Cálculo de Vigas Longitudinais em Pontes

Prévia do material em texto

Aula 07 Data 03/04/2018 
Matéria Pontes e Grandes Estruturas Professor (a) Luiz Antônio 
 
Feito por: Fernando Malaquias
 1 
𝑃 = 100,0𝑘𝑁 
𝐶𝐼𝑉 = 1,25𝑘𝑁 
𝐶𝑁𝐹 = 0,90𝑘𝑁 
Exercício 5 – Para as vigas longitudinais, correspondentes a Ponte 
indicada abaixo, determinar: 
𝑅𝑚á𝑥𝑃1 𝑒 𝑅𝑚𝑖𝑛𝑃1 
𝑀𝑚á𝑥𝑆 𝑒 𝑀𝑚𝑖𝑛𝑆 
𝑉𝑚á𝑥𝑆 𝑒 𝑉𝑚𝑖𝑛𝑆 
Sequência de Cálculo 
1 – Determinação de 𝑔 (1,0) 
2 – Determinação de 𝑅𝑔1, 𝑅𝑔2, 𝑀𝑔𝑆 𝑒 𝑉𝑔𝑠 (2,0) 
3 – Determinação do trem tipo longitudinal (3,0) 
4 – Determinação de 𝑅𝑞
(+)
, 𝑅𝑞
(−)
, 𝑀𝑞
(+)
, 𝑀𝑞
(−)
, 𝑉𝑞
(+)
 𝑒 𝑉𝑞
(−) (3,0) 
5 – Determinação de 𝑅𝑚á𝑥𝑃1, 𝑅𝑚𝑖𝑛𝑃1, 𝑀𝑚á𝑥, 𝑀𝑚𝑖𝑛, 𝑉𝑚á𝑥 𝑒 𝑉𝑚𝑖𝑛 (1,0) 
 
 
 
Aula 07 Data 03/04/2018 
Matéria Pontes e Grandes Estruturas Professor (a) Luiz Antônio 
 
Feito por: Fernando Malaquias
 2 
 
 
 
Aula 07 Data 03/04/2018 
Matéria Pontes e Grandes Estruturas Professor (a) Luiz Antônio 
 
Feito por: Fernando Malaquias
 3 
1 – Determinação de 𝑔 (1,0) 
 
𝐴 = (9,0 ∙ 0,30) + (2,7 ∙ 0,6) ∴ 𝑨 = 𝟒, 𝟑𝟐𝒎𝟐 
𝑔 = (4,32 ∙ 25,0) + 6,0 ∴ 𝒈 = 𝟏𝟏𝟒, 𝟎𝒌𝑵/𝒎 
 
2 – Determinação de 𝑅𝑔1, 𝑅𝑔2, 𝑀𝑔𝑆 𝑒 𝑉𝑔𝑠 (2,0) 
 
𝑅 = 114,0
𝑘𝑁
𝑚
∙ 40𝑚 = 𝟒𝟓𝟔𝟎, 𝟎𝒌𝑵 
𝑅𝑔1 =
4560,0 ∙ 20,0
40,0
= 𝟐𝟐𝟖𝟎, 𝟎𝒌𝑵 
𝑅𝑔2 =
4560,0 ∙ 20,0
40,0
= 𝟐𝟐𝟖𝟎, 𝟎𝒌𝑵 
Devido simetria valores são iguais. 
Aula 07 Data 03/04/2018 
Matéria Pontes e Grandes Estruturas Professor (a) Luiz Antônio 
 
Feito por: Fernando Malaquias
 4 
 
Análise pela esquerda 
M𝑔 = 2280,0 ∙ 10 − (114,0 ∙ 10) ∙
10
2
 
∴ 𝐌𝒈 = 𝟏𝟕𝟏𝟎𝟎, 𝟎𝒌𝑵 ∙ 𝒎 
 
ou Análise pela direita 
M𝑔 = 2280,0 ∙ 30 − (114,0 ∙ 30) ∙
30
2
 
∴ 𝐌𝒈 = 𝟏𝟕𝟏𝟎𝟎, 𝟎𝒌𝑵 ∙ 𝒎 
 
Análise pela esquerda 
𝑉𝑔 = +2280,0 − 114,0 ∙ 10 ∴ 𝑽𝒈 = +𝟏𝟏𝟒𝟎, 𝟎𝟎𝒌𝑵 
 
ou Análise pela direita 
𝑉𝑔 = −2280,0 + 114,0 ∙ 30 ∴ 𝑽𝒈 = +𝟏𝟏𝟒𝟎, 𝟎𝟎𝒌𝑵 
 
Obs: Para determinar a cortante é o mesmo procedimento do Momento, 
porém sem os braços de alavanca. 
Aula 07 Data 03/04/2018 
Matéria Pontes e Grandes Estruturas Professor (a) Luiz Antônio 
 
Feito por: Fernando Malaquias
 5 
3 – Determinação do trem tipo longitudinal (3,0) 
 
1,0 − − − 12,0 
𝑥1 − − − 14,6 
𝑥1 =
1,0 ∙ 14,6
12,0
 
𝒙𝟏 = 𝟏, 𝟐𝟐 
 
1,0 − − − 12,0 
𝑥2 − − − 12,10 
𝑥2 =
1,0 ∙ 12,1
12,0
 
𝒙𝟐 = 𝟏, 𝟎𝟏 
𝑄 = 𝑃 ∙ 𝐶𝐼𝑉 ∙ 𝐶𝑁𝐹 
𝑄 = 100,0 ∙ 1,25 ∙ 0,90 = 112,5 𝑘𝑁 
𝑅 = 𝑄 ∙ (𝑥1 + 𝑥2) → 𝑅 = 112,5 ∙ (1,22 + 1,01) 
𝑹 = 𝟐𝟓𝟎, 𝟖𝟖𝒌𝑵 
Aula 07 Data 03/04/2018 
Matéria Pontes e Grandes Estruturas Professor (a) Luiz Antônio 
 
Feito por: Fernando Malaquias
 6 
4 – Determinação de 𝑅𝑞
(+)
, 𝑅𝑞
(−)
, 𝑀𝑞
(+)
, 𝑀𝑞
(−)
, 𝑉𝑞
(+)
 𝑒 𝑉𝑞
(−) (3,0) 
No P1 a Linha de Influência terá apenas 𝑅𝑞
(+), pois não gera LIR negativo 
em P2. 
 
