ZADATAK 3
PAGE
MAINSKI FAKULTET
U TUZLI
NAUKA O VRSTOI II
Zadatak broj: 1
List broj: 5
Odrediti sile, napone i deformacije u tapovima reetke. Svi
tapovi su od istog materijala i jednakih poprenih presjeka.
Spoljanji tapovi su zagriani za t1 a jedan unutranji za t2.
4
cm
F (kN)
l (m)
t 1(C)
t 2(C)
E (Mpa)
(1/C)
I (
)
5
0,3
30
50
5
10
2
7
10
55
,
1
-
40
S=2n-3
s-br. tapova
n-br. vorova
25-3=7 1x stat.N.O.U.S
67
4W-3S.U.R=1x stat.N.O.V.S
x=0
Xb+Xa-F=0
Xb+Xa=F
y=0
Yb+Ya-F=0
Yb+Ya=F
Ms=0
Xbl
2
EMBED Equation.3
2
0
2
2
2
2
l
l
F
l
F
=
-
-
Xb-
2
F
-
2
F
=0
Xb=F=5 kN
Xb+Xa=F
Xa=0
vor 1
x=0
Xb- s1 cos45 + s4 cos45 = 0
y=0
Yb - s5=0
Yb= s5
vor 2
x=0
-F- s2 cos45 + s1 cos45 +S6 = 0
y=0
-s1 cos45 + s2 cos45 = 0
s1= s2
vor 3
x=0
Xa - s2 cos45 +s3 cos45 = 0
y=0
Ya - s5= 0
Ya= s5
vor 4
x=0
- s3 cos45 - s4 cos45 = 0
s3 = -s4
y=0
- F + s3 cos45 - s4 cos45 = 0
s3= -s4
s2= s3
Yb - s5= 0
Ya - s5= 0
Ya = Yb
Yab=
2
F
= 2,5 kN Ya = 2,5 kN Yb= 2,5 kN
iz vora 4
- F + s3 cos45 - s4 cos45 = 0
2 s3 cos45 = F
s3=
45
cos
2
F
= 3,53 kN = - s4
iz vora 1
Xb s1 cos45 - s4 cos45 = 0
S1 cos45 = 2,5 kN
s1 =
45
cos
5
,
2
kN
= 3,53 kN = s2
Yb - s5 = 0
s5 = 2,5 kN
iz vora 2 dobijemo s6
- F s2 cos45 + s1 cos45 + s6 = 0
s6 = F
s6 = 5 kN
s1= 3,53 kN
s2= 3,53 kN
s3= 3,53 kN
s4= -3,53 kN
s5= 2,5 kN
s6= 5 kN
I=40
4
cm
I=
4
4
4
p
r
7
,
3
4
3
=
=
p
I
r
cm
A=
2
2
2
66
,
4298
43
mm
cm
r
=
=
p
Izduenje
U= U1+ U2+ U3+ U4+ U5= 0,06854 mm
01688
,
0
2
1
1
2
1
1
=
D
+
=
l
t
s
EA
l
s
U
a
mm
01688
,
0
2
1
2
2
2
2
=
D
+
=
l
t
s
EA
l
s
U
a
mm
01688
,
0
2
1
3
2
3
3
=
D
+
=
l
t
s
EA
l
s
U
a
mm
0159
,
0
2
1
4
2
4
4
-
=
D
+
=
l
t
s
EA
l
s
U
a
mm
0338
,
0
2
2
2
2
5
2
5
5
=
D
+
=
l
t
s
EA
l
s
U
a
mm
l
t
AE
l
s
AE
l
s
F
U
D
-
-
=
2
2
2
2
5
2
2
a
023
,
0
2
-
=
=
F
U
X
mm
l
t
AE
l
s
AE
l
s
F
U
D
-
-
=
2
2
2
2
5
2
3
a
0018
,
0
4
-
=
Y
mm
Naponi u tapovima
82
,
0
1
1
=
=
A
s
s
MPa
A
s
5
5
=
s
= 0,58 MPa
82
,
0
2
2
=
=
A
s
s
MPa
16
,
1
6
6
=
=
A
s
s
MPa
82
,
0
3
3
=
=
A
s
s
MPa
82
,
0
4
4
-
=
=
A
s
s
MPa
RADIO
BROJ INDEXA
DATUM
PREGLEDAO
Husagi Adnan
II-122/04
10.01.2008.
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