Capitulo 5 Ecuacion de Los Tres Momentos y Metodos de Angulos de Giro y Deflexion
DEFLEXION
1
d) Determinar la deflexión en el centro del tramo 4-5 Empleando el método de Castigliano: M 4 a= ( 28.357+ Q 2 ) X−40.81−2.5 X 2 → ∂M ∂Q = X / 2 Ma b= ( 28.357+ Q 2 ) X−40.81−2.5 X 2 −5 ( x−3) → ∂M ∂Q =X / 2 M 5 c= ( 26.643+ Q 2 ) X−33.100−2.5 X 2 → ∂M ∂Q =X / 2 Mcb = ( 26.643 + Q 2 ) X−33.100−2.5 X 2 − 5( x−3 ) → ∂M ∂Q =X / 2 δb= δcl ( 45) = 1 3.24EI ∫ 0 3 ( 28.357 X−40.81−2.5 X 2 ) Xdx 2 + 1 3.24 EI ∫ 3 4.5 (28.357 X−40.81−2.5 2 dx δb= δcl ( 45) =56.28212 / EI↓
-
Upload
gilbert-baique-camacho -
Category
Documents
-
view
219 -
download
0
description
uhufhd ahle k hjaff
Transcript of DEFLEXION
d) Determinar la deflexión en el centro del tramo 4-5
Empleando el método de Castigliano:
M 4a=(28.357+Q2 )X−40.81−2.5 X 2→∂M∂Q
=X /2
Mab=(28.357+Q2 )X−40.81−2.5 X 2−5(x−3)→∂M∂Q
=X /2
M 5 c=(26.643+Q2 )X−33.100−2.5 X2→ ∂M∂Q
=X /2
M cb=(26.643+Q2 )X−33.100−2.5 X2−5(x−3)→∂M∂Q
=X /2
δb=δcl (45 )= 13.24 EI
∫0
3 (28.357 X−40.81−2.5 X2 ) Xdx2
+ 13.24 EI
∫3
4.5 (28.357 X−40.81−2.5 X2−5 ( x−3 )) Xdx2dx
+¿ 13.24 EI
∫0
3 (26.643 X−33.100−2.5 X2 ) Xdx2
+ 13.24 EI
∫3
4.5 (26.643 X−33.100−2.5 X2−5 ( x−3 ) ) Xdx2
+¿¿¿
δb=δcl (45 )=56.28212/EI ↓