1,0 − − − 40,0 
𝑥1 − − − 35,0 
𝑥1 =
1,0 ∙ 35,0
40,0
 
𝒙𝟏 = 𝟎, 𝟖𝟕𝟓 
 
1,0 − − − 40,0 
𝑥2 − − − 30,0 
𝑥2 =
1,0 ∙ 30,0
40,0
 
𝒙𝟐 = 𝟎, 𝟕𝟓 
 
1,0 − − − 40,0 
𝑥3 − − − 25,0 
𝑥3 =
1,0 ∙ 25,0
40,0
 
𝒙𝟑 = 𝟎, 𝟔𝟐𝟓 
𝑅𝑞
(+)′ = 250,88 ∙ 1,0 + 250,88 ∙ 0,875 + 250,88 ∙ 0,75 + 250,88 ∙ 0,625 
𝑹𝒒
(+)′ = 𝟖𝟏𝟓, 𝟑𝟔𝒌𝑵 
 
Aula 07 Data 03/04/2018 
Matéria Pontes e Grandes Estruturas Professor (a) Luiz Antônio 
 
Feito por: Fernando Malaquias
 7 
 
1,0 − − − 40,0 
𝑥1 − − − 35,0 
𝑥1 =
1,0 ∙ 35,0
40,0
 
𝒙𝟏 = 𝟎, 𝟖𝟕𝟓 
 
1,0 − − − 40,0 
𝑥2 − − − 30,0 
𝑥2 =
1,0 ∙ 30,0
40,0
 
𝒙𝟐 = 𝟎, 𝟕𝟓 
 
1,0 − − − 40,0 
𝑥3 − − − 25,0 
𝑥3 =
1,0 ∙ 25,0
40,0
 
𝒙𝟑 = 𝟎, 𝟔𝟐𝟓 
 
1,0 − − − 40,0 
𝑥4 − − − 20,0 
𝑥4 =
1,0 ∙ 20,0
40,0
 
𝒙𝟒 = 𝟎, 𝟓𝟎 
𝑅𝑞
(+)′′ = 250,88 ∙ 0,875 + 250,88 ∙ 0,75 + 250,88 ∙ 0,625 + 250,88 ∙ 0,5 
𝑹𝒒
(+)′′ = 𝟔𝟖𝟗, 𝟗𝟐𝒌𝑵 
 
 
Adotaremos a hipótese 1, pois foi a pior. 
∴ 𝑹𝒒
(+) = 𝟖𝟏𝟓, 𝟑𝟔𝒌𝑵 
 
Analisando tanto do P1 p/ P2 quanto P2 p/ P1 não há reação negativa. 
∴ 𝑹𝒒
(−) = 𝟎 𝒌𝑵 
 
Aula 07 Data 03/04/2018 
Matéria Pontes e Grandes Estruturas Professor (a) Luiz Antônio 
 
Feito por: Fernando Malaquias
 8 
 
𝑀 =
𝑎 ∙ 𝑏
𝐿
=
10,0 ∙ 30,0
40,0
= 7,5 
7,5 − − − 10,0 
𝑥1 − − − 5,0 
𝑥1 =
7,5 ∙ 5,0
10,0
 
𝒙𝟏 = 𝟑, 𝟕𝟓 
 
7,5 − − − 30,0 
𝑥2 − − − 25,0 
𝑥2 =
7,5 ∙ 25,0
30,0
 
𝒙𝟐 = 𝟔, 𝟐𝟓 
𝑀𝑞
(+)′ = 250,88 ∙ 0 + 250,88 ∙ 3,75 + 250,88 ∙ 7,5 + 250,88 ∙ 6,25 
𝑀𝑞
(+)′ = 4390,4 𝑘𝑁 ∙ 𝑚 
 
Aula 07 Data 03/04/2018 
Matéria Pontes e Grandes Estruturas Professor (a) Luiz Antônio 
 
Feito por: Fernando Malaquias
 9 
 
7,5 − − − 10,0 
𝑥1 − − − 5,0 
𝑥1 =
7,5 ∙ 5,0
10,0
 
𝒙𝟏 = 𝟑, 𝟕𝟓 
 
7,5 − − − 30,0 
𝑥2 − − − 25,0 
𝑥2 =
7,5 ∙ 25,0
30,0
 
𝒙𝟐 = 𝟔, 𝟐𝟓 
 
7,5 − − − 30,0 
𝑥3 − − − 20,0 
𝑥3 =
7,5 ∙ 20,0
30,0
 
𝒙𝟑 = 𝟓, 𝟎 
𝑀𝑞
(+)′′ = 250,88 ∙ 3,75 + 250,88 ∙ 7,5 + 250,88 ∙ 6,25 + 250,88 ∙ 5,0 
𝑀𝑞
(+)′′ = 5644,8 𝑘𝑁 ∙ 𝑚 
 
Aula 07 Data 03/04/2018 
Matéria Pontes e Grandes Estruturas Professor (a) Luiz Antônio 
 
Feito por: Fernando Malaquias
 10 
 
7,5 − − − 10,0 
𝑥1 − − − 5,0 
𝑥1 =
7,5 ∙ 5,0
10,0
 
𝒙𝟏 = 𝟑, 𝟕𝟓 
 
7,5 − − − 30,0 
𝑥2 − − − 25,0 
𝑥2 =
7,5 ∙ 25,0
30,0
 
𝒙𝟐 = 𝟔, 𝟐𝟓 
 
7,5 − − − 30,0 
𝑥3 − − − 20,0 
𝑥3 =
7,5 ∙ 20,0
30,0
 
𝒙𝟑 = 𝟓, 𝟎 
 
7,5 − − − 30,0 
𝑥4 − − − 15,0 
𝑥4 =
7,5 ∙ 15,0
30,0
 
𝒙𝟒 = 𝟑, 𝟕𝟓 
𝑀𝑞
(+)′′′ = 250,88 ∙ 7,5 + 250,88 ∙ 6,25 + 250,88 ∙ 5,0 + 250,88 ∙ 3,75 
𝑀𝑞
(+)′′′ = 5644,8 𝑘𝑁 ∙ 𝑚 
 
Tanto a hipótese 2 quanto a hipótese 3 são as piores e dão o mesmo 
valor, portanto basta escolher entre 2 ou 3. 
∴ 𝑴𝒒
(+) = 𝟓𝟔𝟒𝟒, 𝟖 𝒌𝑵 ∙ 𝒎 
Não há momento negativo. 
∴ 𝑴𝒒
(−) = 𝟎 𝒌𝑵 ∙ 𝒎 
Aula 07 Data 03/04/2018 
Matéria Pontes e Grandes Estruturas Professor (a) Luiz Antônio 
 
