LISTA RM II 02
6 pág.

LISTA RM II 02

Pré-visualização2 páginas
ANDRADINA 
2017 
 
 
1 
 
 
Engenharia Mecânica \u2013 5º Período 
 
Discente: Eduardo Moreira Bezerra RA: 1530096890 
Docente: Prof. Juliano Torteli de Godoi Zucato Disciplina: Resistência dos Materiais II Data: 17/04/2017 
 
 1) Determine a coordenada do centroide do perfil sólido abaixo. 
 
 
 
Solução. Temos as seguintes componentes: 
 
I II III 
 
\ud835\udc361(\ufffd\u305\ufffd; \ufffd\u305\ufffd) = \ud835\udc361(50; 10) \ud835\udc362(\ufffd\u305\ufffd; \ufffd\u305\ufffd) = \ud835\udc362(10; 100) \ud835\udc363(\ufffd\u305\ufffd; \ufffd\u305\ufffd) = \ud835\udc363(50; 190) 
 
Componente Área (mm2) \ud835\udc99 \u305(mm) \ud835\udc9a \u305(mm) \ufffd\u305\ufffd\ud835\udc68 (mm3) \ufffd\u305\ufffd\ud835\udc68 (mm3) 
I 2.000 50 10 100.000 20.000 
II 3.200 10 100 32.000 320.000 
III 2.000 50 190 100.000 380.000 
 \u2211 \ud835\udc34 = 7.200 \u2211 \ufffd\u305\ufffd\ud835\udc34 = 232.000 \u2211 \ufffd\u305\ufffd\ud835\udc34 = 720.000 
 
Logo, temos que a centroide pode ser encontrada por 
 
\ufffd\u305\ufffd \u2211 \ud835\udc34 = \u2211 \ufffd\u305\ufffd\ud835\udc34 \u27f9 \ufffd\u305\ufffd =
\u2211 \ufffd\u305\ufffd\ud835\udc34
\u2211 \ud835\udc34
=
232.000
7.200
=
290
9
\u27f9 \ufffd\u305\ufffd = \ud835\udfd1\ud835\udfd0, \ud835\udfd0\ud835\udfd0 mm 
 
 
\ufffd\u305\ufffd \u2211 \ud835\udc34 = \u2211 \ufffd\u305\ufffd\ud835\udc34 \u27f9 \ufffd\u305\ufffd =
\u2211 \ufffd\u305\ufffd\ud835\udc34
\u2211 \ud835\udc34
=
720.000
7.200
\u27f9 \ufffd\u305\ufffd = \ud835\udfcf\ud835\udfce\ud835\udfce mm 
 
Portanto as coordenadas da centroide é o ponto \ud835\udc6a(\ufffd\u305\ufffd; \ufffd\u305\ufffd) = \ud835\udc6a(\ud835\udfd1\ud835\udfd0, \ud835\udfd0\ud835\udfd0; \ud835\udfcf\ud835\udfce\ud835\udfce) \ud835\udc26\ud835\udc26. 
 
 
 
 
 
 
 
 
 
 
 
\ud835\udc65 
\ud835\udc361 
100 
\ud835\udc65 \ud835\udc65 
160 
\ud835\udc362 
\ufffd\u305\ufffd 
\ud835\udc363 
20 100 
20 
20 
\ufffd\u305\ufffd 
\ufffd\u305\ufffd 
\ufffd\u305\ufffd 
\ud835\udc66 
\ufffd\u305\ufffd 
\ufffd\u305\ufffd 
\ud835\udc66 \ud835\udc66 
 
 ANDRADINA 
2017 
 
 
2 
 
 
Engenharia Mecânica \u2013 5º Período 
 
Discente: Eduardo Moreira Bezerra RA: 1530096890 
Docente: Prof. Juliano Torteli de Godoi Zucato Disciplina: Resistência dos Materiais II Data: 17/04/2017 
 
 2) A figura abaixo é feita de um pedaço de arame fino e homogêneo. Determine a localização do 
centro de gravidade. 
 
