Opracował: dr in

ż

. Mariusz Leus

- 1 -

T: Wytrzymało

ść

zło

ż

ona – zginanie i skr

ę

canie

Zadanie 1.

Wykorzystuj

ą

c hipotez

ę

Hubera obliczy

ć

ś

rednic

ę

d wału na którym osadzone s

ą

dwa koła pasowe

1 i 2 o

ś

rednicy D

1

= 100 mm i D

2

= 400 mm oraz ci

ęż

arze G

1

= 200 N i G

2

= 500 N. Wał przenosi

moc N = 52.36 kW przy pr

ę

dko

ś

ci n = 500 obr/min. Wykona

ć

wykresy momentów gn

ą

cych,

skr

ę

caj

ą

cych i zredukowanych. Przyj

ąć

k

z

= 120 N/mm

2

. Rozmieszczenie sił przyj

ąć

jak na rys.

Dane: D

1

= 100 mm, D

2

= 400 mm, G

1

= 200 N, G

2

= 500 N, N = 52.36 kW, n = 500 obr/min,

a = 0.1 m, k

z

= 120 N/mm

2

Szukane: d = ?

Opracował: dr in

ż

. Mariusz Leus

- 2 -

1. Moment skr

ę

caj

ą

cy M

S

n

N

M

S

⋅

=

3

,

9549

Nm

1000

500

36

.

52

3

,

9549

=

⋅

=

S

M

Nm

1000

=

S

M

2. Sił P

1

i P

2

na kole 1 i 2

2

2

2

2

1

1

D

P

D

P

M

S

⋅

=

⋅

=

kN

20

N

20000

1

.

0

1000

2

2

1

1

=

=

⋅

=

=

D

M

P

S

kN

20

1

=

P

kN

5

N

5000

4

.

0

1000

2

2

2

2

=

=

⋅

=

=

D

M

P

S

kN

5

2

=

P

3. Zginanie w płaszczy

ź

nie Axy

Reakcje w ło

ż

yskach A i B w płaszczy

ź

nie Axy:

a)

∑

=

0

A

M

;

(

)

0

4

3

2

2

1

=

⋅

−

⋅

+

+

⋅

a

R

a

G

P

a

G

By

(

)

(

)

4

.

0

3

.

0

500

5000

1

.

0

200

4

3

2

2

1

⋅

+

+

⋅

=

⋅

+

+

⋅

=

a

a

G

P

a

G

R

By

N

4175

=

By

R

b)

∑

=

0

B

M

;

(

)

0

3

4

2

2

1

=

⋅

+

−

⋅

−

⋅

a

G

P

a

G

a

R

Ay

(

)

(

)

4

.

0

1

.

0

500

5000

3

.

0

200

4

3

2

2

1

⋅

+

+

⋅

=

⋅

+

+

⋅

=

a

a

G

P

a

G

R

Ay

N

1525

=

Ay

R

Momenty gn

ą

ce w płaszczy

ź

nie Axy:

Nm

0

=

A

y

g

M

Nm

5

.

152

1

0

1525

1

.

0

1

=

⋅

=

⋅

=

.

R

M

Ay

y

g

Nm

5

.

417

1

.

0

4175

1

.

0

2

=

⋅

=

⋅

=

By

y

g

R

M

Nm

0

=

B

y

g

M

4. Zginanie w płaszczy

ź

nie Axz

Reakcje w ło

ż

yskach A i B w płaszczy

ź

nie Axz:

a)

∑

=

0

A

M

;

0

4

1

=

⋅

−

⋅

a

R

a

P

Bz

4

.

0

1

.

0

20000

4

1

⋅

=

⋅

=

a

a

P

R

Bz

N

5000

=

Bz

R

b)

∑

=

0

B

M

;

0

3

4

1

=

⋅

−

⋅

a

P

a

R

Az

4

.

0

3

.

0

20000

4

3

1

⋅

=

⋅

=

a

a

P

R

Az

N

15000

=

Az

R

Opracował: dr in

ż

. Mariusz Leus

- 3 -

Momenty gn

ą

ce w płaszczy

ź

nie Axz:

Nm

0

=

A

z

g

M

Nm

1500

1

.

0

15000

1

.

0

1

=

⋅

=

⋅

=

Az

z

g

R

M

Nm

500

1

.

0

5000

1

.

0

2

=

⋅

=

⋅

=

Bz

z

g

R

M

Nm

0

=

B

z

g

M

5. Momenty gn

ą

ce wypadkowe

2

2

z

g

y

g

w

g

M

M

M

+

=

Nm

0

0

0

2

2

2

2

=

+

=

+

=

A

z

g

A

y

g

A

w

g

M

M

M

Nm

7

.

1507

1500

5

.

152

2

2

2

1

2

1

1

=

+

=

+

=

z

g

y

g

w

g

M

M

M

Nm

4

.

651

500

5

.

417

2

2

2

2

2

2

2

=

+

=

+

=

z

g

y

g

w

g

M

M

M

Nm

0

0

0

2

2

2

2

=

+

=

+

=

B

z

g

B

y

g

B

w

g

M

M

M

6. Momenty zredukowane

2

2

75

.

0

S

w

g

red

M

M

M

⋅

+

=

Nm

0

0

75

.

0

0

75

.

0

2

2

2

2

=

⋅

+

=

⋅

+

=

A

S

A

w

g

A

red

M

M

M

Nm

7

.

1507

0

75

.

0

7

.

1507

75

.

0

2

2

2

1

2

1

L

1

=

⋅

+

=

⋅

+

=

S

w

g

red

M

M

M

Nm

7

.

1738

1000

75

.

0

7

.

1507

75

.

0

2

2

2

1

2

1

P

1

=

⋅

+

=

⋅

+

=

S

w

g

red

M

M

M

Nm

7

.

1083

1000

75

.

0

4

.

651

75

.

0

2

2

2

2

2

2

L

2

=

⋅

+

=

⋅

+

=

S

w

g

red

M

M

M

Nm

4

.

651

0

75

.

0

4

.

651

75

.

0

2

2

2

2

2

2

P

2

=

⋅

+

=

⋅

+

=

S

w

g

red

M

M

M

Nm

0

0

75

.

0

0

75

.

0

2

2

2

2

=

⋅

+

=

⋅

+

=

B

S

B

w

g

B

red

M

M

M

7.

Ś

rednica wału d

g

z

red

red

k

W

M

≤

=

max

σ

;

32

3

d

W

z

⋅

=

π

;

Nm

7

.

1738

max

=

red

M

g

red

red

k

d

M

≤

⋅

=

32

3

max

π

σ

mm

85

.

52

120

1000

7

.

1738

32

32

3

3

max

=

⋅

⋅

⋅

=

⋅

⋅

=

π

π

g

red

k

M

d

mm

85

.

52

=

d