Centro de Gravedad y Momento de Inercia (Ejercicio)

November 19, 2023 | Author: Anonymous | Category: N/A
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Hallar el centro de gravedad de la figura y momento de inercia con respecto a un eje que pase por su centro de gravedad y que es paralelo al eje.

CG (0, Y)

FIGURA 1 A= b x h = 5 cm x 2cm = 10cm^2 XΜ…= Θ²=

𝑏 2 β„Ž 2

= =

5 2

= 2,5 cm2

2 = 1cm 2

FIGURA 2 A= b x h= 1cm x 10 cm = 10 cm ^2 XΜ…= Θ²=

𝑏 2 β„Ž 2

= =

1 2

= 0,5 cm2

10 = = 5 cm +2 cm = 7cm 2

FIGURA 3 A= b x h = 5cm x 2cm = 10 cm^2 XΜ…= Θ²=

𝑏 2 β„Ž 2

= =

5 2

= 2,5 cm2

2 = 1cm + 12cm = 13cm 2

FIGURA 4 A= b x h = 5cm x 2cm = 10 cm^2 XΜ…= Θ²=

𝑏 2 β„Ž 2

= =

5 2

= 2,5 cm2

2 2

= 1cm FIGURA 5

A= b x h = 1cm x 10cm = 10 cm^2 XΜ…= Θ²=

𝑏 2 β„Ž 2

= =

1 2

= 0,5 cm2

10 = = 5 cm +2 cm = 7cm 2

FIGURA 6 A= b x h = 5cm x 2cm = 10 cm^2

𝑏

XΜ…= Θ²=

2 β„Ž 2

= =

5 2 2 2

= 2,5 cm2

= 1cm + 12cm = 13cm

FIG

A(cm2)

X (cm)

Y (cm)

AXΜ… (cm3)

AΘ² (cm3)

1

10

-2,5

1

-25

10

2

10

-0,5

7

-5

70

3

10

-2,5

13

-25

130

4

10

-2,5

1

25

10

5

10

-0,5

7

5

70

6

10

-2,5

13

25

βˆ‘A=60

βˆ‘ 𝑨𝑿̅

XΜ… = βˆ‘ =

0

=0

𝑨

60

βˆ‘ 𝑨𝒀̅

420

YΜ… = βˆ‘ = 𝑨

60

= 7cm

βˆ‘AXΜ…=0

βˆ‘AΘ²=420

CG (0, 7)

MOMENTO DE INERCIA FIGURA 1

𝐼π‘₯

=

𝑏π‘₯β„ŽΒ³ 12

=

5(2)Β³ 12

= 3,333

𝑏 3 β„Ž (5)3 (2) 𝐼𝑦 = = = 20,833 12 12 FIGURA 2

1(10)3 𝐼π‘₯ = = 83.333 12 (1)3 (10) Iy = = 0,833 12 FIGURA 3

5(2)3 𝐼π‘₯ = = 3.333 12 (5)3 (2) Iy = = 20,833 12 FIGURA 4

5(2)3 𝐼π‘₯ = = 3.333 12 (5)3 (2) Iy = = 20,833 12

FIGURA 5

1(10)3 𝐼π‘₯ = = 83.333 12 (1)3 (10) Iy = = 0,833 12 FIGURA 6

5(2)3 𝐼π‘₯ = = 3.333 12 (5)3 (2) Iy = = 20,833 12 dy= 1-7 = -6 cm

dx= -2,5-0= -2,5

dy= 7-7= 0

dx= -0,5-0= -0,5

dy= 13-7= 6

dx= -2,5-0= -2,5

FIG A(cm2)

dy

Ady2

dx

Adx2

IX

IY

1

10

-6

3600

-2,5

62,5

3,333

20,833

2

10

0

0

-0,5

2,5

83,333

0,833

3

10

6

3600

-2,5

62,5

3,333

20,833

4

10

-6

3600

2,5

62,5

3,333

20,833

5

10

0

0

0,5

2,5

83,333

0,833

6

10

6

3600

2,5

62,5

3,333

20,833

βˆ‘=30

βˆ‘=179,998

βˆ‘=84,998

βˆ‘=14400

Ix= βˆ‘Ix + βˆ‘Ady2 Ix= 179,998 + 14400 = 1479,998 cm4 Iy= βˆ‘Iy + βˆ‘Adx2 Iy= 84,998 + 30 =114,998 cm4

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