 
Solução. Podemos considerar os seguintes seguimentos de reta no arame: 
 
\ud835\udc34\ud835\udc36 \ud835\udc34\ud835\udc35 \ud835\udc36\ud835\udc35 
 
\ud835\udc361(\ufffd\u305\ufffd; \ufffd\u305\ufffd) = \ud835\udc361(0; 12,5) \ud835\udc362(\ufffd\u305\ufffd; \ufffd\u305\ufffd) = \ud835\udc362(30; 0) \ud835\udc363(\ufffd\u305\ufffd; \ufffd\u305\ufffd) = \ud835\udc363 (
60
2
;
25
2
) 
 
Seguimento Comprimento \ud835\udc8d (cm) \ufffd\u305\ufffd (cm) \ufffd\u305\ufffd (cm) \ufffd\u305\ufffd\ud835\udc8d (cm2) \ufffd\u305\ufffd\ud835\udc8d (cm2) 
\ud835\udc34\ud835\udc36 25 0 12,5 0 312,5 
\ud835\udc34\ud835\udc35 60 30 0 1.800 0 
\ud835\udc36\ud835\udc35 65 30 12,5 1.950 812,5 
 \u2211 \ud835\udc59 = 150 \u2211 \ufffd\u305\ufffd\ud835\udc59 = 3.750 \u2211 \ufffd\u305\ufffd\ud835\udc59 = 1.125 
 
Assim podemos determinar o centro de gravidade de uma linha composta, ou seja, 
 
\ufffd\u305\ufffd \u2211 \ud835\udc59 = \u2211 \ufffd\u305\ufffd\ud835\udc59 \u27f9 \ufffd\u305\ufffd =
\u2211 \ufffd\u305\ufffd\ud835\udc59
\u2211 \ud835\udc59
=
3.750
150
\u27f9 \ufffd\u305\ufffd = \ud835\udfd0\ud835\udfd3 cm 
 
 
\ufffd\u305\ufffd \u2211 \ud835\udc59 = \u2211 \ufffd\u305\ufffd\ud835\udc59 \u27f9 \ufffd\u305\ufffd =
\u2211 \ufffd\u305\ufffd\ud835\udc59
\u2211 \ud835\udc59
=
1.125
150
\u27f9 \ufffd\u305\ufffd = \ud835\udfd5, \ud835\udfd3 cm 
 
Portanto as coordenadas do centro de gravidade é o ponto \ud835\udc6a(\ufffd\u305\ufffd; \ufffd\u305\ufffd) = \ud835\udc6a(\ud835\udfd0\ud835\udfd3; \ud835\udfd5, \ud835\udfd3) \ud835\udc1c\ud835\udc26. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
\ud835\udc66 \ud835\udc66 \ud835\udc66 
\ud835\udc65 \ud835\udc65 \ud835\udc65 
\ufffd\u305\ufffd 
\ufffd\u305\ufffd \ufffd\u305\ufffd 
\ufffd\u305\ufffd \ud835\udc361 
\ud835\udc362 
\ud835\udc363 
\ud835\udc36 
\ud835\udc34 \ud835\udc34 \ud835\udc35 
\ud835\udc36 
\ud835\udc35 
 
 ANDRADINA 
2017 
 
 
3 
 
 
Engenharia Mecânica \u2013 5º Período 
 
Discente: Eduardo Moreira Bezerra RA: 1530096890 
Docente: Prof. Juliano Torteli de Godoi Zucato Disciplina: Resistência dos Materiais II Data: 17/04/2017 
 
 3) Encontre as coordenadas do centroide da figura abaixo: 
a) por Integração Direta; 
b) pelo Teorema de Pappus\u2010Guldinus. 
 
 
 
Solução. Na reta (função \ud835\udc53(\ud835\udc65)) temos que quando \ud835\udc65 = 2, \ud835\udc53(\ud835\udc65) = 4. \ud835\udc4f é o coeficiente linear da reta, isto é, \ud835\udc4f = 0. Logo 
o coeficiente angular é da do por 
 
\ud835\udc66 = \ud835\udc53(\ud835\udc65) = \ud835\udc4e\ud835\udc65 + \ud835\udc4f \u27fa 4 = \ud835\udc4e \u22c5 2 + 0 \u27fa \ud835\udc4e =
4
2
= 2 
 
Temos que a função \ud835\udc66 = \ud835\udc53(\ud835\udc65) = 2\ud835\udc65. 
 
