑰) 𝒆𝒏 ( 𝑰) 𝒕𝒆𝒏𝒆𝒎𝒐𝒔
𝐴 𝑦 = 5𝐾𝑛 − 4𝐾𝑛 𝑨 𝒚 = 𝟏𝑲𝒏
20m
A
B
5kN
C
5m
- 2. Luego
∑ 𝐹𝑦 = 0
1𝐾𝑛 − 𝑉1 = 0 𝑽 𝟏 = 𝟎
∑ 𝑀1 = 0
1𝐾𝑛 ∙ 0𝑚 + 𝑀1 = 0 𝑴 𝟏 = 𝟎
∑ 𝐹𝑦 = 0
−𝑉2 + 1𝐾𝑛 = 0 𝑽 𝟐 = 𝟏𝑲𝒏
∑ 𝑀2 = 0
-1𝐾𝑛 ∙ 20𝑚 + 𝑀2 = 0 𝑴 𝟐 = 𝟐𝟎𝑲𝒏
∑ 𝐹𝑦 = 0
1𝐾𝑛 − 5𝐾𝑛 − 𝑉3 = 0 𝑽 𝟑 = −𝟒𝑲𝒏
∑ 𝑀3 = 0
-1𝐾𝑛 ∙ 20𝑚 + 5𝐾𝑛(0) + 𝑀3 = 0 𝑴 𝟑 = 𝟐𝟎𝑲𝒏
∑ 𝐹𝑦 = 0
1𝐾𝑛 − 5𝐾𝑛 − 𝑉4 = 0 𝑽 𝟒 = −𝟒𝑲𝒏
∑ 𝑀4 = 0
-1𝐾𝑛 ∙ 25𝑚 + 5𝐾𝑛 ∙ 5𝑚 + 𝑀4 = 0 𝑴 𝟒 = 𝟎
- 3. 2) Determine Reacciones en los APOYOS y Dibuje Diagrama de Fuerza Cortante y
Momento Flector de la viga mostrada a continuación, si F= 4650 Lb
Respuesta Diagrama de Cuerpo Libre
𝐹1 =
3
2
∙ 𝑊
𝐹1 =
3
2
∙ (400𝐿𝑏) = 𝟔𝟎𝟎𝑳𝒃
∑ 𝑀𝐴 = 0
−4650𝐿𝑏 ∙ (3𝑝𝑖𝑒𝑠) + 𝑅 𝐵 ∙ (6𝑝𝑖𝑒𝑠) + 1.5 = 0
𝑅 𝐵 =
4650(3) − 600
6
𝑹 𝑩 = 𝟐𝟐𝟐𝟓𝑳𝒃
∑ 𝐹𝑦 = 0
𝑅 𝐴 + 𝑅 𝐵 − 𝐹1 − 4650 = 0
𝑅 𝐴 = 𝐹1 + 4650 − 𝑅 𝐵 𝑆𝑢𝑠𝑡𝑖𝑡𝑢𝑦𝑒𝑛𝑑𝑜 𝑦 𝑒𝑣𝑎𝑙𝑢𝑎𝑛𝑑𝑜
𝑹 𝑨 = 𝟑𝟎𝟐𝟓𝑳𝒃
F
3pie 3pie 3pie
w = 400x
- 4. ∑ 𝐹𝑦 = 0
−600 − 𝑉1 = 0 𝑽 𝟏 = 𝟔𝟎𝟎𝑳𝒃
∑ 𝑀1 = 0
600 ∙ 0𝑚 + 𝑀1 = 0 𝑴 𝟏 = 𝟎
∑ 𝐹𝑦 = 0
−600 − 𝑉2 = 0 𝑽 𝟐 = −𝟔𝟎𝟎𝑳𝒃
∑ 𝑀2 = 0
-600 ∙ 1𝑝𝑖𝑒 + 𝑀2 = 0 𝑴 𝟐 = −𝟔𝟎𝟎𝒑𝒊𝒆𝒔
∑ 𝐹𝑦 = 0
−600 + 3025 − 𝑉3 = 0 𝑽 𝟑 = 𝟐𝟒𝟐𝟓𝑳𝒃
∑ 𝑀3 = 0
-600 ∙ 1𝑝𝑖𝑒 + 3025(0)+ 𝑀3 = 0 𝑴 𝟑 = −𝟔𝟎𝟎𝒑𝒊𝒆𝒔
∑ 𝐹𝑦 = 0
−600 + 3025 − 𝑉4 = 0 𝑽 𝟒 = 𝟐𝟒𝟐𝟓𝑳𝒃
∑ 𝑀4 = 0
-600 ∙ 4𝑚 − 3025 ∙ 3𝑚 + 𝑀4 = 0
𝑀4 = 600 ∙ 4 + 3025 ∙ 3𝑚 𝑴 𝟒 = 𝟔𝟔𝟕𝟓𝑳𝒃/𝑷𝒊𝒆
- 5. ∑ 𝐹𝑦 = 0
−600 + 3025 − 4650 − 𝑉5 = 0
𝑉5 = 600 − 3025 + 4650 𝑽 𝟓 = −𝟐𝟐𝟐𝟓𝑳𝒃
∑ 𝑀5 = 0
600 ∙ 4 − 3025 ∙ 3𝑚 + 𝑀5 = 0 𝑴 𝟓 = 𝟔𝟔𝟕𝟓𝑳𝒃/𝑷𝒊𝒆
∑ 𝐹𝑦 = 0
−600 − 4650 − 𝑉6 + 3025 = 0 𝑽 𝟔 = −𝟐𝟐𝟓𝑳𝒃
∑ 𝑀6 = 0
600 ∙ 7𝑚 − 3025 ∙ 6𝑚 + 4650 ∙ 3𝑚 + 𝑀6 = 0
𝑀6 = −4200 − 13950 + 18150 𝑴 𝟔 = 𝟎