2do Teorema de Castigliano Viga Hiperestática Reacciones y Diagramas

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Curso: Análisis Estructural I Tema: 2do Teorema de Castigliano
Diagrama de Cuerpo Libre [D.C.L.]
𝐷𝑒 𝑙𝑎 𝑠𝑖𝑔𝑢𝑖𝑒𝑛𝑡𝑒 𝑣𝑖𝑔𝑎, 𝑑𝑒𝑡𝑒𝑚𝑖𝑛𝑒 𝑙𝑎𝑠 𝑟𝑒𝑎𝑐𝑐𝑖𝑜𝑛𝑒𝑠 𝑒𝑛 𝑡𝑜𝑑𝑜𝑠 𝑙𝑜𝑠 𝑎𝑝𝑜𝑦𝑜𝑠. 𝐶𝑜𝑛𝑠𝑖𝑑𝑒𝑟𝑎𝑟 𝐸𝐼 = 𝑐𝑡𝑒.
𝐵𝑦
Corte 1-1
𝐴𝑦
𝑀𝐴
𝐴𝑥
Física con JoseAsesorías al Wsp: +51 902 188 071
1
1
𝐵𝑦
𝑀1
𝑉1
10
𝑥
𝑥/2
10𝑥
෍𝑀1 = 0
−𝑀1 − 10𝑥
𝑥
2
+ 𝐵𝑦(𝑥) = 0
𝑀1 = −5𝑥² + 𝐵𝑦𝑥
𝑑𝑀1
𝐵𝑦
= −5𝑥² + 𝐵𝑦𝑥
Derivada Parcial de M1 con respecto a By
𝑑𝑀1
𝐵𝑦
= 𝑥
Ecuación del Momento
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∆ = න
0
𝐿
𝑀 ∗
𝑑𝑀
𝑑𝑃
𝑑𝑥
𝐸𝐼
2do Teorema de Castigliano
0 =
1
𝐸𝐼
න
0
6
−5𝑥² + 𝐵𝑦𝑥 ∗ 𝑥 𝑑𝑥
0 =
1
𝐸𝐼
න
0
6
−5𝑥3 + 𝐵𝑦𝑥² 𝑑𝑥
0 = −5
𝑥4
4
+ 𝐵𝑦
𝑥3
3
0
6
0 = −5 ∗
64
4
+ 𝐵𝑦
63
3
− −5 ∗
04
4
+ 𝐵𝑦
03
3
0 = −1620 + 72 𝐵𝑦
𝐵𝑦 = 22.50 𝐾𝑁
Física con Jose
0 ≤ 𝑥 ≤ 6𝑚.
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https://www.youtube.com/@fisicaconjose/videos
Curso: Análisis Estructural I Tema: 2do Teorema de Castigliano
Diagrama de Cuerpo Libre [D.C.L.]
𝐷𝑒 𝑙𝑎 𝑠𝑖𝑔𝑢𝑖𝑒𝑛𝑡𝑒 𝑣𝑖𝑔𝑎, 𝑑𝑒𝑡𝑒𝑚𝑖𝑛𝑒 𝑙𝑎𝑠 𝑟𝑒𝑎𝑐𝑐𝑖𝑜𝑛𝑒𝑠 𝑒𝑛 𝑡𝑜𝑑𝑜𝑠 𝑙𝑜𝑠 𝑎𝑝𝑜𝑦𝑜𝑠. 𝐶𝑜𝑛𝑠𝑖𝑑𝑒𝑟𝑎𝑟 𝐸𝐼 = 𝑐𝑡𝑒.
22.50
𝐴𝑦
𝑀𝐴
𝐴𝑥
Física con JoseAsesorías al Wsp: +51 902 188 071
1
1
෍𝑀𝐴 = 0 𝑀𝐴 − 60(3) + 22.50(6) = 0 𝑀𝐴 = 45 𝐾𝑁 ∗ 𝑚
Ecuaciones del Equilibrio
10𝐾𝑁
𝑚
∗ 6𝑚 = 60 𝐾𝑁
3𝑚.
෍𝐹𝑦 = 0 𝐴𝑦 − 60 + 22.50 = 0 𝐴𝑦 = 37.50 𝐾𝑁
Corte 1-1
𝑀1 = −5𝑥² + 𝐵𝑦𝑥
𝑀1 = −5𝑥² + (22.50)𝑥
𝑑𝑀1 = −𝑉1
𝑑𝑀1 = −(−5𝑥2 + 22.50𝑥)
𝑉1 = 10𝑥 − 22.50
𝑉
+
−
𝑀
+
−
37.50
22.50
25.31
45
+
−
+
−
Física con Jose
0 ≤ 𝑥 ≤ 6𝑚.
2.25 𝑚.
3.75 𝑚.
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https://www.facebook.com/fisicaconjose
https://www.instagram.com/fisica_con_jose/
https://www.youtube.com/@fisicaconjose/videos
https://wa.me/51902188071
https://www.youtube.com/@fisicaconjose/videos
https://www.youtube.com/@fisicaconjose/videos
https://wa.me/51902188071
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