exercício
DESCRIPTION
1TRANSCRIPT
A) APLICABILIDADE DO MÉTODO:
Vão da viga longitudinal (m ): ≔l 32
Vão da viga transversal (m ): ≔t 8.5
≔ΣEI →6 ((0.187)) Ε ⋅1.122 Ε
≔ΣEI' →3 ((0.0629)) Ε ⋅0.1887 Ε
≔λ →――t
2 l
⎛⎜⎝
‾‾‾‾‾‾‾4
―l
t――ΣEI
ΣEI'
⎞⎟⎠
0.28888630499105875344 < 0.3 logo, OK
B) Trem -tipo para a viga lateral 1:
B.1) Inércia vigas principais (m 4):≔I1 0.187 ≔I2 0.187 ≔I3 0.187 ≔I4 0.187 ≔I5 0.187 ≔I6 0.187
B.2) Posição das vigas principais
x1=-x6=-4.25mx2=-x5=-2.575mx3=-x4=-0.90m
=++⋅2 ⎛⎝0.187 4.252 ⎞⎠ ⋅2 ⎛⎝0.187 2.575
2 ⎞⎠ ⋅2 ⎛⎝0.187 0.902 ⎞⎠ 9.538 m
6
-Coeficiente de distribuição transversal para a viga 1: ≔x1 −4.25 m
≔r11 =0.187⎛⎜⎝
+――1
1.122――――――((−4.25)) ((−4.25))
9.538
⎞⎟⎠
0.521
≔r12 =0.187⎛⎜⎝
+――1
1.122――――――((−2.575)) ((−4.25))
9.538
⎞⎟⎠
0.381
≔r13 =0.187⎛⎜⎝
+――1
1.122――――――((−0.90)) ((−4.25))
9.538
⎞⎟⎠
0.242
≔r14 =0.187⎛⎜⎝
+――1
1.122―――――((0.9)) ((−4.25))
9.538
⎞⎟⎠
0.092
≔r15 =0.187⎛⎜⎝
+――1
1.122――――――((2.575)) ((−4.25))
9.538
⎞⎟⎠
−0.048
≔r16 =0.187⎛⎜⎝
+――1
1.122―――――((4.25)) ((−4.25))
9.538
⎞⎟⎠
−0.187
-Coeficiente de distribuição transversal para a viga 4: ≔x4 0.90 m
≔r11 =0.187⎛⎜⎝
+――1
1.122―――――((−4.25)) ((0.90))
9.538
⎞⎟⎠
0.092
≔r12 =0.187⎛⎜⎝
+――1
1.122――――――((−2.575)) ((0.90))
9.538
⎞⎟⎠
0.121
≔r13 =0.187⎛⎜⎝
+――1
1.122―――――((−0.90)) ((0.90))
9.538
⎞⎟⎠
0.151
≔r14 =0.187⎛⎜⎝
+――1
1.122――――((0.9)) ((0.90))
9.538
⎞⎟⎠
0.183
≔r15 =0.187⎛⎜⎝
+――1
1.122―――――((2.575)) ((0.90))
9.538
⎞⎟⎠
0.212
≔r16 =0.187⎛⎜⎝
+――1
1.122―――――((4.25)) ((0.90))
9.538
⎞⎟⎠
0.242
-Coeficiente de distribuição transversal para a viga 6: ≔x6 4.25 m
≔r11 =0.187⎛⎜⎝
+――1
1.122―――――((−4.25)) ((0.90))
9.538
⎞⎟⎠
0.092
≔r12 =0.187⎛⎜⎝
+――1
1.122――――――((−2.575)) ((0.90))
9.538
⎞⎟⎠
0.121
≔r13 =0.187⎛⎜⎝
+――1
1.122―――――((−0.90)) ((0.90))
9.538
⎞⎟⎠
0.151
≔r14 =0.187⎛⎜⎝
+――1
1.122――――((0.9)) ((0.90))
9.538
⎞⎟⎠
0.183
≔r15 =0.187⎛⎜⎝
+――1
1.122―――――((2.575)) ((0.90))
9.538
⎞⎟⎠
0.212
≔r16 =0.187⎛⎜⎝
+――1
1.122―――――((4.25)) ((0.90))
9.538
⎞⎟⎠
0.242