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EJERCICIO 13-57 Y 13-113
13-57
El eje macizo de aleación de aluminio 6061-T6 está fijo en un extremo
pero libre en el otro. Si el eje tiene un diámetro de 100 mm, determine su
máximo permitido longitud L si se somete a la fuerza excéntrica P = 80 kN
13-57
El eje macizo de aleación de aluminio 6061-T6 está fijo en un extremo
pero libre en el otro. Si el eje tiene un diámetro de 100 mm, determine su
máximo permitido longitud L si se somete a la fuerza excéntrica P = 80 kN
𝐴 = 𝜋 0.052
= 2.5 10−3
𝜋𝑚2
𝐼 =
𝜋
4
0.054
= 1.5625 10−6
𝜋𝑚4
𝑟 =
𝐼
𝐴
=
1.5625 10−6
2.5 10−3 = 0.025𝑚
𝑒 = 0.1𝑚 𝑐 = 0.05𝑚
𝑆𝑎𝑏𝑒𝑚𝑜𝑠
𝐾 = 2𝐿
𝑃𝑐𝑟 =
𝜋2
𝐸𝐼
𝐾𝐿 2
80 103
=
𝜋2
68.9 109
1.5625 10−6
𝜋
2𝐿 2
𝐿 = 3.230𝑚
13-57
El eje macizo de aleación de aluminio 6061-T6 está fijo en un extremo
pero libre en el otro. Si el eje tiene un diámetro de 100 mm, determine su
máximo permitido longitud L si se somete a la fuerza excéntrica P = 80 kN
𝜎𝑐𝑟 =
𝑃𝑐𝑟
𝐴
=
80 103
2.5 10−3 𝜋
= 10.19𝑀𝑃𝐴 < 𝜎𝑦 = 255𝑀𝑃𝐴
𝜎𝑚𝑎𝑥 =
𝑃
𝐴
1 +
𝑒𝑐
𝑟2
𝑠𝑒𝑐
𝐾𝐿
2𝑟
𝑃
𝐸𝐴
255(106
) =
80 103
2.5 10−3 𝜋
1 +
0.1(0.05)
0.0252 𝑠𝑒𝑐
2𝐿
2(0.0252)
80 103
68.9(109)(2.5(10−3)𝜋
𝐿 = 2.532𝑚 = 2.53𝑚
13-113
La columna de acero estructural A992 W8 * 15 es fija en su parte superior
e inferior. Si admite momentos finales de M = 23 kip · ft, determine la
fuerza axial P que puede ser aplicado. La flexión se trata del eje x - x. Usa
la interacción fórmula con (s b) = 24 ksi.
13-113
La columna de acero estructural A992 W8 * 15 es fija en su parte superior e inferior. Si
admite momentos finales de M = 23 kip · ft, determine la fuerza axial P que puede ser
aplicado. La flexión se trata del eje x - x. Usa la interacción fórmula con (s b) = 24 ksi.
𝐴 = 4.44𝑖𝑛2 𝐼𝑋 = 48𝑖𝑛4 𝑟𝑦 = 0.876𝑖𝑛 𝑑 = 8.11𝑖𝑛
𝐾𝐿
𝑟
=
2𝜋2𝐸
𝜎𝑦
=
2𝜋2(29)(103)
50
= 107
𝜎𝑝𝑒𝑟𝑚𝑖𝑡𝑖𝑑𝑜 =
12𝜋2
(29)(103
)
23 109.59 2 = 12.434𝑘𝑠𝑖
𝜎𝑎 =
𝑃
𝐴
= 0.2252𝑃
𝜎𝑏 =
𝑀𝑐
𝐼
= 23.316
𝜎𝑎
𝜎𝑎𝑝𝑒𝑟𝑚𝑖𝑡𝑖𝑑𝑜
+
𝜎𝑏
𝜎𝑏 𝑝𝑒𝑟𝑚𝑖𝑡𝑖𝑑𝑜
= 1
0.2252𝑃
12.434
+
23.316
24
= 1 𝑃 = 1.57𝑘𝑖𝑝
𝜎𝑎
𝜎𝑎𝑝𝑒𝑟𝑚𝑖𝑡𝑖𝑑𝑜
=
0.2252(1.52)
12.434
= 0.0284 < 0.15

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RESISTENCIA DE MATERIALES 13 57 y 13-113

  • 2. 13-57 El eje macizo de aleación de aluminio 6061-T6 está fijo en un extremo pero libre en el otro. Si el eje tiene un diámetro de 100 mm, determine su máximo permitido longitud L si se somete a la fuerza excéntrica P = 80 kN
  • 3. 13-57 El eje macizo de aleación de aluminio 6061-T6 está fijo en un extremo pero libre en el otro. Si el eje tiene un diámetro de 100 mm, determine su máximo permitido longitud L si se somete a la fuerza excéntrica P = 80 kN 𝐴 = 𝜋 0.052 = 2.5 10−3 𝜋𝑚2 𝐼 = 𝜋 4 0.054 = 1.5625 10−6 𝜋𝑚4 𝑟 = 𝐼 𝐴 = 1.5625 10−6 2.5 10−3 = 0.025𝑚 𝑒 = 0.1𝑚 𝑐 = 0.05𝑚 𝑆𝑎𝑏𝑒𝑚𝑜𝑠 𝐾 = 2𝐿 𝑃𝑐𝑟 = 𝜋2 𝐸𝐼 𝐾𝐿 2 80 103 = 𝜋2 68.9 109 1.5625 10−6 𝜋 2𝐿 2 𝐿 = 3.230𝑚
  • 4. 13-57 El eje macizo de aleación de aluminio 6061-T6 está fijo en un extremo pero libre en el otro. Si el eje tiene un diámetro de 100 mm, determine su máximo permitido longitud L si se somete a la fuerza excéntrica P = 80 kN 𝜎𝑐𝑟 = 𝑃𝑐𝑟 𝐴 = 80 103 2.5 10−3 𝜋 = 10.19𝑀𝑃𝐴 < 𝜎𝑦 = 255𝑀𝑃𝐴 𝜎𝑚𝑎𝑥 = 𝑃 𝐴 1 + 𝑒𝑐 𝑟2 𝑠𝑒𝑐 𝐾𝐿 2𝑟 𝑃 𝐸𝐴 255(106 ) = 80 103 2.5 10−3 𝜋 1 + 0.1(0.05) 0.0252 𝑠𝑒𝑐 2𝐿 2(0.0252) 80 103 68.9(109)(2.5(10−3)𝜋 𝐿 = 2.532𝑚 = 2.53𝑚
  • 5. 13-113 La columna de acero estructural A992 W8 * 15 es fija en su parte superior e inferior. Si admite momentos finales de M = 23 kip · ft, determine la fuerza axial P que puede ser aplicado. La flexión se trata del eje x - x. Usa la interacción fórmula con (s b) = 24 ksi.
  • 6. 13-113 La columna de acero estructural A992 W8 * 15 es fija en su parte superior e inferior. Si admite momentos finales de M = 23 kip · ft, determine la fuerza axial P que puede ser aplicado. La flexión se trata del eje x - x. Usa la interacción fórmula con (s b) = 24 ksi. 𝐴 = 4.44𝑖𝑛2 𝐼𝑋 = 48𝑖𝑛4 𝑟𝑦 = 0.876𝑖𝑛 𝑑 = 8.11𝑖𝑛 𝐾𝐿 𝑟 = 2𝜋2𝐸 𝜎𝑦 = 2𝜋2(29)(103) 50 = 107 𝜎𝑝𝑒𝑟𝑚𝑖𝑡𝑖𝑑𝑜 = 12𝜋2 (29)(103 ) 23 109.59 2 = 12.434𝑘𝑠𝑖 𝜎𝑎 = 𝑃 𝐴 = 0.2252𝑃 𝜎𝑏 = 𝑀𝑐 𝐼 = 23.316 𝜎𝑎 𝜎𝑎𝑝𝑒𝑟𝑚𝑖𝑡𝑖𝑑𝑜 + 𝜎𝑏 𝜎𝑏 𝑝𝑒𝑟𝑚𝑖𝑡𝑖𝑑𝑜 = 1 0.2252𝑃 12.434 + 23.316 24 = 1 𝑃 = 1.57𝑘𝑖𝑝 𝜎𝑎 𝜎𝑎𝑝𝑒𝑟𝑚𝑖𝑡𝑖𝑑𝑜 = 0.2252(1.52) 12.434 = 0.0284 < 0.15