kolokwium II semestr geometria mas rozwiązania

background image

Zadanie 1

Wyznaczyć główne, centralne osie i momenty bezwładności

33

4

6

R 2

y

A

x

A

10 6

1

2

6

⋅ 3

π 2

2

2

+

:=

A

57.283

=

Sx

1

2

− 6

⋅ 3

2

(

)

π 2

2

2

3

4

3

2
π

+

⎛⎜

⎞⎟

+

:=

Sx 42.183

=

Sy

10 6

⋅ 1

1

2

6

⋅ 3

⋅ 2

π 2

2

2

2

(

)

+

:=

Sy 29.434

=

xc

Sy

A

:=

xc 0.514

=

yc

Sx

A

:=

yc 0.736

=

background image

α

1

α

2

+

90.000

=

α

2

5.246

=

I2c 200.556

=

α

2

atan

Ixcyc

Iyc I2c

180

π

:=

I2c

Ixc Iyc

+

2

Ixc Iyc

2

2

Ixcyc

2

+

:=

α

1

84.754

=

I1c 525.004

=

α

1

atan

Ixcyc

Iyc I1c

180

π

:=

I1c

Ixc Iyc

+

2

Ixc Iyc

2

2

Ixcyc

2

+

+

:=

Ixcyc

29.54

=

Iyc 522.292

=

Ixc 203.269

=

Ixcyc

10 6

1

xc

(

)

yc

( )

6

2

3

2

72

1

2

6

⋅ 3

2

xc

(

)

2

yc

(

)

+

π 2

2

2

2

xc

(

)

3

4

3

2
π

+

yc

⎛⎜

⎞⎟

+

:=

Iyc

10

3

6

12

10 6

1

xc

(

)

2

+

6

3

3

36

1

2

6

⋅ 3

2

xc

(

)

2

+

π 2

4

8

+

π 2

2

2

2

xc

(

)

2

+

:=

Ixc

10 6

3

12

10 6

yc

( )

2

+

6 3

3

36

1

2

6

⋅ 3

2

yc

(

)

2

+

π 2

4

8

+

π 2

2

2

4

3

2
π

⎛⎜

⎞⎟

2

π 2

2

2

3

4

3

2
π

+

yc

⎛⎜

⎞⎟

2

+

:=

Sprawdzenie:

Ixcyc

29.54

=

Ixcyc

Ixy A xc

yc

:=

Iyc 522.292

=

Iyc

Iy A xc

2

:=

Ixc 203.269

=

Ixc

Ix A yc

2

:=

Ixy

7.866

=

Ixy

6

2

3

2

72

1

2

6

⋅ 3

⋅ 2

2

(

)

+

π 2

2

2

2

(

)

3

4

3

2
π

+

⎛⎜

⎞⎟

+

:=

Iy 537.416

=

Iy

10

3

6

12

10 6

⋅ 1

2

+

6

3

3

12

π 2

4

8

+

π 2

2

2

2

(

)

2

+

:=

Ix 234.332

=

Ix

10 6

3

12

6 3

3

36

1

2

6

⋅ 3

2

(

)

2

+

π 2

4

8

+

π 2

2

2

4

3

2
π

⎛⎜

⎞⎟

2

π 2

2

2

3

4

3

2
π

+

⎛⎜

⎞⎟

2

+

:=

Zadanie 1 - s2

background image

Zadanie 1 - s3

33

4

6

R 2

yc

x

x

x

y

xc

A

x

C

1c

2c

α1

α2

background image

Zadanie 2

Wyznaczyć główne, centralne osie i momenty bezwładności

4

6

42

3

R 2

y

A

x

x

x

x

A

10 6

1

2

6

⋅ 3

+

π 2

2

2

:=

A

62.717

=

Sx

10 6

1

(

)

1

2

6

⋅ 3

⋅ 3

+

:=

Sx

33

=

Sy

10 6

⋅ 1

1

2

6

⋅ 3

⋅ 2

+

π 2

2

2

4

4

3

2
π

+

⎛⎜

⎞⎟

:=

Sy 97.799

=

xc

Sy

A

:=

xc 1.559

=

yc

Sx

A

:=

yc

0.526

=

background image

α

1

α

2

+

90.000

=

α

2

20.312

=

I2c 286.692

=

α

2

atan

Ixcyc

Iyc I2c

180

π

:=

I2c

Ixc Iyc

+

2

Ixc Iyc

2

2

Ixcyc

2

+

:=

α

1

69.688

=

I1c 412.507

=

α

1

atan

Ixcyc

Iyc I1c

180

π

:=

I1c

Ixc Iyc

+

2

Ixc Iyc

2

2

Ixcyc

2

+

+

:=

Ixcyc 40.96

=

Iyc 397.346

=

Ixc 301.853

=

Ixcyc

10 6

1

xc

(

)

1

yc

(

)

6

2

3

2

72

1

2

6

⋅ 3

2

xc

(

)

3

yc

(

)

+

π 2

2

2

4

4

3

2
π

+

xc

⎛⎜

⎞⎟

yc

( )

:=

Iyc

10

3

6

12

10 6

1

xc

(

)