Feito por: Fernando Malaquias
 11 
 
𝑎
𝐿
=
10,0
40,0
= 0,25 
 
𝑏
𝐿
=
30,0
40,0
= 0,75 
0,75 − − − 30,0 
𝑥1 − − − 25,0 
𝑥1 =
0,75 ∙ 25,0
30,0
 
𝒙𝟏 = 𝟎, 𝟔𝟐𝟓 
 
0,75 − − − 30,0 
𝑥2 − − − 20,0 
𝑥2 =
0,75 ∙ 20,0
30,0
 
𝒙𝟐 = 𝟎, 𝟓𝟎 
 
0,25 − − − 10,0 
𝑥3 − − − 5,0 
𝑥3 =
0,25 ∙ 5,0
10,0
 
𝒙𝟑 = 𝟎, 𝟏𝟐𝟓 
𝑉𝑞
(+)′ = −250,88 ∙ 0,125 + 250,88 ∙ 0,75 + 250,88 ∙ 0,625 + 250,88 ∙ 0,50 
𝑉𝑞
(+)′ = +439,04 𝑘𝑁 
 
Aula 07 Data 03/04/2018 
Matéria Pontes e Grandes Estruturas Professor (a) Luiz Antônio 
 
Feito por: Fernando Malaquias
 12 
 
𝑎
𝐿
=
10,0
40,0
= 0,25 
 
𝑏
𝐿
=
30,0
40,0
= 0,75 
0,75 − − − 30,0 
𝑥1 − − − 25,0 
𝑥1 =
0,75 ∙ 25,0
30,0
 
𝒙𝟏 = 𝟎, 𝟔𝟐𝟓 
 
0,75 − − − 30,0 
𝑥2 − − − 20,0 
𝑥2 =
0,75 ∙ 20,0
30,0
 
𝒙𝟐 = 𝟎, 𝟓𝟎 
 
0,25 − − − 10,0 
𝑥3 − − − 5,0 
𝑥3 =
0,25 ∙ 5,0
10,0
 
𝒙𝟑 = 𝟎, 𝟏𝟐𝟓 
 
0,75 − − − 30,0 
𝑥4 − − − 15,0 
𝑥4 =
0,75 ∙ 15,0
30,0
 
𝒙𝟒 = 𝟎, 𝟑𝟕𝟓 
𝑉𝑞
(+)′′ = 250,88 ∙ 0,75 + 250,88 ∙ 0,625 + 250,88 ∙ 0,50 + 250,88 ∙ 0,375 
𝑉𝑞
(+)′′ = +564,48 𝑘𝑁 
 
Adotaremos a hipótese 2, pois foi a pior. 
∴ 𝑽𝒒
(+) = 𝟓𝟔𝟒, 𝟒𝟖 𝒌𝑵 
Aula 07 Data 03/04/2018 
Matéria Pontes e Grandes Estruturas Professor (a) Luiz Antônio 
 
Feito por: Fernando Malaquias
 13 
 
 
𝑎
𝐿
=
10,0
40,0
= 0,25 
 
𝑏
𝐿
=
30,0
40,0
= 0,75 
0,75 − − − 30,0 
𝑥1 − − − 25,0 
𝑥1 =
0,75 ∙ 25,0
30,0
 
𝒙𝟏 = 𝟎, 𝟔𝟐𝟓 
 
0,75 − − − 30,0 
𝑥2 − − − 20,0 
𝑥2 =
0,75 ∙ 20,0
30,0
 
𝒙𝟐 = 𝟎, 𝟓𝟎0,25 − − − 10,0 
𝑥3 − − − 5,0 
𝑥3 =
0,25 ∙ 5,0
10,0
 
𝒙𝟑 = 𝟎, 𝟏𝟐𝟓 
 
0,75 − − − 30,0 
𝑥4 − − − 15,0 
𝑥4 =
0,75 ∙ 15,0
30,0
 
𝒙𝟒 = 𝟎, 𝟑𝟕𝟓 
𝑉𝑞
(−)′ = −250,88 ∙ 0 − 250,88 ∙ 0,125 − 250,88 ∙ 0,25 + 250,88 ∙ 0,625 
𝑉𝑞
(−)′ = +62,72 𝑘𝑁 
 
Alocando o trem tipo dessa forma o valor positivo fica maior que o 
negativo. 
Aula 07 Data 03/04/2018 
Matéria Pontes e Grandes Estruturas Professor (a) Luiz Antônio 
 
Feito por: Fernando Malaquias
 14 
 
0,75 − − − 30,0 
𝑥1 − − − 25,0 
𝑥1 =
0,75 ∙ 25,0
30,0
 
𝒙𝟏 = 𝟎, 𝟔𝟐𝟓 
 
0,75 − − − 30,0 
𝑥2 − − − 20,0 
𝑥2 =
0,75 ∙ 20,0
30,0
 
𝒙𝟐 = 𝟎, 𝟓𝟎 
 
0,25 − − − 10,0 
𝑥3 − − − 5,0 
𝑥3 =
0,25 ∙ 5,0
10,0
 
𝒙𝟑 = 𝟎, 𝟏𝟐𝟓 
 
0,75 − − − 30,0 
𝑥4 − − − 15,0 
𝑥4 =
0,75 ∙ 15,0
30,0
 
𝒙𝟒 = 𝟎, 𝟑𝟕𝟓 
𝑉𝑞
(−)′′ = −250,88 ∙ 0 − 250,88 ∙ 0,125 − 250,88 ∙ 0,25 
𝑉𝑞
(−)′′ = −94,08𝑘𝑁 
 
Adotaremos a hipótese 2, pois foi a pior. 
∴ 𝑽𝒒
(−) = −𝟗𝟒, 𝟎𝟖𝒌𝑵 
 
Aula 07 Data 03/04/2018 
Matéria Pontes e Grandes Estruturas Professor (a) Luiz Antônio 
 
Feito por: Fernando Malaquias
 15 
5 – Determinação de 𝑅𝑚á𝑥, 𝑅𝑚𝑖𝑛, 𝑀𝑚á𝑥, 𝑀𝑚𝑖𝑛, 𝑉𝑚á𝑥 𝑒 𝑉𝑚𝑖𝑛 (1,0) 
𝑹𝒈𝟏 = 𝟐𝟐𝟖𝟎, 𝟎𝒌𝑵 
∴ 𝑹𝒒
(+) = 𝟖𝟏𝟓, 𝟑𝟔𝒌𝑵 
∴ 𝑹𝒒
(−) = 𝟎 𝒌𝑵 
 
