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Teorija Povrsinskih nosaca,Navierovo resenje
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7/18/2019 TPN - Seminarski Rad
http://slidepdf.com/reader/full/tpn-seminarski-rad 1/7
DRŽAVNI UNIVERZITET U NOVOM PAZARU
DEPARTMAN: TEHNIČKE NAUKE
Studijski program: GRAĐEVINARSTVO
TEORIJA POVRŠINSKIHNOSAČA
I Grafički Rad
Mentor:
Petar Knežević, dip!ing!gra"!
Student:
#a$tijarević Emir 08-001/12
7/18/2019 TPN - Seminarski Rad
http://slidepdf.com/reader/full/tpn-seminarski-rad 2/7
I GRAFIČKI RAD %NA&'ER(ovo re)enje*
Potre+ni podai -a prora.un:
λ=1m
a= λ=1m
b= λ+1.5m=2.5m
p1=5+ λ kN /m2=6kN /m2
p2=3+ λ kN /m2=4 kN /m2
X A=a
4m=0.25m
Y A=3 · b
4m=1.875m
E=30GPa
h=25 cm
ν=0.2
7/18/2019 TPN - Seminarski Rad
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Dekomponovanjem ukupnog opterećenja, do+ijamo:
Re)enje di/erenijane jedna.ine:
ω( x , y )=∑m
∑n
Amn · sin m·π·x
a ·sin
n·π·y
b
Koe/iijent A mn se de/ini)e kao:
A mn= Z mn
k· π 4(
m2
a2 +
n2
b2)
2
0a usvojen prvi .an 1urijeovog reda% m=n=1
*, imamo:
ω= A11
· sin π·x
a ·sin
π·y
b
7/18/2019 TPN - Seminarski Rad
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A11=
Z 11
k· π 4 ( 1
a2+ 1
b2)2
Pri .emu je ukupno opterećenje: Z 11=Z 11
1
+Z 11
2
Prvi dekomponovani deo opterećenja se odre"uje prema o+rasu:
Z 11
1 =4 · p
1
a·b ∫
0
1
sin π·x
a dx ·∫
0
2.5
sin π·y
b dy
Z 11
1 =4 · p
1
a·b ·(−a
π )cos π·x
1 |1
0·(−b
π ) cos π·y
2.5|2.5
0
Z 11
1 =4 · p1
π 2
· (cos π −1 ) · (cos π −1 )=16 · p1
π 2
Drugi dekomponovani deo opterećenja se odre"uje kao:
Z 11
2 =4 · p
2
a·b ·∫
0
2.5
sin π·y
b dy ·∫
0
a
b y
sin π·x
a dx
'ntegra I
1 se odre"uje:
I 1=∫
0
a
b y
sin π·x
a dx=
−a
π cos
π·x
a |a
b y
0
=−a
π ·(cos
π·y
b −1)
Pa -amenom do+ijamo da je Z 11
2
:
Z 11
2 =4 · p
2
a·b ·∫
0
2.5
sin π·y
b · I
1dy=
4 · p1
a·b ·∫
0
2.5
sin π·y
b ·(−a
π ) ·(cos π·y
b −1)dy
Z 11
2 =−4 · p
2
b·π · [∫
0
2.5
sin π·y
b · cos
π·y
b dy−∫
0
2.5
sin π·y
b dy]
Z 11
2 =−4 · p
2
b·π · [∫
0
2.5
sin π·y
b · cos
π·y
b dy−∫
0
2.5
sin π·y
b dy]
7/18/2019 TPN - Seminarski Rad
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Z 11
2 =−4 · p
2
b·π · [ 1
2∫
0
2.5
sin(2· π·y
b )dy+
b
π cos
π·y
2.5|2.5
0 ]
Z 11
2 =−4 · p
2
b·π
·
[−1
2·
b
2· π
