Baixe Cálculo de Forças em Construções: Exemplo de Forças devido a Ventos e outras Exercícios em PDF para Engenharia Civil, somente na Docsity!
Grupo 01 - Pilar P1 - Tramo Garagem
Localidade: Araruna-PB; Terreno plano , com poucos obstáculos (arvores e edificações baixas)
Isopetas - V
S1 - Terreno Plano
S3 - Edif. Residencial
S2 - Rugosidade do terreno
V0 ≔ 30 ―
m
s
S3 ≔ 1 S1 ≔ 1 S2 ≔b ⋅Fr ⋅
z
p
menor lado - vento à 0°
maior lado - vento a 90°
pdir ≔2.75 m h ≔ 15 ⋅pdir =41.25 m
l ≔ 566 cm + 466 cm + 353 cm + 466 cm + 566 cm =24.17 m
c ≔ 453 cm + 243 cm + 453 cm =11.49 m
De acordo com a NBR 6123 -
P/ vento a 90° e a 0°, temos:
Rugosidade Categoria II - Classe B
Considerando
b ≔1.
z1 ≔pdir =2.75 m p ≔0.
Fr ≔0.
Para os demais:
S2.1 ≔b ⋅Fr ⋅ =
z
p
0.873 m
――
9
100
z2 ≔ 2 ⋅pdir =5.5 m
z3 ≔ 3 ⋅pdir =8.25 m
z4 ≔ 4 ⋅pdir = 11 m S2.2 ≔b ⋅Fr ⋅ =
z
p
0.929 m
――
9
100
z5 ≔ 5 ⋅pdir =13.75 m
z6 ≔ 6 ⋅pdir =16.5 m
z7 ≔ 7 ⋅pdir =19.25 m S2.3 ≔b ⋅Fr ⋅ =
z
p
0.963 m
――
9
100
z8 ≔ 8 ⋅pdir = 22 m
z9 ≔ 9 ⋅pdir =24.75 m
z10 ≔ 10 ⋅pdir =27.5 m S2.4 ≔b ⋅Fr ⋅ =
z
p
0.988 m
――
9
100
z11 ≔ 11 ⋅pdir =30.25 m
z12 ≔ 12 ⋅pdir = 33 m
z13 ≔ 13 ⋅pdir =35.75 m S2.5 ≔b ⋅Fr ⋅ =
z
p
1.008 m
――
9
100
z14 ≔ 14 ⋅pdir =38.5 m
z15 ≔ 15 ⋅pdir =41.25 m S2.6 ≔b ⋅Fr ⋅ =
z
p
1.025 m
――
9
100
S2.7 ≔b ⋅Fr ⋅ =
z
p
1.04 m
――
9
100
S2.8 ≔b ⋅Fr ⋅ =
z
p
1.052 m
――
9
100
S2.12 ≔b ⋅Fr ⋅ =
z
p
1.091 m
――
9
100
S2.8 ≔b ⋅Fr ⋅ =
z
p
1.052 m
――
9
100
S2.12 ≔b ⋅Fr ⋅ =
z
p
1.091 m
――
9
100
S2.9 ≔b ⋅Fr ⋅ =
z
p
1.063 m
――
9
100
S2.13 ≔b ⋅Fr ⋅ =
z
p
1.099 m
――
9
100
S2.10 ≔b ⋅Fr ⋅ =
z
p
1.073 m
――
9
100
S2.14 ≔b ⋅Fr ⋅ =
z
p
1.106 m
――
9
100
S2.11 ≔b ⋅Fr ⋅ =
z
p
1.083 m
――
9
100
S2.15 ≔b ⋅Fr ⋅ =
z
p
1.113 m
――
9
100
Para o cálculo da Velocidade Característica (Vk): {m/s} Pressão Dinâmica (q): {N/m²}
Vk1 ≔V0 ⋅S1 ⋅S3 ⋅0.873 =26.19 ―
m
s
qmin ≔ 25 ――
Kg
m²
q ≔0.613 Vk²
Vk2 ≔V0 ⋅S1 ⋅S3 ⋅0.929 =27.87 ―
m
s
q1 ≔0.631 Vk1 =
2
――
m
2
s
2
Vk3 ≔V0 ⋅S1 ⋅S3 ⋅0.963 =28. ―
m
s
q2 ≔0.631 Vk2 =
2
490.121 ――
m
2
s
2
