tugas analisa
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CONTOH STRUKTUR RANGKA BATANG13 t
31 t
1000 cm1000 cm 600 cm Hitung Gaya Batang yang terjadi pada bentuk
struktur di atas
36.869898 36.869898
0
1600 cm
Dimana :E = 2100000 t/cm2 E.A
= 143640E.A = 143640000
A = 68.4 cm2 L
2 Vektor gaya-lendutan di nodal dalam sistem koordinat global4
32
2 6 Y
1 3 5 XElemen 2
3 vektor gaya-lendutan elemen batang dalam sistem koordinat lokal
4
2 4
elemen 21
4 vektor gaya-lendutan elemen batang dalam sistem koordinat global
4
2 4
elemen 21
a =
5 Matrik Kekakuan Elemena Tinjuan elemen 1 ( I = 1 dan j = 2 ) Batang AB- Transformasi koordinat lokal
1 0 -1 0EA 0 0 0 0
kl = L -1 0 1 00 0 0 0
1 0 -1 0kl = 143640000 0 0 0 0
1000 -1 0 1 00 0 0 0
143640 0 -143640 0kl = 143640 0 0 0 0
-143640 0 143640 00 0 0 0
Matrik Transpose
ui 0 0 Uivi = 0 0 x Viuj 0 0 Ujvj 0 0 -sin a Vj
0 0Te = 0 0
0 00 0 -sin a
0.8 0.6 0.0 0.0T1 = -0.6 0.8 0.0 0.0
0.0 0.0 0.8 0.60.0 0.0 -0.6 0.8
0.8 -0.6 0.0 0.00.6 0.8 0.0 0.0
= 0.0 0.0 0.8 -0.60.0 0.0 0.6 0.8
cos a sin a-sin a cos a
cos a sin acos a
cos a sin a-sin a cos a
cos a sin acos a
T1 T
Kekauan Batang elemen 1 transformasi global (kg)
kg = kl
0.80 -0.60 0.00 0.00 143640 0 -143640 0 0.8 0.6 0.0 0.0EA 0.60 0.80 0.00 0.00 0 0 0 0 -0.6 0.8 0.0 0.0
= 1000 0.00 0.00 0.80 -0.60 x -143640 0 143640 0 x 0.0 0.0 0.8 0.60.00 0.00 0.60 0.80 0 0 0 0 0.0 0.0 -0.6 0.8
m4x4 m4x4 m4x4
= EA 114912 0 -114912 0 0.8 0.6 0.0 0.01000 86184 0 -86184 0 x -0.6 0.8 0.0 0.0
-114912 0 114912 0 0.0 0.0 0.8 0.6-86184 0 86184 0 0.0 0.0 -0.6 0.8
91929.6 68947.2 -91930 -68947 367718.4 275788.8 -367718.4 -275789= EA 68947.2 51710.4 -68947 -51710 275788.8 206841.6 -275788.8 -206842
1000 -91930 -68947 91929.6 68947 -367718.4 -275788.8 367718.4 275788.8-68947 -51710 68947.2 51710 -275788.8 -206841.6 275788.8 206841.6
Matrik Kekauan :2 Tinjuan elemen 2 (I = 2 dan j =3)
- Transformasi koordinat lokal
1 0 -1 0EA 0 0 0 0
kl = L -1 0 1 00 0 0 0
1 0 -1 0kl = EA 0 0 0 0
1600 -1 0 1 00 0 0 0
89775 0 -89775 00 0 0 0
kl = 89775 -89775 0 89775 00 0 0 00 0 0 0
T1 T T1
Matrik Transpor
ui 0 0 Uivi = 0 0 x Viuj 0 0 Ujvj 0 0 -sin a Vj
0 0Te = 0 0
0 00 0 -sin a
1 0 0 0T2 = 0 1 0 0
0 0 1 00 0 0 1
1 0 0 00 1 0 0
= 0 0 1 00 0 0 1
Kekauan Batang elemen 2 transformasi global (kg)
kg = kl
1 0 0 0 89775 0 -89775 0 1 0 0 0= 0 1 0 0 0 0 0 0 0 1 0 0
0 0 1 0 x -89775 0 89775 0 x 0 0 1 00 0 0 1 0 0 0 0 0 0 0 1
m4x4 m4x4 m4x4
= EA 89775 0 -89775 0 1 0 0 00 0 0 0 0 x 0 1 0 0
-89775 0 89775 0 0 0 0 00 0 0 0 0 0 1 1
89775 0 0 0 359100 0 0 0= EA 0 0 0 0 0 0 0 0
0 -89775 0 0 0 -359100 0 0 00 0 0 0 0 0 0 0
Matrik Kekauan :3 Tinjuan elemen 3 (I = 1 dan j =3)
- Transformasi koordinat lokal
1 0 -1 0EA 0 0 0 0