a) Integração Direta 
\uf0b7 \ufffd\u305\ufffd\ud835\udc52\ud835\udc59 = \ud835\udc65 
\uf0b7 \ufffd\u305\ufffd\ud835\udc52\ud835\udc59 = \ud835\udc66 2\u2044 
\uf0b7 \ud835\udc51\ud835\udc34 = \ud835\udc66 \ud835\udc51\ud835\udc65 
\uf0b7 \ud835\udc66 = \ud835\udc53(\ud835\udc65) = 2\ud835\udc65 
 
 
\ud835\udc51\ud835\udc34 = \ud835\udc66 \ud835\udc51\ud835\udc65 \u27f9 \u222b \ud835\udc51\ud835\udc34 = \u222b \ud835\udc66 \ud835\udc51\ud835\udc65 \u27f9 \ud835\udc34 = \u222b 2\ud835\udc65
2
0
\ud835\udc51\ud835\udc65 = 2 \u222b \ud835\udc65
2
0
\ud835\udc51\ud835\udc65 = 2 \u22c5 [
\ud835\udc652
2
]
0
2
\u27f9 \ud835\udc34 = 4 m2 
 
i) Momento de primeira ordem em relação ao eixo \ud835\udc66 (\ud835\udc44\ud835\udc66) 
 
\ud835\udc44\ud835\udc66 = \ufffd\u305\ufffd\ud835\udc34 = \u222b \ufffd\u305\ufffd\ud835\udc52\ud835\udc59 \ud835\udc51\ud835\udc34 = \u222b \ud835\udc65 \u22c5 \ud835\udc66
2
0
\ud835\udc51\ud835\udc65 = \u222b \ud835\udc65 \u22c5 2\ud835\udc65
2
0
\ud835\udc51\ud835\udc65 = 2 \u222b \ud835\udc652
2
0
\ud835\udc51\ud835\udc65 = 2 \u22c5 [
\ud835\udc653
3
]
0
2
=
16
3
\u27f9 \ud835\udc44\ud835\udc66 = 5,33 m
3 
 
Logo \ufffd\u305\ufffd\ud835\udc34 = 5,33 \u27f9 \ufffd\u305\ufffd \u22c5 4 = 5,33 \u27f9 \ufffd\u305\ufffd = \ud835\udfcf, \ud835\udfd1\ud835\udfd1 m. 
 
ii) Momento de primeira ordem em relação ao eixo \ud835\udc65 (\ud835\udc44\ud835\udc65) 
 
\ud835\udc44\ud835\udc65 = \ufffd\u305\ufffd\ud835\udc34 = \u222b \ufffd\u305\ufffd\ud835\udc52\ud835\udc59 \ud835\udc51\ud835\udc34 = \u222b
\ud835\udc66
2
\u22c5 \ud835\udc66
2
0
\ud835\udc51\ud835\udc65 =
1
2
\u222b \ud835\udc662
2
0
\ud835\udc51\ud835\udc65 =
1
2
\u222b (2\ud835\udc65)2
2
0
\ud835\udc51\ud835\udc65 =
1
2
\u222b 4\ud835\udc652
2
0
\ud835\udc51\ud835\udc65 = 2 \u222b \ud835\udc652
2
0
\ud835\udc51\ud835\udc65 = 2 \u22c5 [
\ud835\udc653
3
]
0
2
 
 
\ud835\udc44\ud835\udc65 = \ufffd\u305\ufffd\ud835\udc34 = 5,33 m
3 
 
Logo \ufffd\u305\ufffd\ud835\udc34 = 5,33 \u27f9 \ufffd\u305\ufffd \u22c5 4 = 5,33 \u27f9 \ufffd\u305\ufffd = \ud835\udfcf, \ud835\udfd1\ud835\udfd1 m. 
 
Portanto as coordenadas da centroide é o ponto \ud835\udc6a(\ufffd\u305\ufffd; \ufffd\u305\ufffd) = \ud835\udc6a(\ud835\udfcf, \ud835\udfd1\ud835\udfd1; \ud835\udfcf, \ud835\udfd1\ud835\udfd1) \ud835\udc26. 
 
b) Teorema de Pappus\u2010Guldinus 
 
 
A área do triângulo é 
 
\ud835\udc34 =
\ud835\udc4f\u210e
2
 
 
onde \ud835\udc4f = 2 m e \u210e = 4 m. 
 