∴ 𝐌𝒈 = 𝟏𝟕𝟏𝟎𝟎, 𝟎𝒌𝑵 ∙ 𝒎 
∴ 𝑴𝒒
(+) = 𝟓𝟔𝟒𝟒, 𝟖 𝒌𝑵 ∙ 𝒎 
∴ 𝑴𝒒
(−) = 𝟎 𝒌𝑵 ∙ 𝒎 
 
∴ 𝑽𝒈 = +𝟏𝟏𝟒𝟎, 𝟎𝟎𝒌𝑵 
∴ 𝑽𝒒
(+) = 𝟓𝟔𝟒, 𝟒𝟖 𝒌𝑵 
∴ 𝑽𝒒
(−) = −𝟗𝟒, 𝟎𝟖𝒌𝑵 
- Reação 
𝑅𝑚á𝑥𝑃1 = 𝑅𝑔1 + 𝑅𝑞
(+) 
𝑅𝑚á𝑥𝑃1 = 2280,0 + 815,36 
∴ 𝑹𝒎á𝒙𝑷𝟏 = 𝟑𝟎𝟗𝟓, 𝟑𝟔 𝒌𝑵 
 
𝑅𝑚𝑖𝑛𝑃1 = 𝑅𝑔1 + 𝑅𝑞
(−) 
𝑅𝑚𝑖𝑛𝑃1 = 2280,0 + 0 
∴ 𝑹𝒎𝒊𝒏𝑷𝟏 = 𝟐𝟐𝟖𝟎, 𝟎 𝒌𝑵 
- Momento 
𝑀𝑚á𝑥 = 𝑀𝑔 + 𝑀𝑞
(+) 
𝑀𝑚á𝑥 = 17100,0 + 5644,8 
∴ 𝑴𝒎á𝒙 = 𝟐𝟐𝟕𝟒𝟒, 𝟖 𝒌𝑵 ∙ 𝒎 
𝑀𝑚𝑖𝑛 = 𝑀𝑔 + 𝑀𝑞
(−) 
𝑀𝑚𝑖𝑛 = 17100,0 + 0 
∴ 𝑴𝒎𝒊𝒏 = 𝟏𝟕𝟏𝟎𝟎, 𝟎 𝒌𝑵 ∙ 𝒎 
 
- Cortante 
𝑉𝑚á𝑥 = 𝑉𝑔 + 𝑉𝑞
(+) 
𝑉𝑚á𝑥 = 1140,0 + 564,48 
∴ 𝑽𝒎á𝒙 = 𝟏𝟕𝟎𝟒, 𝟒𝟖 𝒌𝑵 
𝑉𝑚𝑖𝑛 = 𝑉𝑔 + 𝑉𝑞
(−) 
𝑉𝑚𝑖𝑛 = 1140,0 − 94,08 
∴ 𝑽𝒎𝒊𝒏 = 𝟏𝟎𝟒𝟓, 𝟗𝟐 𝒌𝑵
 
Aula 06 Data 27/03/2018 
Matéria Pontes e Grandes Estruturas Professor (a) Luiz Antônio 
 
Feito por: Fernando Malaquias
 16 
𝑃 = 100,0𝑘𝑁 
𝐶𝐼𝑉 = 1,25𝑘𝑁 
𝐶𝑁𝐹 = 0,90𝑘𝑁 
Exercício 6 – Para as vigas longitudinais, correspondentes a Ponte 
indicada abaixo, determinar: 
𝑅𝑚á𝑥𝑃2 𝑒 𝑅𝑚𝑖𝑛𝑃2 
𝑀𝑚á𝑥𝑆 𝑒 𝑀𝑚𝑖𝑛𝑆 
𝑉𝑚á𝑥𝑆 𝑒 𝑉𝑚𝑖𝑛𝑆 
Sequência de Cálculo 
1 – Determinação de 𝑔 (1,0) 
2 – Determinação de 𝑅𝑔1, 𝑅𝑔2, 𝑀𝑔𝑆 𝑒 𝑉𝑔𝑠 (2,0) 
3 – Determinação do trem tipo longitudinal (3,0) 
4 – Determinação de 𝑅𝑞
(+)
, 𝑅𝑞
(−)
, 𝑀𝑞
(+)
, 𝑀𝑞
(−)
, 𝑉𝑞
(+)
 𝑒 𝑉𝑞
(−) (3,0) 
5 – Determinação de 𝑅𝑚á𝑥𝑃2, 𝑅𝑚𝑖𝑛𝑃2, 𝑀𝑚á𝑥, 𝑀𝑚𝑖𝑛, 𝑉𝑚á𝑥 𝑒 𝑉𝑚𝑖𝑛 (1,0) 
 
 
 
Aula 06 Data 27/03/2018 
Matéria Pontes e Grandes Estruturas Professor (a) Luiz Antônio 
 
Feito por: Fernando Malaquias
 17 
 
 
 
Aula 06 Data 27/03/2018 
Matéria Pontes e Grandes Estruturas Professor (a) Luiz Antônio 
 
Feito por: Fernando Malaquias
 18 
1 – Determinação de 𝑔 (1,0) 
 
𝐴 = (9,0 ∙ 0,30) + (2,7 ∙ 0,6) ∴ 𝑨 = 𝟒, 𝟑𝟐𝒎𝟐 
𝑔 = (4,32 ∙ 25,0) + 6,0 ∴ 𝒈 = 𝟏𝟏𝟒, 𝟎𝒌𝑵/𝒎 
 
2 – Determinação de 𝑅𝑔1, 𝑅𝑔2, 𝑀𝑔𝑆 𝑒 𝑉𝑔𝑠 (2,0) 
 
Análise pela esquerda
𝑅 = 114,0
𝑘𝑁
𝑚
∙ 50𝑚 = 𝟓𝟕𝟎𝟎, 𝟎𝒌𝑵 
 
R𝐴 + R𝐵 − 5700,0 = 0 
𝐑𝑨 + 𝐑𝑩 = 𝟓𝟕𝟎𝟎, 𝟎 𝒌𝑵 
Σ𝑀𝐴 = 0 
R𝐴 ∙ 38 − 5700 ∙ 18,0 = 0 
R𝐴 =
5700 ∙ 18,0
38
 
∴ 𝐑𝑨 = 𝟐𝟕𝟎𝟎, 𝟎 𝒌𝑵 
R𝐴 + R𝐵 = 5700,0 
2700,0 + R𝐵 = 5700,0 
∴ 𝐑𝑩 = 𝟑𝟎𝟎𝟎, 𝟎𝒌𝑵 
 
Aula 06 Data 27/03/2018 
Matéria Pontes e Grandes Estruturas Professor (a) Luiz Antônio 
 
