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# LISTA RM II 02

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```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

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

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

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

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

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```