Pelo segundo teorema de Pappus, podemos obter: 
 
\ud835\udc49\ud835\udc66 = 2\ud835\udf0b\ufffd\u305\ufffd\ud835\udc34 \u27f9 \ufffd\u305\ufffd =
\ud835\udc49\ud835\udc66
2\ud835\udf0b\ud835\udc34
 
 
\ud835\udc49\ud835\udc65 = 2\ud835\udf0b\ufffd\u305\ufffd\ud835\udc34 \u27f9 \ufffd\u305\ufffd =
\ud835\udc49\ud835\udc65
2\ud835\udf0b\ud835\udc34
 
\ud835\udc65 
\ud835\udc65 
\ufffd\u305\ufffd\ud835\udc52\ud835\udc59 
\ud835\udc66 
2 m 
\ufffd\u305\ufffd 
4 m 
\ufffd\u305\ufffd 
\ufffd\u305\ufffd\ud835\udc52\ud835\udc59 \ud835\udc51\ud835\udc65 
\ud835\udc66 
\ud835\udc66 
\ud835\udc65 
(\ud835\udc65; \ud835\udc66) 
\ud835\udc53(\ud835\udc65) \u27f6 
\u2199 
\ud835\udc52\ud835\udc59 
 
 ANDRADINA 
2017 
 
 
4 
 
 
Engenharia Mecânica \u2013 5º Período 
 
Discente: Eduardo Moreira Bezerra RA: 1530096890 
Docente: Prof. Juliano Torteli de Godoi Zucato Disciplina: Resistência dos Materiais II Data: 17/04/2017 
 
 i) Centroide \ufffd\u305\ufffd 
 
 
\ud835\udc49\ud835\udc36\ud835\udc56\ud835\udc59\ud835\udc56\ud835\udc5b\ud835\udc51\ud835\udc5f\ud835\udc5c = \ud835\udc34\ud835\udc4f\u210e = \ud835\udf0b\ud835\udc4f
2\u210e \ud835\udc49\ud835\udc36\ud835\udc5c\ud835\udc5b\ud835\udc52 =
1
3
\ud835\udc34\ud835\udc4f\u210e =
1
3
\ud835\udf0b\ud835\udc4f2\u210e 
 
\ud835\udc49\ud835\udc66 = \ud835\udc49\ud835\udc36\ud835\udc56\ud835\udc59\ud835\udc56\ud835\udc5b\ud835\udc51\ud835\udc5f\ud835\udc5c \u2212 \ud835\udc49\ud835\udc36\ud835\udc5c\ud835\udc5b\ud835\udc52 
\ud835\udc49\ud835\udc66 = \ud835\udf0b\ud835\udc4f
2\u210e \u2212
1
3
\ud835\udf0b\ud835\udc4f2\u210e =
2
3
 \ud835\udf0b\ud835\udc4f2\u210e 
 
A centroide \ufffd\u305\ufffd é igual a: 
 
\ufffd\u305\ufffd =
\ud835\udc49\ud835\udc66
2\ud835\udf0b\ud835\udc34
=
2
3 \ud835\udf0b\ud835\udc4f
2\u210e
2\ud835\udf0b\ud835\udc34
=
2
3 \ud835\udf0b\ud835\udc4f
2\u210e
2\ud835\udf0b
\ud835\udc4f\u210e
2
=
2
3
\ud835\udf0b\ud835\udc4f2\u210e \u22c5
1
\ud835\udf0b\ud835\udc4f\u210e
\u27f9 \ufffd\u305\ufffd =
\ud835\udfd0
\ud835\udfd1
\ud835\udc83 
 
ii) Centroide \ufffd\u305\ufffd 
 
 
 
 
\ud835\udc49\ud835\udc65 =
1
3
\ud835\udc34\ud835\udc4f\ud835\udc4f =
1
3
\ud835\udf0b\u210e2\ud835\udc4f 
 
A centroide \ufffd\u305\ufffd é igual a: 
 