Feito por: Fernando Malaquias
 19 
Análise pela direita 
𝑅 = 114,0
𝑘𝑁
𝑚
∙ 50𝑚 = 𝟓𝟕𝟎𝟎, 𝟎𝒌𝑵 
R𝐴 + R𝐵 − 5700,0 = 0 
𝐑𝑨 + 𝐑𝑩 = 𝟓𝟕𝟎𝟎, 𝟎 𝒌𝑵 
 
Σ𝑀𝐵 = 0 
R𝐵 ∙ 38 − 5700 ∙ 20,0 = 0 
R𝐵 =
5700 ∙ 20,0
38
 
∴ 𝐑𝑩 = 𝟑𝟎𝟎𝟎, 𝟎 𝒌𝑵 
 
R𝐴 + R𝐵 = 5700,0 
R𝐴 + 3000 = 5700,0 
∴ 𝐑𝑨 = 𝟐𝟕𝟎𝟎, 𝟎𝒌𝑵 
 
∴ 𝐑𝒈𝟐 = 𝟑𝟎𝟎𝟎, 𝟎𝒌𝑵 
 
Análise pela esquerda 
M𝑔 = 2700,0 ∙ 30 − (114,0 ∙ 35) ∙
35
2
 
∴ 𝐌𝒈 = 𝟏𝟏𝟏𝟕𝟓, 𝟎𝒌𝑵 ∙ 𝒎 
ou Análise pela direita 
M𝑔 = 3000,0 ∙ 8 − (114,0 ∙ 15) ∙
15
2
 
∴ 𝐌𝒈 = 𝟏𝟏𝟏𝟕𝟓, 𝟎𝒌𝑵 ∙ 𝒎 
 
Análise pela esquerda 
𝑉𝑔 = +2700,0 − 114,0 ∙ 35 ∴ 𝑽𝒈 = −𝟏𝟐𝟗𝟎, 𝟎𝒌𝑵 
ou Análise pela direita 
𝑉𝑔 = −3000,0 + 114,0 ∙ 15 ∴ 𝑽𝒈 = −𝟏𝟐𝟗𝟎, 𝟎𝒌𝑵 
Obs: Para determinar a cortante é o mesmo procedimento do Momento, porém sem 
os braços de alavanca. 
Aula 06 Data 27/03/2018 
Matéria Pontes e Grandes Estruturas Professor (a) Luiz Antônio 
 
Feito por: Fernando Malaquias
 20 
3 – Determinação do trem tipo longitudinal (3,0) 
 
1,0 − − − 12,0 
𝑥1 − − − 12,6 
𝑥1 =
1,0 ∙ 14,6
12,0
 
𝒙𝟏 = 𝟏, 𝟐𝟐 
 
1,0 − − − 12,0 
𝑥2 − − − 0,10 
𝑥2 =
1,0 ∙ 12,1
12,0
 
𝒙𝟐 = 𝟏, 𝟎𝟏 
𝑄 = 𝑃 ∙ 𝐶𝐼𝑉 ∙ 𝐶𝑁𝐹 
𝑄 = 100,0 ∙ 1,25 ∙ 0,90 = 112,5 𝑘𝑁 
𝑅 = 𝑄 ∙ (𝑥1 + 𝑥2) → 𝑅 = 112,5 ∙ (1,22 + 1,01) 
𝑹 = 𝟐𝟓𝟎, 𝟖𝟖𝒌𝑵 
Aula 06 Data 27/03/2018 
Matéria Pontes e Grandes Estruturas Professor (a) Luiz Antônio 
 
Feito por: Fernando Malaquias
 21 
4 – Determinação de 𝑅𝑞
(+)
, 𝑅𝑞
(−)
, 𝑀𝑞
(+)
, 𝑀𝑞
(−)
, 𝑉𝑞
(+)
 𝑒 𝑉𝑞
(−) (3,0) 
 
1,0 − − − 38,0 
𝑥1 − − − 45,0 
𝑥1 =
1,0 ∙ 45,0
38,0
 
𝒙𝟏 = 𝟏, 𝟏𝟖 
 
1,0 − − − 38,0 
𝑥2 − − − 40,0 
𝑥2 =
1,0 ∙ 40,0
38,0
 
𝒙𝟐 = 𝟏, 𝟎𝟓 
 
1,0 − − − 38,0 
𝑥3 − − − 35,0 
𝑥3 =
1,0 ∙ 35,0
38,0
 
𝒙𝟑 = 𝟎, 𝟗𝟐 
 
1,0 − − − 38,0 
𝑥4 − − − 30,0 
𝑥4 =
1,0 ∙ 30,0
38,0
 
𝒙𝟒 = 𝟎, 𝟕𝟗 
1,0 − − − 38,0 
𝑥5 − − − 10,0 
𝑥5 =
1,0 ∙ 10,0
38,0
 
𝒙𝟓 = 𝟎, 𝟐𝟔 
 
1,0 − − − 38,0 
𝑥6 − − − 5,0 
𝑥6 =
1,0 ∙ 5,0
38,0
 
𝒙𝟔 = 𝟎, 𝟏𝟑 
 
1,0 − − − 38,0 
𝑥7 − − − 5,0 
𝑥7 =
1,0 ∙ 5,0
38,0
 
𝒙𝟕 = 𝟎, 𝟏𝟑 
 
 
Aula 06 Data 27/03/2018 
Matéria Pontes e Grandes Estruturas Professor (a) Luiz Antônio 
 
Feito por: Fernando Malaquias
 22 
𝑅𝑞
(+)′ = 250,88 ∙ 1,18 + 250,88 ∙ 1,05 + 250,88 ∙ 0,92 + 250,88 ∙ 0,79 
𝑅𝑞
(+)′ = 988,47𝑘𝑁 
 
𝑅𝑞
(−)′ = −250,88 ∙ 0,13 − 250,88 ∙ 0 + 250,88 ∙ 0,13 + 250,88 ∙ 0,26 
𝑅𝑞
(−)′ = +65,23𝑘𝑁 
 
