Przykład 2.1.

3U]\NáDG%HONDZROQRSRGSDUWD]HZVSRUQLNLHP
3ROHFHQLH Z\]QDF]\ü UHDNFMH SRGSRURZH GOD SRQL*V]HM EHONL REFL*RQHM REFL*HQLHP
FLJá\PRVWDá\PQDW
*HQLX
q RUD]VLáVNXSLRQP = ql.
P = ql
q
l
4l
l
Przyjmujemy oznaczenia podpór: A i B.
P = ql
q
B
A
4l
l
l
5R]ZL]\ZDQLH ]DGDQLD UR]SRF]\QDP\ RG RVZRERG]HQLD EHONL RG ZL
]yZ ]DVW
SXMF
SRGSRU\
UHDNFMDPL
/HZD
SRGSRUD
R]QDF]RQD OLWHU A MHVW SRGSRU SU]HJXERZ
A G]LDáDM GZLH QLH]DOH*QH RG VLHELH VNáDGRZH UHDNFML
pionowa VA i pozioma HA 3UDZD SRGSRUD R]QDF]RQD OLWHU B MHVW SRGSRU SU]HJXERZ
SU]HVXZQ : SXQNFLH B Z\VW
SXMH UHDNFMD RB NWyUHM OLQLD G]LDáDQLD MHVW SLRQRZD
QLHSU]HVXZQ =DWHP Z SXQNFLH
SURVWRSDGáD GR NLHUXQNX PR*OLZHJR SU]HVXZX : FHOX XSURV]F]HQLD ]DSLVX UyZQD
UyZQRZDJL ]DVWSLP\ REFL*HQLH FLJáH
q VLá Z\SDGNRZ NWyUHM NLHUXQHN L ]ZURW MHVW
]JRGQ\ ] NLHUXQNLHP L ]ZURWHP REFL*HQLD FLJáHJR :DUWRü Z\SDGNRZHM MHVW UyZQD SROX
ILJXU\ SRG Z\NUHVHP UR]NáDGX QDW
*HQLD REFL*HQLD FLJáHJR QDWRPLDVW OLQLD G]LDáDQLD VLá\
Z\SDGNRZHMSU]HFKRG]LSU]H]URGHNFL
*NRFLWHMILJXU\:UR]SDWU\ZDQ\PSU]\NáDG]LHVLáD
wypadkowa jest równa W = q Âl = 4ql2GOHJáRüOLQLLG]LDáDQLDZ\SDGNRZHMRGSRGSRU\ A
wynosi 2l.
W = 4ql
P = ql
y
A
HA
x
B
RB
VA
2l
3l
l
%HONDREFL*RQDMHVWGZLHPDVLáDPL]HZQ
WU]Q\PLNWyU\FKZLHONRFLV]QDQHR]QDF]RQ\PL
NRORUHPF]DUQ\PRUD]WU]HPDUHDNFMDPLR]QDF]RQ\PLNRORUHPSRPDUDF]RZ\P1LH]QDQH
ZDUWRFL UHDNFML Z\]QDF]\P\ ] WU]HFK UyZQD UyZQRZDJL GOD SáDVNLHJR GRZROQHJR XNáDGX
VLá =DSLV]HP\ GZD UyZQDQLD U]XWyZ VLá QD RVLH XNáDGX ZVSyáU]
GQ\FK RUD] UyZQDQLH
PRPHQWyZ2VWDWQLH]Z\PLHQLRQ\FKUyZQDQDMNRU]\VWQLHMMHVW]DSLVDüZ]JO
GHPSXQNWX
A
lub B /LQLH G]LDáDQLD UHDNFML VA i HA SU]HFKRG] SU]H] SXQNW A, zatem te niewiadome nie
Z\VWSL Z UyZQDQLX PRPHQWyZ Z]JO
GHP SXQNWX A PRPHQW VLá\ Z]JO
GHP SXQNWX
OH*FHJRQDOLQLLG]LDáDQLDWHMVLá\MHVWUyZQ\]HUX:WDN]DSLVDQ\PUyZQDQLXZ\VWSLW\ONR
jedna niewiadoma RB3RGREQLHZUyZQDQLXPRPHQWyZZ]JO
GHPSXQNWX BQLHZ\VWSLá\E\
reakcje RB i HA JG\* LFK OLQLH G]LDáDQLD SU]HFKRG] SU]H] SXQNW B .RU]\VWDMF ] WHJR
UyZQDQLDPR*QDE\Z\]QDF]\üUHDNFM
VA.
14
RB Âl íW Âl íP Âl = 0
⇒
RB =
ql = 2,8 ql
∑i M iA = 0 :
5
:QDVW
SQ\PUyZQDQLXUyZQRZDJLNRU]\VWDP\]Z\]QDF]RQHMZDUWRFLUHDNFMLRB.
11
VA + RB íW íP = 0
⇒
VA =
ql = 2,2 ql
∑i Piy = 0 :
5
=UyZQDQLDU]XWyZVLáQDRSR]LRPZ\]QDF]DP\QLHZLDGRPHA.
∑ Pix = 0 : HA = 0
i
: FHOX VSUDZG]HQLD SRSUDZQRFL Z\NRQDQ\FK REOLF]H ]DSLV]HP\ UyZQDQLH UyZQRZDJL ]
NWyUHJRZF]HQLHMQLHNRU]\VWDOLP\
W = 4ql
P = ql
A
B
HA = 0
RB = 2,8 ql
VA = 2,2 ql
2l
∑M
iB
=0:
3l
l
W Âl íP Âl íVA Âl = 4ql Âl íql Âl íql Âl = 12 ql2 íql2 íql2 = 0
i
5yZQDQLHVSHáQLRQHMHVWWR*VDPRFLRZR
2