\ufffd\u305\ufffd =
\ud835\udc49\ud835\udc65
2\ud835\udf0b\ud835\udc34
=
1
3 \ud835\udf0b\u210e
2\ud835\udc4f
2\ud835\udf0b\ud835\udc34
=
1
3 \ud835\udf0b\u210e
2\ud835\udc4f
2\ud835\udf0b
\ud835\udc4f\u210e
2
=
1
3
\ud835\udf0b\u210e2\ud835\udc4f \u22c5
1
\ud835\udf0b\ud835\udc4f\u210e
\u27f9 \ufffd\u305\ufffd =
\ud835\udfcf
\ud835\udfd1
\ud835\udc89 
 
Portanto, a centroide é \ud835\udc36 (
2
3
\ud835\udc4f; 
1
3
\u210e) = \ud835\udc6a(\ud835\udfcf, \ud835\udfd1\ud835\udfd1; \ud835\udfcf, \ud835\udfd1\ud835\udfd1) m 
 
 
 
 
 
\ud835\udc66 
\ud835\udc65 
\ud835\udc67 
\ud835\udc4f 
\u210e 
\ud835\udc66 
\ud835\udc65 
\ufffd\u305\ufffd =
2
3
\ud835\udc4f 
\ufffd\u305\ufffd =
1
3
\u210e 
 
 ANDRADINA 
2017 
 
 
5 
 
 
Engenharia Mecânica \u2013 5º Período 
 
Discente: Eduardo Moreira Bezerra RA: 1530096890 
Docente: Prof. Juliano Torteli de Godoi Zucato Disciplina: Resistência dos Materiais II Data: 17/04/2017 
 
 
4) Encontre o centroide da área plana mostrada abaixo. 
 
 
 
Solução. Na curva (função \ud835\udc66 = \ud835\udc53(\ud835\udc65)) temos que quando \ud835\udc65 = 480, \ud835\udc53(\ud835\udc65) = 200. Por hipótese a curva é uma parábola 
passando pela origem. Então seja \ud835\udc58 coeficiente angular, temos que: 
 
\ud835\udc53(\ud835\udc65) = \ud835\udc58\ud835\udc652 \u27fa \ud835\udc53(480) = \ud835\udc58(480)2 \u27fa 200 = 230.400\ud835\udc58 \u27fa \ud835\udc58 =
200
230.400
\u27fa \ud835\udc58 =
1
1.152
 
 
Por integração direta, temos 
 
 
\uf0b7 \ufffd\u305\ufffd\ud835\udc52\ud835\udc59 = \ud835\udc65 
\uf0b7 \ufffd\u305\ufffd\ud835\udc52\ud835\udc59 = \ud835\udc66 2\u2044 
\uf0b7 \ud835\udc51\ud835\udc34 = \ud835\udc66 \ud835\udc51\ud835\udc65 
\uf0b7 \ud835\udc66 = \ud835\udc53(\ud835\udc65) =
1
1.152
\ud835\udc652 
 
\ud835\udc51\ud835\udc34 = \ud835\udc66 \ud835\udc51\ud835\udc65 \u27f9 \u222b \ud835\udc51\ud835\udc34 = \u222b \ud835\udc66 \ud835\udc51\ud835\udc65 \u27f9 \ud835\udc34 = \u222b
1
1.152
\ud835\udc652
480
240
\ud835\udc51\ud835\udc65 =
1
1.152
\u222b \ud835\udc652
480
240
\ud835\udc51\ud835\udc65 =
1
1.152
\u22c5 [
\ud835\udc653
3
]
240
480
 
 
\ud835\udc34 =
1
1.152
\u22c5 [
4803
3
\u2212
2403
3
] = 28 \u22c5 103 mm2 
 
i) Momento de primeira ordem em relação ao eixo \ud835\udc66 (\ud835\udc44\ud835\udc66) 
 
\ud835\udc44\ud835\udc66 = \ufffd\u305\ufffd\ud835\udc34 = \u222b \ufffd\u305\ufffd\ud835\udc52\ud835\udc59 \ud835\udc51\ud835\udc34 = \u222b \ud835\udc65 \u22c5 \ud835\udc66
480
240
\ud835\udc51\ud835\udc65 = \u222b \ud835\udc65 \u22c5
1
1.152
\ud835\udc652
480
240
\ud835\udc51\ud835\udc65 =
1
1.152
\u222b \ud835\udc653
480
240
\ud835\udc51\ud835\udc65 =
1
1.152
\u22c5 [
\ud835\udc654
4
]
240
480