𝑅𝑞
(+)′′ = 250,88 ∙ 1,18 + 250,88 ∙ 1,05 + 250,88 ∙ 0,92 + 250,88 ∙ 0,79 
𝑅𝑞
(+)′′ = 988,47𝑘𝑁 
∴ 𝑹𝒒
(+) = 𝟗𝟖𝟖, 𝟒𝟕𝒌𝑵 
 
𝑅𝑞
(−)′′ = −250,88 ∙ 0,13 − 250,88 ∙ 0 
∴ 𝑹𝒒
(−) = −𝟑𝟐, 𝟔𝟏 𝒌𝑵 
Aula 06 Data 27/03/2018 
Matéria Pontes e Grandes Estruturas Professor (a) Luiz Antônio 
 
Feito por: Fernando Malaquias
 23 
 
𝑀 =
𝑎 ∙ 𝑏
𝐿
=
30,0 ∙ 8,0
38,0
= 6,32 
6,32 − − − 30,0 
𝑥1 − − − 25,0 
𝑥1 =
6,32 ∙ 25,0
30,0
 
𝒙𝟏 = 𝟓, 𝟐𝟕 
 
6,32 − − − 30,0 
𝑥2 − − − 20,0 
𝑥2 =
6,32 ∙ 20,0
30,0
 
𝒙𝟐 = 𝟒, 𝟐𝟏 
 
6,32 − − − 30,0 
𝑥3 − − − 15,0 
𝑥3 =
6,32 ∙ 15,0
30,0
 
𝒙𝟑 = 𝟑, 𝟏𝟔 
𝑀𝑞
(+)′ = 250,88 ∙ 6,32 + 250,88 ∙ 5,27 + 250,88 ∙ 4,21 + 250,88 ∙ 3,16 
𝑀𝑞
(+)′ = 4756,68 𝑘𝑁 ∙ 𝑚 
 
 
Aula 06 Data 27/03/2018 
Matéria Pontes e Grandes Estruturas Professor (a) Luiz Antônio 
 
Feito por: Fernando Malaquias
 24 
 
6,32 − − − 30,0 
𝑥1 − − − 25,0 
𝑥1 =
6,32 ∙ 25,0
30,0
 
𝒙𝟏 = 𝟓, 𝟐𝟕 
 
6,32 − − − 30,0 
𝑥2 − − − 20,0 
𝑥2 =
6,32 ∙ 20,0
30,0
 
𝒙𝟐 = 𝟒, 𝟐𝟏 
 
6,32 − − − 30,0 
𝑥3 − − − 15,0 
𝑥3 =
6,32 ∙ 15,0
30,0
 
𝒙𝟑 = 𝟑, 𝟏𝟔 
 
6,32− − − 8,0 
𝑥4 − − − 3,0 
𝑥4 =
6,32 ∙ 3,0
8,0
 
𝒙𝟒 = 𝟐, 𝟑𝟕 
𝑀𝑞
(+)′′ = 250,88 ∙ 4,21 + 250,88 ∙ 5,27 + 250,88 ∙ 6,32 + 250,88 ∙ 2,37 
𝑀𝑞
(+)′′ = 4558,49 𝑘𝑁 ∙ 𝑚 
 
Utilizaremos a hipótese 1, pois é a pior. 
∴ 𝑴𝒒
(+) = 𝟒𝟕𝟓𝟔, 𝟔𝟖 𝒌𝑵 ∙ 𝒎 
 
Aula 06 Data 27/03/2018 
Matéria Pontes e Grandes Estruturas Professor (a) Luiz Antônio 
 
Feito por: Fernando Malaquias
 25 
 
6,32 − − − 30,0 
𝑥1 − − − 25,0 
𝑥1 =
6,32 ∙ 25,0
30,0
 
𝒙𝟏 = 𝟓, 𝟐𝟕 
 
6,32 − − − 30,0 
𝑥2 − − − 20,0 
𝑥2 =
6,32 ∙ 20,0
30,0
 
𝒙𝟐 = 𝟒, 𝟐𝟏 
 
6,32 − − − 30,0 
𝑥3 − − − 15,0 
𝑥3 =
6,32 ∙ 15,0
30,0
 
𝒙𝟑 = 𝟑, 𝟏𝟔 
6,32 − − − 8,0 
𝑥4 − − − 3,0 
𝑥4 =
6,32 ∙ 3,0
8,0
 
𝒙𝟒 = 𝟐, 𝟑𝟕 
 
6,32 − − − 8,0 
𝑥5 − − − 2,0 
𝑥5 =
6,32 ∙ 2,0
8,0
 
𝒙𝟓 = 𝟏, 𝟓𝟖 
 
6,32 − − − 8,0 
𝑥6 − − − 7,0 
𝑥6 =
6,32 ∙ 7,0
8,0
 
𝒙𝟔 = 𝟓, 𝟓𝟑 
 
𝑀𝑞
(−)′ = 250,88 ∙ 6,32 + 250,88 ∙ 2,37 − 250,88 ∙ 1,58 − 250,88 ∙ 5,53 
𝑀𝑞
(−)′ = 396,39 𝑘𝑁 ∙ 𝑚 
 
 
Aula 06 Data 27/03/2018 
Matéria Pontes e Grandes Estruturas Professor (a) Luiz Antônio 
 
Feito por: Fernando Malaquias
 26 
 
6,32 − − − 30,0 
𝑥1 − − − 25,0 
𝑥1 =
6,32 ∙ 25,0
30,0
 
𝒙𝟏 = 𝟓, 𝟐𝟕 
 
6,32 − − − 30,0 
𝑥2 − − − 20,0 
𝑥2 =
6,32 ∙ 20,0
30,0
 
𝒙𝟐 = 𝟒, 𝟐𝟏 
 
6,32 − − − 30,0 
𝑥3 − − − 15,0 
𝑥3 =
6,32 ∙ 15,0
30,0
 
𝒙𝟑 = 𝟑, 𝟏𝟔 
6,32 − − − 8,0 
𝑥4 − − − 3,0 
𝑥4 =
6,32 ∙ 3,0
8,0
 
𝒙𝟒 = 𝟐, 𝟑𝟕 
 
6,32 − − − 8,0 
𝑥5 − − − 2,0 
𝑥5 =
6,32 ∙ 2,0
8,0
 
𝒙𝟓 = 𝟏, 𝟓𝟖 
 
6,32 − − − 8,0 
𝑥6 − − − 7,0 
𝑥6 =
6,32 ∙ 7,0
8,0
 
𝒙𝟔 = 𝟓, 𝟓𝟑 
𝑀𝑞
(−)′′ = −250,88 ∙ 1,58 − 250,88 ∙ 5,53 
𝑀𝑞
(−)′′ = −1783,76 𝑘𝑁 ∙ 𝑚 
 
