L 1
A.
PRORAČUN REZERVOARA ZA KOMPRIMIRANI VAZDUH
Pritisak vazduha….…..
p=15
⋅
10
5
Pa = 1,5 N/mm
2
Usvojeni materijal…..
Č 0371
Prečnik rezervoara…...
D=780 mm
Granica tečenja….…..
R
e
=235 N/mm
2
Dužina rezervoara……
L=3800 mm
Faktor slabljenja…….
ϕ
=0,7
Radijus danceta……...
.R=
0,8
⋅
D
=625mm
Stepen sigurnosti……
S=1,5
Širina otvora cjevi……
d
1
=0,15
⋅
D=
117 mm
Proračun debljine limova:
σ
doz
=
=
=
5
,
1
/
235
2
mm
N
S
R
e
157 N/mm
2
Debljina lima :
δ
=
=
+
⋅
⋅
⋅
=
+
⋅
⋅
1
/
157
7
,
0
2
/
5
,
1
780
1
2
2
2
mm
N
mm
N
mm
koroziju
za
mm
p
D
doz
ϕσ
6,3 mm
Debljina danceta :
δ
d
=
=
⋅
⋅
⋅
=
+
⋅
⋅
⋅
2
2
/
157
7
,
0
2
/
5
,
1
625
1
2
mm
N
mm
N
mm
p
R
doz
σ
ϕ
5,3 mm
Debljina cjevi :
δ
c
=
=
+
⋅
⋅
⋅
=
+
⋅
⋅
⋅
1
/
157
7
,
0
2
/
5
,
1
117
1
2
2
2
1
mm
N
mm
N
mm
p
d
doz
σ
ϕ
1,8
≈
2 mm
Spoljašnji prečnik cjevi :
d
2
=
d
1
+2
δ
c
=117mm+2
⋅
2mm
=121 mm
Proračun zavarenih sastavaka :
Detalj A:
τ
s
=
F
s
/A
s
<
τ
s doz
; F
s
=
p
⋅
d
2
2
⋅π
/4
; A
s
=
d
2
⋅π ⋅
a ;
a
(debljina vara)=0,7
⋅δ
d
τ
s
=
≈
⋅
⋅
⋅
=
⋅
⋅
⋅
=
⋅
⋅
⋅
⋅
⋅
mm
mm
mm
N
d
p
a
d
d
p
d
3
,
5
7
,
0
4
121
/
5
,
1
7
,
0
4
4
2
2
2
2
2
δ
π
π
12,23 N/mm
2
Detalj B:
σ
z
=
F
z
/A
z
<
σ
z doz
; F
z
=
[D+2(
δ
-
δ
d
)]
2
⋅π ⋅
p/4 ;
A
z
=
[D+2(
δ
-
δ
d
)]
⋅πδ
d
σ
z
=
≈
⋅
⋅
−
+
=
⋅
+
+
=
−
+
−
+
mm
mm
N
mm
mm
p
D
D
p
D
d
d
d
d
d
3
,
5
4
/
5
,
1
]
)
3
,
5
3
,
6
(
2
780
[
4
)]
(
2
[
)]
(
2
[
4
)]
(
2
[
2
2
δ
δ
δ
πδ
δ
δ
π
δ
δ
55,3N/mm
2
Detalj C:
σ
z
=
F
z
/A
z
<
σ
z doz
; F
z
=
p
⋅
D
2
⋅π
/4
; A
z
=
D
⋅π ⋅δ
σ
z
=
=
⋅
⋅
=
⋅
⋅
=
⋅
⋅
⋅
⋅
⋅
mm
mm
mm
N
D
p
D
D
p
3
,
6
4
780
/
5
,
1
4
4
2
2
δ
δ
π
π
46,42 N/mm
2
B.1 ZAVARENI SPOJ RUČICE I VRATILA (sl.2_)
Prečnik vratila………….….
d
=
110 mm
Sila ručice……………….…
F=
±
9900 N
Krak ručice………………....
R=145 mm
Računska debljina vara.…….
a=5 mm
Zavarivanje………………….
N
Materijal…………………….
Č 0545
Dozvoljeni
τ
napon………….
