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Centroide es el punto C, lo cual coincide con el centro de gravedad , punto donde actúa la fuerza resultante.
Primeros momentos Q:
…….Por simetría
…….Relación
Del grafico se cumple:
𝑑𝑢
2𝜋𝑢
𝑇𝑒𝑜𝑟𝑒𝑚𝑎 𝑑𝑒 𝑒𝑗𝑒𝑠 𝑃𝑎𝑟𝑎𝑙𝑒𝑙𝑜𝑠 𝑜 𝑆𝑡𝑒𝑖𝑛𝑒𝑟:
𝑑𝐼𝑥𝑦 = 𝑑𝐼𝑥 ′𝑦′ + 𝑥 𝑒𝑙
𝑦𝑒𝑙
𝑑𝐴
Solución
Integramos para la región
𝑑𝑥
𝑑𝐼𝑥 ′𝑦′ = 0 (𝑠𝑖𝑚𝑒𝑡𝑟𝑖𝑎 𝑑𝑒𝑙 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑜 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑐𝑖𝑎𝑙)
𝑥 ′
𝑦′
𝑦𝑒𝑙
𝑥 𝑒𝑙
𝑥 𝑒𝑙
= 𝑥
𝑦𝑒𝑙
=
𝑦
2
𝑑𝐴 = 𝑦 𝑑𝑥
𝑑𝐼𝑥𝑦 = 𝑥
𝑦
2
𝑦 𝑑𝑥 =
1
2
𝑥𝑦2
𝑑𝑥
𝑦2
=
4𝑎4
− 𝑥2
𝑎2
4𝑎2
𝑑𝐼𝑥𝑦 =
1
2
𝑥(
4𝑎4
− 𝑥2
𝑎2
4𝑎2
)𝑑𝑥
𝐴
𝑑𝐼𝑥𝑦 =
0
2𝑎
1
2
𝑥
4𝑎4
− 𝑥2
𝑎2
4𝑎2
𝑑𝑥 =
1
8𝑎2
0
2𝑎
(4𝑎4
𝑥 − 𝑥3
𝑎2
)𝑑𝑥
1
8𝑎2
2𝑎4
𝑥2
−
1
4
𝑥4
𝑎2
0
2𝑎
=
1
2
𝑎4
𝑑𝑦
𝑥 ′
𝑦′
𝑦𝑒𝑙
𝑥 𝑒𝑙
𝑑𝐼𝑥𝑦 = 𝑑𝐼𝑥 ′𝑦′ + 𝑥 𝑒𝑙
𝑦𝑒𝑙
𝑑𝐴
Integramos para la región
𝑦𝑒𝑙
= 𝑦
𝑥 𝑒𝑙
=
𝑥
2
𝑑𝐴 = 𝑥 𝑑𝑦
𝑑𝐼𝑥𝑦 =
𝑥
2
𝑥 𝑑𝑦 =
1
2
𝑦𝑥2
𝑑𝑦
𝑎, 𝑏 𝑝𝑒𝑟𝑡𝑒𝑛𝑒 𝑎 𝑙𝑎 𝑝𝑎𝑟𝑎𝑏𝑜𝑙𝑎
𝑥 = 𝑘𝑦2
→ 𝑥 =
𝑎
𝑏2
𝑦2
→ 𝑥2
=
𝑎2
𝑏4
𝑦4
𝐴
𝑑𝐼𝑥𝑦 =
0
𝑏
1
2
𝑦
𝑎2
𝑏4
𝑦4
𝑑𝑦 =
𝑎2
2𝑏4
0
𝑏
𝑦5
𝑑𝑦 =
𝑎2
𝑏2
12
Solución
𝑑𝐼𝑥 ′𝑦′ = 0 (𝑠𝑖𝑚𝑒𝑡𝑟𝑖𝑎 𝑑𝑒𝑙 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑜 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑐𝑖𝑎𝑙)
𝐼𝑚𝑎𝑥 = 𝐶 + 𝑅
𝐼𝑚𝑖𝑛 = 𝐶 − 𝑅
𝐼𝑚𝑎𝑥 = 𝐶 + 𝑅
𝐼𝑚𝑖𝑛 = 𝐶 − 𝑅
𝐶 =
𝐼𝑥 + 𝐼𝑦
2
𝑅 =
𝐼𝑥 − 𝐼𝑦
2
2
+ (𝐼𝑥𝑦)2
𝐼𝑥 − 𝐼𝑦
2
El circulo de Mohr queda definido con el radio y el centro:
El vector par es un vector libre, por lo tanto, resulta lógico situarlo en un lugar conveniente
para los cálculos.
Dirección de F:
(vector unitario)
Se cumple:
Parcial
Parcial
Parcial
Parcial

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Parcial

  • 1. Centroide es el punto C, lo cual coincide con el centro de gravedad , punto donde actúa la fuerza resultante.
  • 2.
  • 3.
  • 4.
  • 5.
  • 8.
  • 9.