Utilizaremos a hipótese 2, pois é a pior. 
∴ 𝑴𝒒
(−) = −𝟏𝟕𝟖𝟑, 𝟕𝟔 𝒌𝑵 ∙ 𝒎 
 
Aula 06 Data 27/03/2018 
Matéria Pontes e Grandes Estruturas Professor (a) Luiz Antônio 
 
Feito por: Fernando Malaquias
 27 
 
𝑎
𝐿
=
30,0
38,0
= 𝟎, 𝟕𝟗 
 
𝑏
𝐿
=
8,0
38,0
= 𝟎, 𝟐𝟏 
0,79 − − − 30,0 
𝑥1 − − − 25,0 
𝑥1 =
0,79 ∙ 25,0
30,0
 
𝒙𝟏 = 𝟎, 𝟔𝟔 
 
0,79 − − − 30,0 
𝑥2 − − − 20,0 
𝑥2 =
0,79 ∙ 20,0
30,0
 
𝒙𝟐 = 𝟎, 𝟓𝟑 
 
0,79 − − − 30,0 
𝑥3 − − − 15,0 
𝑥3 =
0,79 ∙ 15,0
30,0
 
𝒙𝟑 = 𝟎, 𝟒𝟎 
𝑉𝑞
(−)′ = −250,88 ∙ 0,79 − 250,88 ∙ 0,66 − 250,88 ∙ 0,53 − 250,88 ∙ 0,40 
𝑉𝑞
(−)′ = −597,09 𝑘𝑁 
 
Aula 06 Data 27/03/2018 
Matéria Pontes e Grandes Estruturas Professor (a) Luiz Antônio 
 
Feito por: Fernando Malaquias
 28 
 
𝑎
𝐿
=
30,0
38,0
= 𝟎, 𝟕𝟗 
 
𝑏
𝐿
=
8,0
38,0
= 𝟎, 𝟐𝟏 
0,79 − − − 30,0 
𝑥1 − − − 25,0 
𝑥1 =
0,79 ∙ 25,0
30,0
 
𝒙𝟏 = 𝟎, 𝟔𝟔 
 
0,79 − − − 30,0 
𝑥2 − − − 20,0 
𝑥2 =
0,79 ∙ 20,0
30,0
 
𝒙𝟐 = 𝟎, 𝟓𝟑 
 
0,79 − − − 30,0 
𝑥3 − − − 15,0 
𝑥3 =
0,79 ∙ 15,0
30,0
 
𝒙𝟑 = 𝟎, 𝟒𝟎 
 
0,21 − − − 8,0 
𝑥4 − − − 3,0 
𝑥4 =
0,21 ∙ 3,0
8,0
 
𝒙𝟒 = 𝟎, 𝟎𝟖 
𝑉𝑞
(−)′′ = −250,88 ∙ 0,53 − 250,88 ∙ 0,66 − 250,88 ∙ 0,79 + 250,88 ∙ 0,08 
𝑉𝑞
(−)′′ = −476,67 𝑘𝑁 
 
Utilizaremos a hipótese 1, pois é a pior. 
∴ 𝑽𝒒
(−) = −𝟓𝟗𝟕, 𝟎𝟗 𝒌𝑵 
 
Aula 06 Data 27/03/2018 
Matéria Pontes e Grandes Estruturas Professor (a) Luiz Antônio 
 
Feito por: Fernando Malaquias
 29 
 
𝑎
𝐿
=
30,0
38,0
= 𝟎, 𝟕𝟗 
 
𝑏
𝐿
=
8,0
38,0
= 𝟎, 𝟐𝟏 
0,79 − − − 30,0 
𝑥1 − − − 25,0 
𝑥1 =
0,79 ∙ 25,0
30,0
 
𝒙𝟏 = 𝟎, 𝟔𝟔 
 
0,79 − − − 30,0 
𝑥2 − − − 20,0 
𝑥2 =
0,79 ∙ 20,0
30,0
 
𝒙𝟐 = 𝟎, 𝟓𝟑 
 
0,79 − − − 30,0 
𝑥3 − − − 15,0 
𝑥3 =
0,79 ∙ 15,0
30,0
 
𝒙𝟑 = 𝟎, 𝟒𝟎 
0,21 − − − 8,0 
𝑥4 − − − 3,0 
𝑥4 =
0,21 ∙ 3,0
8,0
 
𝒙𝟒 = 𝟎, 𝟎𝟖 
 
0,21 − − − 8,0 
𝑥5 − − − 2,0 
𝑥5 =
0,21 ∙ 2,0
8,0
 
𝒙𝟓 = 𝟎, 𝟎𝟓 
 
0,21 − − − 8,0 
𝑥6 − − − 7,0 
𝑥6 =
0,21 ∙ 7,0
8,0
 
𝒙𝟔 = 𝟎, 𝟏𝟖 
𝑉𝑞
(+)′ = 250,88 ∙ 0,21 + 250,88 ∙ 0,08 − 250,88 ∙ 0,05 − 250,88 ∙ 0,18 
𝑉𝑞
(+)′ = 15,05 𝑘𝑁 
 
Aula 06 Data 27/03/2018 
Matéria Pontes e Grandes Estruturas Professor (a) Luiz Antônio 
 
Feito por: Fernando Malaquias
 30 
 
𝑎
𝐿
=
30,0
38,0
= 𝟎, 𝟕𝟗 
 
𝑏
𝐿
=
8,0
38,0
= 𝟎, 𝟐𝟏 
0,79 − − − 30,0 
𝑥1 − − − 25,0 
𝑥1 =
0,79 ∙ 25,0
30,0
 
𝒙𝟏 = 𝟎, 𝟔𝟔 
 
0,79 − − − 30,0 
𝑥2 − − − 20,0 
𝑥2 =
0,79 ∙ 20,0
30,0
 
𝒙𝟐 = 𝟎, 𝟓𝟑 
 
0,79 − − − 30,0 
𝑥3 − − − 15,0 
𝑥3 =
0,79 ∙ 15,0
30,0
 
𝒙𝟑 = 𝟎, 𝟒𝟎 
0,21 − − − 8,0 
𝑥4 − − − 3,0 
𝑥4 =
0,21 ∙ 3,0
8,0
 
𝒙𝟒 = 𝟎, 𝟎𝟖 
 
0,21 − − − 8,0 
𝑥5 − − − 2,0 
𝑥5 =
0,21 ∙ 2,0
8,0
 
𝒙𝟓 = 𝟎, 𝟎𝟓 
 
0,21 − − − 8,0 
𝑥6 − − − 7,0 
𝑥6 =
0,21 ∙ 7,0
8,0
 
𝒙𝟔 = 𝟎, 𝟏𝟖 
𝑉𝑞
(+)′′ = −250,88 ∙ 0,66 + 250,88 ∙ 0,21 + 250,88 ∙ 0,08 − 250,88 ∙ 0,05 
𝑉𝑞
(+)′′ = −105,37 𝑘𝑁 
 