Vk4 ≔V0 ⋅S1 ⋅S3 ⋅0.988 =29.64 ―
m
s
q3 ≔0.631 Vk3 =
2
526.653 ――
m
2
s
2
Vk5 ≔V0 ⋅S1 ⋅S3 ⋅1.008 =30.24 ―
m
s
q4 ≔0.631 Vk4 =
2
――
m
2
s
2
Vk6 ≔V0 ⋅S1 ⋅S3 ⋅1.025 =30. ―
m
s
q5 ≔0.631 Vk5 =
2
577.023 ――
m
2
s
2
Vk7 ≔V0 ⋅S1 ⋅S3 ⋅1.040 =31.2 ―
m
s
q6 ≔0.631 Vk6 =
2
596.65 ――
m
2
s
2
Vk8 ≔V0 ⋅S1 ⋅S3 ⋅1.052 =31.56 ―
m
s
q7 ≔0.631 Vk7 =
2
――
m
2
s
2
Vk9 ≔V0 ⋅S1 ⋅S3 ⋅1.063 =31.89 ―
m
s
q8 ≔0.631 Vk8 =
2
628.497 ――
m
2
s
2
Vk10 ≔V0 ⋅S1 ⋅S3 ⋅1.073 =32.19 ―
m
s
q9 ≔0.631 Vk9 =
2
641.709 ――
m
2
s
2
Vk11 ≔V0 ⋅S1 ⋅S3 ⋅1.083 =32. ―
m
s
q10 ≔0.631 Vk10 =
2
――
m
2
s
2
Vk12 ≔V0 ⋅S1 ⋅S3 ⋅1.091 =32.73 ―
m
s
q11 ≔0.631 Vk11 =
2
666.084 ――
m
2
s
2
Vk13 ≔V0 ⋅S1 ⋅S3 ⋅1.099 =32.97 ―
m
s
q12 ≔0.631 Vk12 =
2
675.961 ――
m
2
s
2
Vk14 ≔V0 ⋅S1 ⋅S3 ⋅1.106 =33. ―
m
s
q13 ≔0.631 Vk13 =
2
――
m
2
s
2
Vk15 ≔V0 ⋅S1 ⋅S3 ⋅1.113 =33.39 ―
m
s
q14 ≔0.631 Vk14 =
2
694.676 ――
m
2
s
2
q15 ≔0.631 Vk15 =
2
703.497 ――
m
2
s
2
Força de Arrasto na região do pavimento (F): {KN}
F15.0 ≔q15 ⋅A0 ⋅Ca.0 ⋅ 1 ―――=
kN ⋅s ⋅s
m
4
22229 kN F15.90 ≔q15 ⋅A90 ⋅Ca.90 ⋅ 1 ―――=
kN ⋅s ⋅s
m
4
64061 kN
Transformando as forças centradas nos vãos em forças nas Lajes, temos:
Fgar.0 ≔ = ――
F1.
2
6837.9 kN Fgar.90 ≔ = ――
F1.
2
19706.1 kN
Fpav1.0 ≔ + = ――
F1.
2
――
F2.
2
14581.2 kN Fpav1.90 ≔ + = ――
F1.
2
――
F2.
2
42021.4 kN
Fpav2.0 ≔――+ =
F2.
2
――
F3.
2
16063.8 kN Fpav2.90 ≔――+ =
F2.
2
――
F3.
2
46294 kN
Fpav3.0 ≔ + = ――
F3.
2
――
F4.
2
17078.5 kN Fpav3.90 ≔ + = ――
F3.
2
――
F4.
2
49218.4 kN
Fpav4.0 ≔――+ =
F4.
2
――
F5.
2
17874.3 kN Fpav4.90 ≔――+ =
F4.
2
――
F5.
2
51511.8 kN
Fpav5.0 ≔ + = ――
F5.
2
――
F6.
2
18542.6 kN Fpav5.90 ≔ + = ――
F5.
2
――
F6.
2
53437.6 kN
Fpav6.0 ≔――+ =
F6.
2
――
F7.
2
19130.6 kN Fpav6.90 ≔――+ =
F6.
2
――
F7.
2
55132.1 kN
Fpav7.0 ≔ + = ――
F7.
2
――
F8.
2
19633.7 kN Fpav7.90 ≔ + = ――
F7.
2
――
F8.
2
56582.1 kN
Fpav8.0 ≔――+ =
F8.
2
――
F9.
2
20067.7 kN Fpav8.90 ≔――+ =
F8.
2
――
F9.
2
57832.8 kN
Fpav9.0 ≔ + = ――
F9.
2
――
F10.
2
20468.1 kN Fpav9.90 ≔ + = ――
F9.
2
―――
F10.
2
58986.7 kN
Fpav10.0 ≔――+ =
F10.
2
――
F11.
