1 GRAĐEVNA STATIKA II
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Zadatak Za zadani sistem metodom pomaka odrediti MTN dijagrame. Greda: Stupovi:
b/h
= 36/72 [cm],
b/h
= E =
= 36/36 [cm],
Student 32 Vedran Slunjski
3 . 107 kN/m2;
= E =
3 . 107 kN/m2.
Zadatak 2
2.8 1.4 3 35 80
-1
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7
Geometrijske i materijalne karakteristike:
Es,Is
80 kN
35 kN/m' 4
3
4 , 4
EG,IG
5
J G
6
=
E G I G 3
Es,Is
Greda: 12 =
=
0,36 ⋅ 0,72 3 12
=
1119,744 ⋅ 10
=
139,968 ⋅ 10
−
5
335923,2kNm 2
b ⋅ h3 12
=
0,36 ⋅ 0,36 12
3 −
5
m4
E S I S = 41990,4kNm 2 3
0,001 m 1 1,4
2 2,8
1,4
Koeficijenti krutosti: E S ⋅ I S 41990,4 = = 6998, 4kNm l14 6 E ⋅ I 41990,4 k 25 = S S = = 6998,4 kNm l 25 6 E ⋅ I 335923,2 k 45 = G G = = 119972 ,57 kNm l 45 2,8 E ⋅ I 41990,4 k 57 = S S = = 9543, 27 kNm l57 4,4 k 14
=
POJEDNOSTAVLJENJE SISTEMA KONZOLA 3-4, djelovanje stavimo u čvor 4 35 kN/m'
MPL 3
1,4
TPL
TPL
M PL
4
T PL
MPL
m4
Stupovi: J S =
80 kN Es,Is
b ⋅ h3
−
35 ⋅ 1,4
2
= −34,3kNm 2 = −35 ⋅ 1,4 = −49,0kN =
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3 GRAĐEVNA STATIKA II
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KONZOLA 5-6, djelovanje stavimo u čvor 5 80 kN
TPD 5
MPD 6
1,4
TPD
MPD
M PD = 80 ⋅ 1,4 = 112kNm T Pd = −80,0kN
POJEDNOSTAVLJENI SISTEM – ZGLOBNA SHEMA 7
s = 2nČ − nŠ s = 4−4=0 4 , 4
49 kN 80kN
35 kN/m'
112,0kNm 34,3kNm
4
5
3
80 kN
3
0,001 m 1 1,4
2 2,8
1,4
zglobna shema ima 1 stupanj slobode NEPOZNANICE: φ4; φ5; u45
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PLAN POMAKA
57
4 , 4
ψ 14
ψ 25
u45
ψ 57
14
u 45 − u 45 = 6 l14 − u 45 − u 45 = = l 25 6 u 45 u 45 = = l57 4,4
=
−
6
25
2,8
PLAN POMAKA OD PRISILNOG POMAKA
4 , 4
ψ 14
=
−
0,001 6
6
14
0,001
2,8
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MOMENTI NA KRAJEVIMA ELEMENATA M 14 = M 14
+
m14
M 14 = 13996,8ϕ 4
6k 14ψ 14
= −
6998,4u 45
+
+
M 41 = M 41 + m 41 = −6k 14ψ 14 M 41 = 27993,6ϕ 4
M 45 = M 45 M 45
6998,4u 45
=
6k 14ψ 14
=
+
m54
= −
+
c m52
=−
P⋅l 8
−
1 P ⋅l 2
3499,2u 45
M 57 = M 57 + m57
=
4k 57 ϕ 5
=
=
38173,09ϕ 5
M 75 = M 75 + m75 =
−
=
19086,55ϕ 5
6k 14 (− 0,001) 6
2k 14ϕ 4
+
−
− u 45 6
6k 14
−
6k 14 (− 0,001) 6
+
4k 14ϕ 4
−
− u 45 6
6k 14
35 ⋅ 2,8 2 12
⋅
−
8
+
3k 25ϕ 5
−
+
4k 45ϕ 4
35 ⋅ 2,8 2
+
12
3k 25ψ 25
+
2k 45ϕ 5
2k 45ϕ 4
+
90 + 3k 25ϕ 5
= −
4k 45ϕ 5
−
− u 45 6
3k 25
90,0
6k 57ψ 57
u 45 4 , 4
=
4k 57ϕ 5
−
6k 57
=
4k 57ϕ 5
−
6k 57
13013,55u 45
2k 57ϕ 5
−
−
−
6,9984
g ⋅l2
+
M 75
−
g ⋅l2
20995,2ϕ 5
M 57
=
6,9984
4k 14ϕ 4
+
6k 14ψ 14
+ 2k 45ϕ 4 + 4 k 45ϕ 5 = − 12 = 239945,14ϕ 4 + 479890,2856ϕ 5 − 22,867
c c M 52 = M 52 c M 52
+
m45
+
−
+ 4k 45ϕ 4 + 2k 45ϕ 5 = 12 = 479890 ,28ϕ 4 + 239945,15ϕ 5 + 22,867
M 54 = M 54 M 54
+
2k 14ϕ 4
+
−
6k 57ψ 57
u 45 4 , 4
13013,55u 45
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JEDNADŽBE RAVNOTEŽE ČVORA 4
∑ M
=
0 ⇒ − M 41 − M 45 + M PL 27993,6ϕ 4
+
507883,89ϕ 4
6998,4u 45 +
=
+
0
6,9984 + 479890,28ϕ 4
239945,15ϕ 5
+
6998,4u 45
=
239945,15ϕ 5
+
−
34,3 = 0
4,435............................................(1)
ČVORA 5
∑ M
=
c − M 54 − M 57 − M Pd 0 ⇒ − M 52
20995,2ϕ 5
+
38173,09ϕ 5
3499,2u 45
−
239945,14ϕ 4
−
0
90,0 + 239945,14ϕ 4
13013,55u 45
+
=
+
+
479890,2856ϕ 5
−
22,867 +
112 = 0
539058,58ϕ 5
−
9514,35u 45
=
0,86& ...............................................(2)
VIRTUALNOG RADA M75 (1) 57