τ
D(-1)
=140 N/mm
2
L 2
Radni napon uvijanja :
τ
t
=M
t
/W
0
;
M
t
= F
⋅
R ,
W
0
=
)
2
(
16
]
)
2
[(
4
4
a
d
d
a
d
+
⋅
−
+
π
τ
t
=
=
⋅
−
⋅
+
⋅
+
⋅
⋅
⋅
=
⋅
−
+
+
⋅
⋅
14
,
3
]
)
110
(
)
5
2
110
[(
)
5
2
110
(
16
145
9900
]
)
2
[(
)
2
(
16
4
4
4
4
mm
mm
mm
mm
mm
mm
N
d
a
d
a
d
R
F
π
14,4 N/mm
Radni napon smicanja :
τ
s
=F
s
/A
s
;
A
s
=
π
⋅
−
+
4
)
2
(
2
2
d
a
d
τ
s
=
=
⋅
−
⋅
+
⋅
=
⋅
−
+
14
,
3
]
)
110
(
)
5
2
110
[(
9900
4
]
)
2
[(
4
2
2
2
2
mm
mm
mm
N
d
a
d
F
s
π
5,48 N/mm
2
Ukupna amplituda napona (oba napona tangencijalna) :
τ
a
=
±
(
τ
s
+
τ
t
)=
±
(14,4+5,48)=
±
19,88 N/mm
2
ξ
1
=0,9
τ
naponi i ugaoni sastavak
ξ
2
=0,5
Normalno zavarivanje
ξ
3
=0,8
Ugaoni sastavak bez zavarenog korena
Dinamički stepen sigurnosti : S
d
=
=
⋅
⋅
⋅
⋅
=
⋅
⋅
⋅
2
2
3
2
1
/
88
,
19
/
140
8
,
0
5
,
0
5
,
0
9
,
0
mm
N
mm
N
a
d
τ
τ
ξ
ξ
ξ
1,268
B.2 ROTOR SA ZAVARENIM RUKAVCEM (sl.4_)
Sila savijanja………………..
F=1580 N
Materijal……………………..
Č. 0545
Računska debljina vara……...
a=7 mm
Prerčnik osovine…………….
d=115 mm
Krak sile……………………..
L=
1,5
⋅
d=
172,5 mm
Moment uvijanja…………….
M
u
=
0,28 kNm
=0,28
⋅
10
6
Nmm
Moment savijanja…………….
M
f
=
F
⋅
L
=272550 Nmm
Spoljašni prečnik zavara….…
d
z
=
d+2a
=129 mm
Površina preseka zavara…….
A=
≈
⋅
+
π
z
z
d
d
d
2
2
2680 mm
2
Polarni otporni moment ……..
W
0
=
=
⋅
−
16
4
4
π
z
z
d
d
d
174103,2 mm
3
Otporni moment za osu………
W
x
=
=
⋅
−
32
4
4
π
z
z
d
d
d
87051,6 mm
3
Napon savijanja na zavaru……
σ
=
=
=
3
0
6
,
87051
272550
mm
Nmm
W
M
f
3,131 N/mm
2
Napon uvijanja u zavaru………
τ
t
=
=
⋅
=
3
6
0
2
,
174103
10
28
,
0
mm
Nmm
W
M
u
1,61 N/mm
2
Napon smicanja zavara………..
τ
s
=
=
=
2
2680
1580
mm
N
A
F
s
s
0,59 N/mm
2
Sabiranje tangencijalnih napona……….
τ
min
=
τ
t
=1,61 N/mm
2
τ
max
=
τ
t
+
τ
s
=1,61N/mm
2
+0,59N/mm
2
=
2,2 N/mm
2
τ
sr
=
(
τ
min
+
τ
max
)/2=
1,905 N/mm
2
Trajna dinamička izdržljivost…………..
τ
d
=140 N/mm
2
Koeficijenti otpornosti za
τ
napone…..
.
ξ
1
=0,4 ,
ξ
2
=1 ,
ξ
3
=0,5 ,
ξ
4
=0,75
Din.stepen sigurnosti za
τ
napone..
S
τ
=
=
⋅
⋅
⋅
⋅
=
⋅
⋅
⋅
⋅
2
,
2
/
140
75
,
0
5
,
0
1
4
,
0
2
4
3
2
1
mm
N
a
d
τ
τ
ξ
ξ
ξ
ξ
9,5