  • 10.
  • 11.
  • 12.
  • 13.
  • 14.
  • 15.
  • 16. 𝑑𝑢 2𝜋𝑢 𝑇𝑒𝑜𝑟𝑒𝑚𝑎 𝑑𝑒 𝑒𝑗𝑒𝑠 𝑃𝑎𝑟𝑎𝑙𝑒𝑙𝑜𝑠 𝑜 𝑆𝑡𝑒𝑖𝑛𝑒𝑟:
  • 17.
  • 18.
  • 19. 𝑑𝐼𝑥𝑦 = 𝑑𝐼𝑥 ′𝑦′ + 𝑥 𝑒𝑙 𝑦𝑒𝑙 𝑑𝐴 Solución Integramos para la región 𝑑𝑥 𝑑𝐼𝑥 ′𝑦′ = 0 (𝑠𝑖𝑚𝑒𝑡𝑟𝑖𝑎 𝑑𝑒𝑙 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑜 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑐𝑖𝑎𝑙) 𝑥 ′ 𝑦′ 𝑦𝑒𝑙 𝑥 𝑒𝑙 𝑥 𝑒𝑙 = 𝑥 𝑦𝑒𝑙 = 𝑦 2 𝑑𝐴 = 𝑦 𝑑𝑥 𝑑𝐼𝑥𝑦 = 𝑥 𝑦 2 𝑦 𝑑𝑥 = 1 2 𝑥𝑦2 𝑑𝑥 𝑦2 = 4𝑎4 − 𝑥2 𝑎2 4𝑎2 𝑑𝐼𝑥𝑦 = 1 2 𝑥( 4𝑎4 − 𝑥2 𝑎2 4𝑎2 )𝑑𝑥 𝐴 𝑑𝐼𝑥𝑦 = 0 2𝑎 1 2 𝑥 4𝑎4 − 𝑥2 𝑎2 4𝑎2 𝑑𝑥 = 1 8𝑎2 0 2𝑎 (4𝑎4 𝑥 − 𝑥3 𝑎2 )𝑑𝑥 1 8𝑎2 2𝑎4 𝑥2 − 1 4 𝑥4 𝑎2 0 2𝑎 = 1 2 𝑎4
  • 20. 𝑑𝑦 𝑥 ′ 𝑦′ 𝑦𝑒𝑙 𝑥 𝑒𝑙 𝑑𝐼𝑥𝑦 = 𝑑𝐼𝑥 ′𝑦′ + 𝑥 𝑒𝑙 𝑦𝑒𝑙 𝑑𝐴 Integramos para la región 𝑦𝑒𝑙 = 𝑦 𝑥 𝑒𝑙 = 𝑥 2 𝑑𝐴 = 𝑥 𝑑𝑦 𝑑𝐼𝑥𝑦 = 𝑥 2 𝑥 𝑑𝑦 = 1 2 𝑦𝑥2 𝑑𝑦 𝑎, 𝑏 𝑝𝑒𝑟𝑡𝑒𝑛𝑒 𝑎 𝑙𝑎 𝑝𝑎𝑟𝑎𝑏𝑜𝑙𝑎 𝑥 = 𝑘𝑦2 → 𝑥 = 𝑎 𝑏2 𝑦2 → 𝑥2 = 𝑎2 𝑏4 𝑦4 𝐴 𝑑𝐼𝑥𝑦 = 0 𝑏 1 2 𝑦 𝑎2 𝑏4 𝑦4 𝑑𝑦 = 𝑎2 2𝑏4 0 𝑏 𝑦5 𝑑𝑦 = 𝑎2 𝑏2 12 Solución 𝑑𝐼𝑥 ′𝑦′ = 0 (𝑠𝑖𝑚𝑒𝑡𝑟𝑖𝑎 𝑑𝑒𝑙 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑜 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑐𝑖𝑎𝑙)
  • 21.
  • 22.
  • 23.
  • 24.
  • 25. 𝐼𝑚𝑎𝑥 = 𝐶 + 𝑅 𝐼𝑚𝑖𝑛 = 𝐶 − 𝑅
  • 26. 𝐼𝑚𝑎𝑥 = 𝐶 + 𝑅 𝐼𝑚𝑖𝑛 = 𝐶 − 𝑅 𝐶 = 𝐼𝑥 + 𝐼𝑦 2 𝑅 = 𝐼𝑥 − 𝐼𝑦 2 2 + (𝐼𝑥𝑦)2 𝐼𝑥 − 𝐼𝑦 2
  • 27. El circulo de Mohr queda definido con el radio y el centro:
  • 28.
  • 29.
  • 30.
  • 31.
  • 32.
  • 33. El vector par es un vector libre, por lo tanto, resulta lógico situarlo en un lugar conveniente para los cálculos.
  • 34.
  • 35.
  • 36.
  • 37.
  • 38.
  • 39.
  • 40.
  • 41.
  • 42. Dirección de F: (vector unitario) Se cumple: