Uploaded by Alexander AchiΓ±a

PREESFUERZO EJE AXIAL OTRO

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PESO PROPIO
𝑀 =𝛾∗𝐴 =
2.4 𝑇
𝑇
π‘˜π‘”
∗ (0.15 π‘š ∗ 0.15 π‘š) = 0.054 = 0.54
3
π‘š
π‘š
π‘π‘š
MOMENTO PESO PROPIO
𝑅𝐴 = −0.54 ∗ 25 = 13.5 π‘˜π‘”
𝑅𝐴 = 𝑅𝐡 = 13.5 π‘˜π‘”
𝑀𝐺 = −0.54 ∗ 25 ∗ 12.5 + 13.5 ∗ 22.5
𝑀𝐺 = 135 π‘˜π‘” − π‘π‘š
CARGA PUNTUAL
𝑃 = 26142 π‘˜π‘”
𝑃 26142
=
= 13071 π‘˜π‘”
2
2
REACCIONES
𝑅𝐴 = −0.54 ∗ 25 − 13071 = 13084.5 π‘˜π‘”
𝑅𝐴 = 𝑅𝐡 = 13084.5 π‘˜π‘”
MOMENTO MÁXIMO
𝑀𝑇 = −0.54 ∗ 25 ∗ 12.5 + 13084.5 ∗ 22.5 − 13071 ∗ 7.5
𝑀𝑇 = 196200 π‘˜π‘” − π‘π‘š
CÁLCULO DE C
𝑐1 = 𝑐2 = 7.5 π‘π‘š
CÁLCULO DE Z
𝐼=
𝑧1 =
15 ∗ 153
= 4218.75 π‘π‘š4
12
𝐼 4218.75
=
= 562.5 π‘π‘š3
𝑐
7.5
CÁLCULO DE K
𝐴 = 15 ∗ 15 = 225 π‘π‘š2
π‘˜2 =
CÁLCULO DE F
𝑧1 562.5
=
= 2.5 π‘π‘š
𝐴
225
𝛼 = 𝑒 + π‘˜1
𝛼 = 0 + 2.5 = 2.5
𝑀𝑇 = 𝐹 ∗ 𝛼
𝐹=
𝑀𝑇 196200
=
= 78480 π‘˜π‘”
𝛼
2.5
CÁLCULO DE Fo
DATOS
𝑏 = 15 π‘π‘š
β„Ž = 15 π‘π‘š
𝐴 = 15 ∗ 15 = 225 π‘π‘š2
𝐢1 = 𝐢2 = 7.5 π‘π‘š
𝐼=
15 ∗ 153
= 4218.75 π‘π‘š4
12
𝑍1 = 𝑍2 =
𝐼
4218.75
=
= 562.5 π‘π‘š3
𝐢1
7. .5
𝐹 = 78480 π‘˜π‘”
𝑀𝐺 = 135 π‘˜π‘” − π‘π‘š
𝑀𝑇 = 196200 π‘˜π‘” − π‘π‘š
ETAPA INICIAL
𝜎1 = −
πΉπ‘œ 𝑀𝐺
−
𝐴
𝑍1
𝜎2 = −
πΉπ‘œ 𝑀𝐺
+
𝐴
𝑍2
ETAPA FINAL
𝜎1 = −
𝐹 𝑀𝑇
−
𝐴 𝑍1
𝜎2 = −
𝐹 𝑀𝑇
+
𝐴 𝑍2
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