2

+

6

3

3

36

+

1

2

6

⋅ 3

2

xc

(

)

2

+

π 2

4

8

π 2

2

2

4

3

2
π

⎛⎜

⎞⎟

2

π 2

2

2

4

4

3

2
π

+

xc

⎛⎜

⎞⎟

2

+

:=

Ixc

10 6

3

12

10 6

1

yc

(

)

2

+

6 3

3

36

+

1

2

6

⋅ 3

3

yc

(

)

2

+

π 2

4

8

π 2

2

2

yc

( )

2

+

:=

Sprawdzenie:

Ixcyc 40.96

=

Ixcyc

Ixy A xc

yc

:=

Iyc 397.346

=

Iyc

Iy A xc

2

:=

Ixc 301.853

=

Ixc

Ix A yc

2

:=

Ixy

10.5

=

Ixy

10 6

⋅ 1

1

(

)

6

2

3

2

72

1

2

6

⋅ 3

⋅ 2

⋅ 3

+

:=

Iy 549.853

=

Iy

10

3

6

12

10 6

⋅ 1

2

+

6

3

3

12

+

π 2

4

8

π 2

2

2

4

3

2
π

⎛⎜

⎞⎟

2

π 2

2

2

4

4

3

2
π

+

⎛⎜

⎞⎟

2

+

:=

Ix 319.217

=

Ix

10 6

3

12

10 6

1

(

)

2

+

6 3

3

36

+

1

2

6

⋅ 3

⋅ 3

2

+

π 2

4

8

:=

Zadanie 2 - s2

background image

Zadanie 2 - s3

4

6

42

3

R 2

y

C

yc

A

α2

α1

1c

2c

x

x

x

x
xc

background image

Zadanie 3

Wyznaczyć główne, centralne osie i momenty bezwładności

4

6

333

R 2

x

x

x

x

y

A

A

10 6

1

2

6

⋅ 3

+

π 2

2

2

:=

A

62.717

=

Sx

1

2

6

⋅ 3

⋅ 4

π 2

2

2

1

(

)

:=

Sx 42.283

=

Sy

10 6

⋅ 1

1

2

6

⋅ 3

⋅ 2

+

π 2

2

2

6

4

3

2
π

⎛⎜

⎞⎟

:=

Sy 45.634

=

xc

Sy

A

:=

xc 0.728

=

yc

Sx

A

:=

yc 0.674

=

background image

α

1

α

2

+

90.000

=

α

2

23.948

=

I2c 256.737

=

α

2

atan

Ixcyc

Iyc I2c

180

π

:=

I2c

Ixc Iyc

+

2

Ixc Iyc

2

2

Ixcyc

2

+

:=

α

1

66.052

=

I1c 443.007

=

α

1

atan

Ixcyc

Iyc I1c

180

π

:=

I1c

Ixc Iyc

+

2

Ixc Iyc

2

2

Ixcyc

2

+

+

:=

Ixcyc 69.1

=

Iyc 412.318

=

Ixc 287.427

=

Ixcyc

10 6

1

xc

(

)

yc

( )

6

2

3

2

72

1

2

6

⋅ 3

2

xc

(

)

4

yc

(

)

+

π 2

2

2

6

4

3

2
π

xc

⎛⎜

⎞⎟

1

yc

(

)

:=

Iyc

10

3

6

12

10 6

1

xc

(

)

2

+

6

3

3

36

+

1

2

6

⋅ 3

2

xc

(

)

2

+

π 2

4

8

π 2

2

2

4

3

2
π

⎛⎜

⎞⎟

2

π 2

2

2

6

4

3

2
π

xc

⎛⎜

⎞⎟

2

+

:=

Ixc

10 6

3

12

10 6

yc

( )

2

+

6 3

3

36

+

1

2

6

⋅ 3

4

yc

(

)

2

+

π 2

4

8

π 2

2

2

1

yc

(

)

2

+

:=

Sprawdzenie:

Ixcyc 69.1

=

Ixcyc

Ixy A xc

yc

:=

Iyc 412.318

=

Iyc

Iy A xc

2

:=

Ixc 287.427

=

Ixc

Ix A yc

2

:=

Ixy 99.866

=

Ixy

6

2

3

2

72

1

2

6

⋅ 3

⋅ 2

⋅ 4

+

π 2

2

2

6

4

3

2
π

⎛⎜

⎞⎟

1

(

)

:=

Iy 445.522

=

Iy

10

3

6

12

10 6

⋅ 1

2

+

6

3

3

12

+

π 2

4

8

π 2

2

2

4

3

2
π

⎛⎜

⎞⎟

2

π 2

2

2

6

4

3

2
π

⎛⎜

⎞⎟

2

+

:=

Ix 315.934

=

Ix

10 6

3

12

6 3

3

36

+

1

2

6

⋅ 3

⋅ 4

2

+

π 2

4

8

π 2

2

2

1

(

)

2

+

:=

Zadanie 3 - s2

background image

Zadanie 3 - s3

4

6

33

3

R 2

y

A

x

xc

yc

α1

α2

2c

C

1c


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