kN

10

3

N

:=

π

3.142

=

E

2.1 10

5



N

mm

2

:=

Q

52kN

:=

H

340mm

:=

h

100 mm



:=

x

5.5

:=

dlugosc zredukowana

β

2

:=

Δ

1.4 h



:=

Δ

140 mm

=

l

H

0.5 h



+

Δ

+

:=

l

530 mm

=

lr

β l



:=

lr

1.06

10

3



mm

=

Wybieramy material na srube

St5 stale konstrukcyjne zwyklej jakosci, stan prety walcowane

Sc

215

N

mm

2

:=

λgr

100

:=

kc

136

N

mm

2

:=

λgra

π

E

Sc



:=

λgra

98.184

=

lE

λgra

2

1

4

π

3



E





Q x





:=

lE

1.01

10

3



mm

=

lr

lE

>

1

=

zakres sprezysty

srednica rdzenia sruby

d1

4

64 Q



x

 lr

2



(

)

π

3

E



:=

d1

42.157 mm

=

D1

77 mm



:=

d

95 mm



:=

d1

74.174 mm



:=

S95x12

sruby sciskane i zginane

P

12 mm



:=

kg

kc

:=

Wg

π d1

3



(

)

32

:=

Wg

4.006

10

4



mm

3

=

kw

Sc

x

:=

kw

3.909

10

7



Pa

=

kE

kw

:=

αg

3 deg



:=

Pv

Q cos

αg

(

)



:=

ψ

kw

kc

:=

Ph

Q sin

αg

(

)



:=

ψ

0.287

=

F1

π d1

2

4

:=

F1

4.321

10

3



mm

2

=

smuklosc sruby

J

π

d1

4

64

:=

J

1.486

10

6

-



m

4

=

i

J

F1

:=

i

18.544 mm

=

λ

lr

i

:=

wspolczynnik smuklosci

λ

57.163

=

Maksymalny moment zginajacy

M

Ph lr



:=

M

2.885

10

3



J

=

Srube zginane i sciskane

M

Wg













Pv

ψ F1















+

kc



1

=

moment skrecajacy

ds

D1

d

+

(

)

2

:=

ds

86 mm

=

γ

atan

P

π ds















:=

γ

2.543 deg

=

β

3deg

:=

μ

0.15

:=

ρ

atan

μ

cos

β

( )













:=

ρ

8.542 deg

=

H

Q tan

ρ

γ

+

(

)



:=

H

1.019

10

4



N

=

Ms

0.5 Q



ds



tan

ρ

γ

+

(

)



:=

Ms

438.096 J

=

naprezenie sciskajace

σc

Q

cos

αg

(

)

F1



:=

σc

1.202

10

7



Pa

=

naprezenie skrecajace

Wo

π

d1

3

16



:=

Wo

0.08 L

=

τs

Ms

Wo

:=

τs

5.467

10

6



Pa

=

naprezenia gnace

σg

Q l



sin

αg

(

)

Wg



:=

σg

3.6

10

7



Pa

=

σz

σc

σg

+

(

)

2

3

τs

2



+

:=

σz

4.894

10

7



Pa

=

σz

kc



1

=