2
20853.1 kN Fpav10.90 ≔―――+ =
F10.
2
―――
F11.
2
60096.4 kN
Fpav11.0 ≔ + = ――
F11.
2
――
F12.
2
21202.6 kN Fpav11.90 ≔ + = ―――
F11.
2
―――
F12.
2
61103.6 kN
Fpav12.0 ≔――+ =
F12.
2
――
F13.
2
21515.9 kN Fpav12.90 ≔―――+ =
F12.
2
―――
F13.
2
62006.3 kN
Fpav13.0 ≔ + = ――
F13.
2
――
F14.
2
21811.5 kN Fpav13.90 ≔ + = ―――
F13.
2
―――
F14.
2
62858.4 kN
Fpav14.0 ≔――+ =
F14.
2
――
F15.
2
22089.4 kN Fpav14.90 ≔―――+ =
F14.
2
―――
F15.
2
63659.1 kN
Fcob15.0 ≔ = ――
F15.
2
11114.4 kN Fcob15.90 ≔ = ―――
F15.
2
32030.4 kN
Mtotal.vento: {kN*m/m}
Mtotal.vento: {kN*m/m}
Mtotal.vento0 ≔ =
Fgar.0 +Fpav1.0 +Fpav2.0 +Fpav3.
Fpav4.0 Fpav5.0 Fpav6.
Fpav7.0 Fpav8.0 Fpav9.0 Fpav10.
Fpav11.0 Fpav12.0 Fpav13.
Fpav14.0 Fcob15.
5
m
kN m
Mtotal.vento90 ≔ =
Fgar.90 +Fpav1.90 +Fpav2.
Fpav3.90 Fpav4.90 Fpav5.90 Fpav6.
Fpav7.90 Fpav8.90 Fpav9.90 Fpav10.
Fpav11.90 Fpav12.90 Fpav13.
Fpav14.90 Fcob15.
5
⎞
m
kN m
Mtotal.vento: {kN*m}
Mtotal.vento0 ≔ =
↲
↲
↲
↲
Fgar.0 ⋅ 0 m +Fpav1.0 ⋅2.75 m +Fpav2.0 ⋅5.5 m +Fpav3.0 ⋅8.25 m
Fpav4.0 ⋅ 11 m Fpav5.0 ⋅13.75 m Fpav6.0 ⋅16.5 m
Fpav7.0 ⋅19.25 m Fpav8.0 ⋅ 22 m Fpav9.0 ⋅24.75 m
Fpav10.0 ⋅27.5 m Fpav11.0 ⋅30.25 m Fpav12.0 ⋅ 33 m
Fpav13.0 ⋅35.75 m Fpav14.0 ⋅38.5 m Fcob15.0 ⋅41.25 m
6376136 kN ⋅m
Mtotal.vento90 ≔ =
↲
↲
↲
↲
↲
Fgar.90 ⋅ 0 m +Fpav1.90 ⋅2.75 m +Fpav2.90 ⋅5.5 m
Fpav3.90 ⋅8.25 m Fpav4.90 ⋅ 11 m Fpav5.90 ⋅13.75 m
Fpav6.90 ⋅16.5 m Fpav7.90 ⋅19.25 m Fpav8.90 ⋅ 22 m
Fpav9.90 ⋅24.75 m Fpav10.90 ⋅27.5 m Fpav11.90 ⋅30.25 m
Fpav12.90 ⋅ 33 m Fpav13.90 ⋅35.75 m
Fpav14.90 ⋅38.5 m Fcob15.90 ⋅41.25 m
18375313 kN ⋅m
Para o desaprumo, temos:
H ≔ 15 ⋅2.75 m =41.25 m
θ1.max ≔――=
1
200
Θ1 ≔――――=
1
100 ⋅
‾‾
H
0.0016 ――
1
m
―
1
θ1.min ≔ = 2 ――
1
300
Θa ≔Θ1⋅
‾‾‾‾‾‾‾
2
―――
⎛
⎜
⎝
1 + ―
1
n
⎞
⎟
⎠
²
Θa.90 ≔0.003⋅ =
‾‾‾‾‾‾‾
2
―――
⎛
⎜
⎝
1 + ―
1
4
⎞
⎟
⎠
2
0.00237 rad
Exemplo: caso da imagem n = 3
n é o número de prumadas (colunas) do pórtico!! ;
Depende da direção do vento!!!
Θa.0 ≔0.003⋅ =
‾‾‾‾‾‾‾
2
―――
⎛
⎜
⎝
1 +―
1
6
⎞
⎟
⎠
2
0.00229 rad
Feq.i ≔Θa ⋅Wi
