3U]\NáDG%HOND]HVNUDWRZDQLHP

3ROHFHQLH Z\]QDF]\ü VLá\ Z SUWDFK GZXSU]HJXERZ\FK GOD SRQL*V]HM EHONL REFL*RQHM VLá

VNXSLRQ

P oraz momentem skupionym M = Pl.

Przyjmujemy oznaczenia dla podpór: A i B

GOD SRáF]H SU]HJXERZ\FK

1, 2, 3 i 4, a dla

SRáF]HQLDWHOHVNRSRZHJR

5.

5R]ZL]DQLH]DGDQLDUR]SRF]\QDP\RGRVZRERG]HQLDXNáDGXRGZL]yZ]DVWSXMFSRGSRU\

UHDNFMDPL/HZDSRGSRUDR]QDF]RQDOLWHU

A

MHVWSRGSRUSU]HJXERZQLHSU]HVXZQ=DWHP

w punkcie A

G]LDáDMGZLHQLH]DOH*QHRGVLHELHVNáDGRZHUHDNFMLSLRQRZD

V

A

i pozioma H

A

.

3UDZD SRGSRUD R]QDF]RQD OLWHU

B

MHVW SRGSRU SU]HJXERZ SU]HVXZQ : SXQNFLH

B

Z\VWSXMHUHDNFMD

R

B

NWyUHMOLQLDG]LDáDQLDMHVWSLRQRZDSURVWRSDGáDGRNLHUXQNXPR*OLZHJR

przesuwu).

:SROHFHQLXGR]DGDQLDQLHVZ\PLHQLRQHUHDNFMHSRGSRURZHMHGQDN*HZ\]QDF]HQLHUHDNFML

V

A

lub R

B

MHVW NRQLHF]QH =DSLV]HP\ UyZQDQLH PRPHQWyZ Z]JOGHP SXQNWX

A

GOD FDáHJR

XNáDGX]NWyUHJRZ\]QDF]\P\UHDNFM

R

B

0RPHQW\VLá

H

A

i V

A

Z]JOGHPSXQNWX

A

VUyZQH

]HUXJG\*OLQLHG]LDáDQLDW\FKVLáSU]HFKRG]SU]H]SXQNW

A.

∑

=

i

iA

M

0 : R

B

Â

l

í

P

Â

l

í

M = 0

⇒

R

B

=

3

2

P

Po wyznaczeniu reakcji R

B

PR*HP\REOLF]\üZDUWRüVLá\

S

3

G]LDáDMFHMZSUFLH

1

í

.

P

2 l

l

2 l

l

l

l

2 l

M = Pl

P

2 l

l

2 l

l

l

l

2 l

M = Pl

B

A

5

4

3

2

1

1

P

2 l

l

2 l

l

l

l

2 l

M = Pl

B

A

5

4

3

2

V

A

H

A

R

B

y

x

2

3UW

1

í ]DVWSXMHP\ VLá

S

3

NWyUD G]LDáD Z W\P SUFLH : SRáF]HQLX WHOHVNRSRZ\P

5

Z\VWSXMH PRPHQW

M

5

L SR]LRPD VLáD

R

5

SURVWRSDGáD GR RNáDGHN WHOHVNRSX 5HDNFMD

R

B

-

MDNRZLHONRü]QDQD±R]QDF]RQDMHVWNRORUHPF]DUQ\P

: FHOX Z\]QDF]HQLD VLá\

S

3

]DSLV]HP\ UyZQDQLH U]XWyZ VLá QD R SLRQRZ GOD

I

F]FL

XNáDGX

∑

=

i

I

iy

P

0 : R

B

í

S

3

ÂVLQ.

= 0

⇒

S

3

=

3

13

P

3R]RVWDáHGZLHQLHZLDGRPHVLá\G]LDáDMFHZSUWDFK

1

í

2 i 1

í

3

Z\]QDF]\P\]GZXUyZQD

UyZQRZDJLGODZ]áD

1.

6Lá

S

1

PR*HP\Z\]QDF]\ü]UyZQDQLDU]XWyZVLáQDRSR]LRPGODZ]áD

1.

0

=

∑

i

1

ix

P

:

í

S

1

ÂFRV

+ S

3

ÂFRV.

= 0

⇒

S

1

= 5 P

2VWDWQLQLHZLDGRPREOLF]\P\]UyZQDQLDU]XWyZVLáQDRSLRQRZGODZ]áD

1.

∑

=

i

1

iy

P

0 : S

1

ÂVLQ

+ S

2

+ S

3

ÂVLQ.

= 0

⇒

S

2

=

í

3

8

P

1

P

2 l

l

2 l

l

l

l

2 l

M = Pl

B

A

5

4

3

2

V

A

H

A

R

B

I

.

1

2 l

2 l

l

l

2 l

M = Pl

B

5

4

M

5

R

5

R

B

I

S

3

.

90

o

3l

2l

13 l

sin

.

=

13

2

cos

.

=

13

3

cos

=

5

1

sin

=

5

2

2 l

l

5 l

1

4

3

2

S

1

S

2

S

3

=

3

13

P

.

3

6Lá\

S

1

i S

3

PDM ZDUWRFL GRGDWQLH F]\OL LFK ]ZURW\ V ]JRGQH ] ]DáR*RQ\PL D ]DWHP V WR

VLá\ UR]FLJDMFH 1DWRPLDVW VLáD

S

2

MHVW XMHPQD D ZLF MHM ]ZURW MHVW SU]HFLZQ\ GR

]DáR*RQHJR3UW

1

í

3

MHVWFLVNDQ\

=H Z]JOGX QD SROHFHQLH Z\]QDF]RQH ]RVWDá\ MHG\QLH VLá\

S

1

, S

2

i S

3

RUD] MDNR ZLHONRü

pomocnicza reakcja R

B

1LH PR*HP\ ]DWHP EH] GRGDWNRZ\FK REOLF]H Z\NRQDü

VSUDZG]HQLDSRSUDZQRFLUR]ZL]DQLD