4 , 4
49 kN
M45
1
ψ 14 1
80kN M57
35 kN/m'
M54 Mc52
M14
(1) 14
1
=
ψ 25
=
ψ 57
=
−
1
l14 −
1
l 25 1
l57
=
=
=
−
1
6 −
1
6 1 4,4
80 kN 6
(1) 25
M14 2,8 ___________________________________________________________________________ GRAĐEVNA STATIKA II VEDRAN SLUNJSKI, 5228 ak.god.2006/07
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∑W
=
0⇒
1 1 1 1 c 1 M 14ψ 14 + M 41ψ 14 + M 57ψ 57 + M 75ψ 57 + M 52ψ 25 + P ⋅ 0,5 = 0
− 1 (13996,8ϕ 4 + 6998,4u 45 + 6,9984 + 27993,6ϕ 4 + 6998,4u 45 + 6,9984) +
6
1 + 4,4
(38173,09ϕ 5 − 13013,55u 45 + 19086,55ϕ 5 − 13013,55u 45 ) − 1 (20995,2ϕ 5 + 3499,2u 45 − 90,0) + 40 = 0
6
−
6998,4ϕ 4
6998,4ϕ 4
−
+
9514,35ϕ 5
9514,35ϕ 5
+
−
8831,25u 45
8831,25u 45
=
52,67
= −
(− 1)
52,67.........................................................(3)
KONAČNI SUSTAV JEDNADŽBI 507883,88 239945,14 239945,14 539058,58 6998,4 − 9514,35
6998,4 ϕ 4
4,439 ϕ = 0,867 − 9514,35 5 9122,85 u 45 52,667
RJEŠENJE SUSTAVA JEDNADŽBI {{φ4 -> -0.000160298,
φ5
-> 0.000180345,
u45 -> 0.00608416}}
KONAČNI MOMENTI NA KRAJEVIMA ELEMENATA M 14 = 13996,8(- 0.000160298) + 6998,4(0.00608416) + 6,9984 = 47,33kNm M 41 = 27993,6(- 0.00016029 8) + 6998,4(0.00608416 ) + 6,9984 = 45,09kNm M 45
=
479890,28(- 0.000160298) + 239945,15(0.00018034 5) + 22,867 = −10,79kNm
M 54
=
239945,14(- 0.000160298) + 479890 ,2856(0.000180345) − 22,867 = 25,22kNm
M c52
=
20995,2(0.00018034 5) + 3499,2(0.00608416) − 90,0 = −64,92kNm
M 57
=
38173,09(0.00018034 5) − 13013,55(0.00608416 ) = −72,29kNm
M 75 = 19086,55(0.00018034 5) − 13013,55(0.00608416 ) = −75,73kNm ___________________________________________________________________________ GRAĐEVNA STATIKA II VEDRAN SLUNJSKI, 5228 ak.god.2006/07
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ODREĐIVANJE UZDUŽNIH SILA RAVNOTEŽA ČVORA 4 54,15kN N45
49kN
N 45 N 41
15,4kN
= −
49 − 54,15 = −103,15kN
= −
15,4kN N52
RAVNOTEŽA ČVORA 5 N57 33,64kN 43,85kN
80kN
15,4kN 50,82kN N52
N 57 N 52
=
N 52 +
EA l 25 6 ⋅ = l57 EA 4,4 6 4,4
N 52
= −
N 57
=
N 52
80 − 43,85
= −
53,40kN
6 4,4
(− 53,4) = −78,6kN
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DIJAGRAMI M,T,N 75,73
33,64
M 3 , 4 9 3 7 , 0 1
T 5 1 , 4 5
112 1 2 , 5 2
45,09
9 4
72,29 64,92
87,54
47,33
80
5 8 , 3 4
50,82
15,4
29,8
78,6
N 4 , 5 1
103,15
53,4
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RJEŠENJE U DIM-u
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Degrees of Freedom: 12
Nodal displacements: nd u_i v_i -------------------------------------------------------1: -0.001 0 2: 0 0 3: 0.00629668 0.000182733 4: 0.00629668 -1.03267e-09 5: 0.00629668 -5.23485e-10 6: 0.00629668 4.3622e-05 7: 0 0
phi_i 0 0 -0.000118612 -0.000166262 0.000186751 -4.66369e-05 0
Element end forces: el N_ij T_ij M_ij N_ji -----------------------------------------------------------------------------------------------1: 103.267 15.8581 48.738 -103.267 2: 52.3485 29.3257 0 -52.3485 3: 0 0 0 0 4: -15.8581 54.2673 -12.1108 15.8581 5: 0 80 112 0 6: -71.3843 -34.8162 -74.8134 71.3843
T_ji
M_ji
-15.8581 50.6743 49 43.7327 -80 34.8162
46.4108 -64.0458 -34.3 26.8591 0 -78.3778
Reactions: nd R_x R_y -------------------------------------------------------1: -15.8581 103.267 2: -29.3257 52.3485 7: -34.8162 71.3843
M 48.738 -78.3778
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