3U]\NáDGURGHNFL *NRFLILJXU\SáDVNLHM±]DGDQLH

=QDOH(üURGHNFL *NRFLILJXU\SU]HGVWDZLRQHMSRQL*HM

Rysunek 1

: FHOX SROLF]HQLD ZVSyáU] GQ\FK URGND FL *NRFL ILJXU\ ] U\VXQNX SU]\M WR XNáDG RVL L

SRG]LHORQRMQDWU]\ILJXU\VNáDGRZHMDNQDU\VXQNX

Rysunek 2

=QDQH V SRáR*HQLD URGNyZ FL *NRFL ILJXU\ L URGHN FL *NRFL üZLDUWNL NRáD OH*\ QD

GZXVLHF]QHM NWD SURVWHJR JG\* MHVW WR R V\PHWULL ILJXU\ QLH]QDQH V MHGQDN MHJR

ZVSyáU] GQH

3ROLF]P\ ]DWHP RJyOQLH ZVSyáU] GQH URGND FL *NRFL Z\FLQND NRáD R NFLH ZHZQ WU]Q\P

3 1D U\VXQNX SU]HGVWDZLRQR Z\FLQHN NRáD ZUD] ] SU]\M W\P XNáDGHP ZVSyáU] GQ\FK

RUD]ZSURZDG]RQ\PLGODXáDWZLHQLDREOLF]HZVSyáU] GQ\PLELHJXQRZ\PL

Rysunek 3

Przy obliczaniu momentów statycznych Sx i SySRGVWDZLDP\QDVW SXMFR

[ = U FRVα \ = U VLQ α

G$ = GU U Gα = U GU Gα

Pole powierzchni G$SROLF]RQH]RVWDáRMDNGODSURVWRNWDRERNDFK dr i UGα GáXJRüERF]QHM

FLDQNLLGáXJRüáXNXFRGODPDá\FKZLHONRFLNWD Gα jest prawdziwe.

3U]HG]LDá\ ]PLHQQRFL GOD ZVSyáU] GQ\FK ELHJXQRZ\FK GOD UR]SDWU\ZDQHJR Z\FLQND NRáD

Z\QRV]

U ∈< 5 > α ∈< −ϕ ϕ > .

/LF]HQLHZVSyáU] GQHMxCPR*QDSRPLQüJG\*URGHNFL *NRFLILJXU\OH*\QDRVL\

3R SRGVWDZLHQLX ZVSyáU] GQ\FK ELHJXQRZ\FK X]\VNDQH FDáN ZH Z]RU]H QD Sx PR*QD

REOLF]\üQLH]DOH*QLHGODRE\GZXZVSyáU] GQ\FK

π +

π

ϕ

5

+

ϕ



π

π



5









6 = π

U VLQ α GU Gα =

U (− FRVα )

= − 5

∫

 FRV

+ ϕ

FRV

ϕ  =

−ϕ ∫

[

π







 −

−







−ϕ





= −

(

5

VLQ ϕ

VLQ ϕ

5 VLQ ϕ

[ − ( )− ( ) ]=

( )

2

3ROHSRZLHU]FKQLILJXU\PR*QDSROLF]\üZVWRVXQNXGRSRZLHU]FKQLFDáHJRNRáD

ϕ

$ = π 5

= ϕ 5

π

:VSyáU] GQDSLRQRZDURGNDFL *NRFLILJXU\ZSU]\M W\FKRVLDFK[\Z\QRVL

5 VLQ(ϕ)

6

5 VLQ ϕ

[

( )

\ =

=

=

F

$

ϕ 5

ϕ

π

π

'ODüZLDUWNLNRáD]QDMGXMHP\ZVSyáU] GQy

=

ϕ =

CSRGVWDZLDMF ϕ

, bo

:

 π

5 VLQ

5

 





5

\ =

=

=

F

π

π

π

:UDFDMFGRILJXU\]U\VXQNXZVSyáU] GQHURGNDFL *NRFLILJXU\SU]\SU]\M W\FKMDNQD

U\VXQNXRVLDFKXNáDGXZVSyáU] GQ\FKZ\QRV]

5

5

[

= \ =

=

&

&

π

π

6WRVXMFPHWRG JUXSRZDQLDILJXURWU]\PXMHP\Z]RU\QDVWDW\F]QHPRPHQW\EH]ZáDGQRFL

π 5 5



6 =

5

5 +

+ 5 − 5

5

5

5

5

[

π





 =

+

−

=

π



5 5

6

= 5 − 5

5

5

5

5

5

5

\





 +

+

= −

+

+

=

π

Pole powierzchni wynosi:

π 5

π

+

$ =

5 +

+ 5 =

5

:VSyáU] GQHURGNDFL *NRFLZ\QRV]

6

5

\

[

=

=

5 ≅ 5

&

+

=

οπ

$

π

( + )

5

5

6

[

\

=

=

5 ≅ 5

&

+

=

οπ

$

π

( + )

5

=QDMGXMFQDU\VXQNXURGHNFL *NRFLX]\VNXMHP\

3

Rysunek 4

4