Centro de Gravedad y Momento de Inercia (Ejercicio)
November 19, 2023 | Author: Anonymous | Category: N/A
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Description
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=
π 2 β 2
= =
5 2
= 2,5 cm2
2 = 1cm 2
FIGURA 2 A= b x h= 1cm x 10 cm = 10 cm ^2 XΜ
= Θ²=
π 2 β 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=
π 2 β 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=
π 2 β 2
= =
5 2
= 2,5 cm2
2 2
= 1cm FIGURA 5
A= b x h = 1cm x 10cm = 10 cm^2 XΜ
= Θ²=
π 2 β 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
π
XΜ
= Θ²=
2 β 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
βA=60
β π¨πΏΜ
XΜ
= β =
0
=0
π¨
60
β π¨πΜ
420
YΜ
= β = π¨
60
= 7cm
βAXΜ
=0
βAΘ²=420
CG (0, 7)
MOMENTO DE INERCIA FIGURA 1
πΌπ₯
=
ππ₯βΒ³ 12
=
5(2)Β³ 12
= 3,333
π 3 β (5)3 (2) πΌπ¦ = = = 20,833 12 12 FIGURA 2
1(10)3 πΌπ₯ = = 83.333 12 (1)3 (10) Iy = = 0,833 12 FIGURA 3
5(2)3 πΌπ₯ = = 3.333 12 (5)3 (2) Iy = = 20,833 12 FIGURA 4
5(2)3 πΌπ₯ = = 3.333 12 (5)3 (2) Iy = = 20,833 12
FIGURA 5
1(10)3 πΌπ₯ = = 83.333 12 (1)3 (10) Iy = = 0,833 12 FIGURA 6
5(2)3 πΌπ₯ = = 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
β=30
β=179,998
β=84,998
β=14400
Ix= βIx + βAdy2 Ix= 179,998 + 14400 = 1479,998 cm4 Iy= βIy + βAdx2 Iy= 84,998 + 30 =114,998 cm4
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