Aula 06 Data 27/03/2018 
Matéria Pontes e Grandes Estruturas Professor (a) Luiz Antônio 
 
Feito por: Fernando Malaquias
 31 
 
0,79 − − − 30,0 
𝑥1 − − − 25,0 
𝑥1 =
0,79 ∙ 25,0
30,0
 
𝒙𝟏 = 𝟎, 𝟔𝟔 
 
0,79 − − − 30,0 
𝑥2 − − − 20,0 
𝑥2 =
0,79 ∙ 20,0
30,0
 
𝒙𝟐 = 𝟎, 𝟓𝟑 
 
0,79 − − − 30,0 
𝑥3 − − − 15,0 
𝑥3 =
0,79 ∙ 15,0
30,0
 
𝒙𝟑 = 𝟎, 𝟒𝟎 
0,21 − − − 8,0 
𝑥4 − − − 3,0 
𝑥4 =
0,21 ∙ 3,0
8,0
 
𝒙𝟒 = 𝟎, 𝟎𝟖 
 
0,21 − − − 8,0 
𝑥5 − − − 2,0 
𝑥5 =
0,21 ∙ 2,0
8,0
 
𝒙𝟓 = 𝟎, 𝟎𝟓 
 
0,21 − − − 8,0 
𝑥6 − − − 7,0 
𝑥6 =
0,21 ∙ 7,0
8,0
 
𝒙𝟔 = 𝟎, 𝟏𝟖 
 
0,79 − − − 30,0 
𝑥7 − − − 5,0 
𝑥7 =
0,79 ∙ 5,0
30,0
 
𝒙𝟕 = 𝟎, 𝟏𝟑 
𝑉𝑞
(+)′′′ = 250,88 ∙ 0,13 + 250,88 ∙ 0 
𝑉𝑞
(+)′′′ = 32,61 𝑘𝑁 
 
Utilizaremos a hipótese 3, pois é a pior. 
∴ 𝑽𝒒
(+) = 𝟑𝟐, 𝟔𝟏 𝒌𝑵 
 
 
Aula 06 Data 27/03/2018 
Matéria Pontes e Grandes Estruturas Professor (a) Luiz Antônio 
 
Feito por: Fernando Malaquias
 32 
5 – Determinação de 𝑅𝑚á𝑥𝑃2, 𝑅𝑚𝑖𝑛𝑃2, 𝑀𝑚á𝑥, 𝑀𝑚𝑖𝑛, 𝑉𝑚á𝑥 𝑒 𝑉𝑚𝑖𝑛 (1,0) 
∴ 𝐑𝒈𝟐 = 𝟑𝟎𝟎𝟎, 𝟎 𝒌𝑵 
∴ 𝑹𝒒
(+) = 𝟗𝟖𝟖, 𝟒𝟕𝒌𝑵 
∴ 𝑹𝒒
(−) = −𝟑𝟐, 𝟔𝟏 𝒌𝑵 
 
∴ 𝐌𝒈 = 𝟏𝟏𝟏𝟕𝟓, 𝟎𝒌𝑵 ∙ 𝒎 
∴ 𝑴𝒒
(+) = 𝟒𝟕𝟓𝟔, 𝟔𝟖 𝒌𝑵 ∙ 𝒎 
∴ 𝑴𝒒
(−) = −𝟏𝟕𝟖𝟑, 𝟕𝟔 𝒌𝑵 ∙ 𝒎 
 
∴ 𝑽𝒈 = −𝟏𝟐𝟗𝟎, 𝟎𝒌𝑵 
∴ 𝑽𝒒
(+) = 𝟏𝟓, 𝟎𝟓 𝒌𝑵 
∴ 𝑽𝒒
(−) = −𝟓𝟗𝟕, 𝟎𝟗 𝒌𝑵 
- Reação 
𝑅𝑚á𝑥𝑃2 = 𝑅𝑔2 + 𝑅𝑞
(+) 
𝑅𝑚á𝑥𝑃2 = 3000,0 + 988,47 
∴ 𝑹𝒎á𝒙𝑷𝟐 = 𝟑𝟗𝟖𝟖, 𝟒𝟕 𝒌𝑵 
 
𝑅𝑚𝑖𝑛𝑃2 = 𝑅𝑔1 + 𝑅𝑞
(−) 
𝑅𝑚𝑖𝑛𝑃2 = 3000,0 − 32,61 
∴ 𝑹𝒎𝒊𝒏𝑷𝟐 = 𝟐𝟗𝟔𝟕, 𝟑𝟗 𝒌𝑵 
- Momento 
𝑀𝑚á𝑥 = 𝑀𝑔 + 𝑀𝑞
(+) 
𝑀𝑚á𝑥 = 11175,0 + 4756,68 
∴ 𝑴𝒎á𝒙 = 𝟏𝟓𝟗𝟑𝟏, 𝟔𝟖 𝒌𝑵 ∙ 𝒎 
𝑀𝑚𝑖𝑛 = 𝑀𝑔 + 𝑀𝑞
(−) 
𝑀𝑚𝑖𝑛 = 11175,0 − 1783,76 
∴ 𝑴𝒎𝒊𝒏 = 𝟗𝟑𝟗𝟏, 𝟐𝟒 𝒌𝑵 ∙ 𝒎 
 
- Cortante 
𝑉𝑚á𝑥 = 𝑉𝑔 + 𝑉𝑞
(+) 
𝑉𝑚á𝑥 = −1290,0 + 32,61 
∴ 𝑽𝒎á𝒙 = −𝟏𝟐𝟓𝟕, 𝟑𝟗𝒌𝑵 
𝑉𝑚𝑖𝑛 = 𝑉𝑔 + 𝑉𝑞
(−) 
𝑉𝑚𝑖𝑛 = −1290,0 − 597,09 
∴ 𝑽𝒎𝒊𝒏 = −𝟏𝟖𝟖𝟕, 𝟎𝟗 𝒌𝑵