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Ploče opterećene na savijanje – izrazi za sile
yxw
1kM
xw
yw
kM
yw
xw
kM
2
xy
2
2
2
2
y
2
2
2
2
x
∂⋅∂∂⋅ν−⋅−=
∂∂⋅ν+
∂∂⋅−=
∂∂⋅ν+
∂∂⋅−=
)(
∂⋅∂∂+
∂∂⋅−=
∂⋅∂∂+
∂∂⋅−=
yxw
yw
kT
yxw
xw
kT
2
3
3
3
y
2
3
3
3
x
∂⋅∂∂⋅ν−+
∂∂⋅−=
∂⋅∂∂⋅ν−+
∂∂⋅−=
2
3
3
3
y
2
3
3
3
x
xyw
2yw
kT
yxw
2xw
kT
)(
)(
)( 2
3
112hE
kν−
⋅=
Navier-ovo rešenje
∑∑∞
=
∞
=
⋅⋅⋅⋅⋅⋅=11 n
mnm b
yna
xmAyxw
ππsinsin),(
∫∫⋅⋅⋅⋅⋅⋅⋅
⋅=
+⋅⋅
=
yxmn
mnmn
dxdyb
yna
xmyxZ
baZ
bn
am
k
ZA
ππ
π
sinsin),(4
2
2
2
2
24
rešenje za jednako podeljeno opterećenje na celoj ploči
∑∑∞
=
∞
=
⋅⋅⋅⋅⋅⋅
+⋅⋅
⋅⋅⋅=
12
2
2
2
21
60 116
nm byn
axm
bn
am
nmk
Zyxw
πππ
sinsin),(
rešenje za jednako podeljeno opterećenje na pravougaoniku 2c×2d
∑∑∞
=
∞
=
⋅⋅⋅⋅⋅⋅
+⋅⋅
⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅
⋅⋅⋅
=1
2
2
2
2
216016
nm byn
axm
bn
am
nm
bdn
bvn
acm
aum
k
Zyxw
ππππππ
πsinsin
sinsinsinsin),(
rešenje za koncentrisanu silu u tački (u, v)
∑∑∞
=
∞
=
⋅⋅⋅⋅⋅⋅
+
⋅⋅⋅⋅⋅
⋅⋅⋅⋅
⋅=1
2
2
2
2
21
4
4
nm byn
axm
bn
am
bvn
aum
bakP
yxwππ
ππ
πsinsin
sinsin),(
M. Levy – jevo rešenje
∑∞
=
⋅⋅⋅
⋅⋅⋅
⋅⋅⋅++⋅⋅⋅
⋅⋅⋅+=
+=
1nnnnnh
ph
axn
ayn
shDa
ynC
ayn
chBa
ynAw
www
πππππsin
∑∞
=
⋅⋅⋅=1
sinn
np axn
wwπ
4
⋅⋅=
πna
kZ
w nn
∑∞
=
⋅⋅⋅=1
sinn
n axn
ZZπ ∫
⋅⋅⋅⋅=x
n dxa
xnxZ
aZ
πsin)(
2
∑
∑
∞
=
∞
=
⋅π⋅⋅
⋅π⋅⋅+⋅π⋅⋅⋅π⋅⋅=
⋅π⋅⋅
⋅π⋅⋅⋅π⋅⋅+⋅π⋅⋅=
1nnnanth
1nnnsimh
axn
ayn
shCa
ynch
ayn
Bw
axn
ayn
sha
ynD
ayn
chAw
sin
sin
,
,
rešenje za slobodno oslonjenu ploču na konturi y=b/2 usled jednako podeljenog opterećenja
a2bn
k
aZ4S
ch2S
n
1DS
ch22th
n
1A
n5
40
n5n
n
nn5n
⋅π⋅=α⋅π
⋅⋅=
α⋅⋅=⋅
α⋅+α⋅α⋅−=
rešenje za uklještenu ploču na konturi y=b/2 usled jednako podeljenog opterećenja
a2bn
k
aZ4S
Schsh
sh
n
1DS
chshshch
n
1A
n5
40
nnn
n5n
nnn
nnn5n
⋅π⋅=α⋅π
⋅⋅=
⋅α⋅α+α
α⋅=⋅α⋅α+αα+α⋅α⋅−=
Metoda konačnih razlika – diferencna metoda
k
xZww
1wwwwww2
ww1
1ww1481
6w
x
y
42k
hm22k2k2
1l1i1l1i
il21k1k2
22
k
∆α)(
α
)(α)(
)(α
)()α(α
α
∆
∆α
⋅⋅=+⋅++⋅++++⋅+
+
+⋅
+++⋅+⋅−
+
+⋅⋅
=
−+++−−
−+
kxZ
wwww
wwww2wwww8w20
1
42k
hm2k2k
1l1i1l1iil1k1kk
∆⋅α⋅=++++
++++⋅++++⋅−⋅
=α
−+
++−−−+
)(
)()(
( )
( )[ ]
( ) [ ]1i1i1l1l2kxy
1kk1k2
ikl2ky
ikl21kk1k2kx
wwww1x4
kM
ww2www2wy
kM
ww2www2wx
kM
−+−+
−+
−+
−++−⋅ν−⋅α⋅∆⋅
=
−⋅+−⋅α⋅ν+−⋅+−⋅∆
=
−⋅+−⋅αν+−⋅+−⋅
∆=
,
,
,
( ) ( )
( ) ( ) ( )[ ]
( )[ ] ( )[ ]m1l1l1i1i22
ilh3ky
2k1l1i1l1i221k1k2k3kx
m1l1l1i1i2
il2
h3ky
2k1l1i1l1i21k1k22k3kx
wwwww2212wwwy2
kT
wwwww22
12wwwx2
kT
wwwwwww12wy2
kT
wwwww1
ww1
12wx2
kT
−−−+⋅ν−α+ν−⋅α+⋅⋅−+⋅∆⋅
=
−−−+⋅α
ν−
+
αν−+⋅⋅−+⋅
∆⋅=
−−−+⋅α+−⋅α+⋅+⋅∆⋅
=
−−−+⋅
α+−⋅
α+⋅+⋅
∆⋅=
−+−+
+++−−−+−
+−+−
+++−−−+−
)()(
)()(
,
,
,
,
xy ∆∆ ≠
( )2114 α+⋅−
( ) 816 22 ++⋅ αα
( )2114 α+⋅−
21 α
( )214 α+⋅−( )214 α+⋅−
21 α
2α
2α
22
2 2
xy ∆∆ =
Kružne ploče opterećene na savijanje
kZ
drrdw
drwd
drrd
drd
2
2
2
2
=
⋅+
⋅+
ar
CCCCww 42
32
21p =ρρ⋅+ρ⋅ρ⋅+ρ⋅++= lnln
∫ ∫∫∫ ⋅⋅⋅⋅⋅⋅⋅= drrZrdr
drrrdr
k1
wp
−+−=
⋅+⋅−=
⋅ν+
⋅−=
⋅⋅ν+−=
φ
drdw
r1
drwd
r1
drwd
kdrr
dwdr
wddrd
kT
drwd
drrdw
kM
drrdw
drwd
kM
22
2
3
3
2
2
r
2
2
2
2
r