3U]\NáDG5DPD]DPNQLWD]HFLJLHP

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l

q

ql

l

l

l

5R]ZL]DQLH

8NáDGVLáRWU]\PDQ\SRXZROQLHQLX]ZL]yZSU]HGVWDZLRQ\MHVWQDU\VXQNXSRQL*HM

B

q

ql

A

1

2

H

A

M

B

V

A

V

B

3

4

x

2EOLF]P\ UHDNFM +

A

MHG\Q SR]LRP Z\NRU]\VWXMF UyZQDQLH

0

P

i

ix

=

∑

. Przyjmuje ono

SRVWDü+

A

- ql = 0

⇒

H

A

= ql.

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roz-

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W\FK SXQNWDFK SU]HDQDOL]XMP\ HOHPHQW\ L 6 WR SUW\ QLHREFL*RQH ]DNRF]RQH

SU]HJXEDPL=ZDUXQNXLFKUyZQRZDJLZ\QLND*HUHDNFMHQDNRFDFKPXV]PLHüNLHUXQHN

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2

2

q

ql

I

S

1

S

1

ql

M

B

V

A

V

B

S

2

S

2

x

Z równania

0

M

i

I

2

i

=

∑

PR*QD Z\]QDF]\ü QLH]QDQ\ PRPHQW UHDNF\MQ\ 0

B

(w punkcie 2

SU]HFLQDM VL OLQLH G]LDáDQLD QLH]QDQHM UHDNFML 9

B

i niewiadomych S

1

i S

2

, co eliminuje je z

równania):

0

M

i

I

2

i

=

∑

⇒

-M

B

–ql l = 0

⇒

M

B

= – ql.

=QDMF 0

B

PR*HP\ WHUD] ] UyZQD

M

iA

i

∑

=

0 i

M

iB

i

∑

=

0

REOLF]\ü UHDNFMH SLRQRZH

5yZQDQLDWHSU]\MPXMSRVWDü

V

B

2l - M

B

+ ql l – ql l/2 = 0

-V

A

2l + ql 3/2l + ql l – M

B

= 0,

co daje V

B

= – 3/4 ql i V

A

=7/4 ql

=HVWDZLHQLHREOLF]RQ\FKZLHONRFLSU]HGVWDZLDU\VXQHN

q

ql

ql

ql

2

7/4 ql

3/4 ql