cos 2 · π·y
2.5
|
2.5
0
+
b
π
(cos π −1)
]Z
11
2 =−4 · p
2
b·π · [ −b
4 · π ·(cos2π −1)−
2 ·b
π ]=8· p
2
π 2
Pa kona.no do+ijamo da jeZ
11 :
Z 11=Z
11
1 +Z 11
2 =16 · p
1
π 2 +
8 · p2
π 2 =
8
π 2 (2· p
1+ p
2 )
Z 11=
8
π 2 (2 · 4+2 )=8.11 kN /m
2
Krutost po.e K se odre"uje prema o+rasu:
K = E· h
3
12·(1−ν2)=
3 · 107· 0.25
3
12 ·(1−0.22)=40690.1 kNm
Potom imamo da je koe/iijent A
11 :
A11=
Z 11
K· π 4(
12
a2+
12
b2)
2=
8.11
40690.1 · π 4 (
1
12+
1
2.52)
2=1.521 ·10
−6
2gi+ ω je de/inisan o+rasem:
ω= A11
· sin π·x
a ·sin
π·y
b
Pa su njegovi parijani i-vodi:
∂ ω
∂ x = A
11· π
a cos
π·x
a · sin
π·y
b
∂ ω
∂ y = A
11· π
b sin
π·x
a · cos
π·y
b
7/18/2019 TPN - Seminarski Rad
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∂2ω
∂ x2= A
11·(−π
2
a2 )sin
π·x
a · sin
π·y
b
∂2ω
∂ y2= A
11·(−π
2
b2 )sin
π·x
a · sin
π·y
b
∂2ω
∂ x ∂ y= A
11· π
2
a·b cos
π·x
a ·cos
π·y
b
∂3ω
∂ x3 = A
11·(−π
3
a3 )cos
π·x
a · sin
π·y
b
∂3ω
∂ y3= A
11·(−π
3
b3 )sin
π·x
a · cos
π·y
b
∂3
ω
∂ x2
∂ y= A
11· (−π
3
a2
· b)sin
π·x
a · cos
π·y
b
∂3
ω
∂ x ∂ y2= A
11· (−π
3
a·b2)cos
π·x
a · sin
π·y
b
Ui! "a #ačk$ %%A&'
ω A=ω x=0.25
y=1.875
=1.521·10−6
·sin π·0.25
1·sin π· 1.875
2.5=7.60510
−7m
Potre+ni podai -a odredjivanje sia u preseima:
sin
π·x
a =sin
π· 0.25
1 =sin
π
4 =√ 2
2
cos π·x
a =cos
π·0.25
1=cos
π
4=√ 2
2
sin π·y
b =sin
π· 1.875
2.5=sin
3 π
4=√ 2
2
x=0.25 m a=1 m
y=1.875m b=2.5m
K· A11= Z 11
π 4 1
2
+1
2 2=0.0618741
7/18/2019 TPN - Seminarski Rad
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cos π·y
b =cos
π· 1.875
2.5=cos
3 π
4=−√ 2
2
Si() $ *r)+),i-a'
M x=− K (∂2
ω
∂ x2 +ν·
∂2
ω
∂ y2 )=− K· A
11(−π 2
a2
sin π·x
a ·sin
π·y
b −ν·
π 2
b2
sin π·x
a · sin
π·y
b )=0.0618741 ·√ 2
2· √ 2
2·( π
1
M y=− K (∂2
ω
∂ y2+ν·
∂2ω
∂ x2 )=− K· A
11(−π 2
b2
sin π·x
a · sin
π·y
b −ν·
π 2
a2
sin π·x
a · sin
π·y
b )=0.0618741 ·√ 2
2·√ 2
2·( π
2
M xy=− K (1−ν ) · ∂
2ω
∂ x ∂ y=− K· A
11· π
2
a·b cos
π·x
a · cos
π·y
b (1−ν )=−0.0618741 ·
π 2
1 ·2.5·√ 2
2·(−√ 2
2 ) · (1−0.2
T x=− K (∂3
ω
∂ x3 +ν
∂3
ω
∂ x ∂ y2 )=− K· A
11·(−π
3
a3
cos π·x
a · sin
π·y
b −ν
π 3
a·b2
cos π·x
a sin
π·y
b )=−0.0618741 · √ 2
2· √
2
T y=− K (∂3
ω
∂ y3+ν
∂3ω
∂ x2∂ y )=− K· A
11(−π 3
b3
sin π·x
a ·cos
π·y
b −ν
π 3
a2· b
sin π·x
a · cos
π·y
b )=−0.0618741 ·√ 2
2·(