kl = L -1 0 1 00 0 0 0
1 0 -1 0kl = EA 0 0 0 0
1000 -1 0 1 00 0 0 0
143640 0 -143640 0kl = 0 0 0 0
-143640 0 143640 00 0 0 0
Matrik Transpor
ui 0 0 Uivi = 0 0 x Viuj 0 0 Ujvj 0 0 -sin a Vj
cos a sin a-sin a cos a
cos a sin acos a
cos a sin a-sin a cos a
cos a sin acos a
T2 T
T2 T T2
cos a sin a-sin a cos a
cos a sin acos a
0 0Te = 0 0
0 00 0 -sin a
0.80 -0.60 0.00 0.00T3 = 0.60 0.80 0.00 0.00
0.00 0.00 0.80 -0.600.00 0.00 0.60 0.80
0.8 0.6 0.0 0.0-0.6 0.8 0.0 0.0
= 0.0 0.0 0.8 0.60.0 0.0 -0.6 0.8
Kekauan Batang elemen 3 transformasi global (kg)
kg = kl
0.8 0.6 0.0 0.0 143640 0 -143640 0 0.8 -0.6 0.0 0.0EA -0.6 0.8 0.0 0.0 0 0 0 0 0.6 0.8 0.0 0.0
= 1000 0.0 0.0 0.8 0.6 x -143640 0 143640 0 x 0.0 0.0 0.8 -0.60.0 0.0 -0.6 0.8 0 0 0 0 0.0 0.0 0.6 0.8
m4x4 m4x4 m4x4
= EA 114912 0 -114912 0 0.8 -0.6 0 01000 -86184 0 86184 0 x 0.6 0.8 0 0
-114912 0 114912 0 0 0 0.8 -0.686184 0 -86184 0 0 0 0.6 0.8
91929.6 -68947.2 -91929.6 68947 91929.6 -68947.2 -91929.6 68947.2= EA -68947.2 51710.4 68947.2 -51710 = EA -68947.2 51710.4 68947.2 -51710.4
1000 -91929.6 68947.2 91929.6 -68947 0.0108779 -91929.6 68947.2 91929.6 -68947.268947.2 -51710.4 -68947.2 51710 68947.2 -51710.4 -68947.2 51710.4
Overall Stiffness matrixGaya Batang
F = Te Kg U
0.8 0.6 0 0 1 0 0 0-0.6 0.8 0 0 0 1 0 0
T1 = 0.0 0 0.8 0.6 T3 = 0 0 1 00.0 0 -0.6 0.8 0 0 0 1
0.8 -0.6 0 00.6 0.8 0 0
T2 = 0 0 0.8 -0.60 0 0.6 0.8
0.8 0.6 0 0 0 0-0.6 1.6 -0.6 0 0 0
Te = 0 0.6 2.6 0.6 0 00 0 -0.6 2.6 -0.6 00 0 0 0.6 0.8 00 0 0 0 0 1
u1 v1 u2 v291929.6 68947.2 -91930 -68947.2 u1
Kg1 = AE 68947.2 51710.4 -68947 -51710.4 v11000 -91929.6 -68947.2 91930 68947.2 u2
-68947.2 -51710.4 68947 51710.4 v2
cos a sin a-sin a cos a
cos a sin acos a
T3 T
T3 T T3
u2 v2 u3 v389775 0 0 0 u2
Kg2 = AE 0 0 0 0 v20 -89775 0 0 0 u3
0 0 0 0 v3
91929.6 -68947.2 -91930 68947.2 u2-68947.2 51710.4 68947 -51710.4 v2
Kg3 = -91929.6 68947.2 91930 -68947.2 u368947.2 -51710.4 -68947 51710.4 v3
u1 v1 u2 v2 u3 v3181704.6 68947.2 -91930 -68947.2 0 0 u168947.2 51710.4 -68947 -51710.4 0 0 v1
Kg = -181704.6 -68947.2 183859 0 -91930 68947.2 u2-68947 -51710.4 0 103420.8 68947 -51710.4 v2-89775 0 -91930 68947 91930 -68947.2 u3
0 0 68947 -51710.4 -68947.2 51710.4 v3
6 jawaban yang ditanyakan :Menghitung displacement yang terjadi :lendutan terjadi pada titik B pada displacement u2 dan v2 :
u1 v1 u2 v2 u3 v3181704.6 68947.2 -91929.6 -68947.2 0 0 u168947.2 51710.4 -68947.2 -51710.4 0 0 v1
Kg = -181704.6 -68947.2 183859.2 0 -91929.6 68947.2 u2-68947.2 -51710.4 0 103420.8 68947.2 -51710.4 v2-89775 0 -91929.6 68947.2 91929.6 -68947.2 u3
0 0 68947.2 -51710.4 -68947.2 51710.4 v3sehingga
Kg = 2E+05 00 103420.8
P = Kg x
x = kg x
x = 5.4389E-06 0 310 9.67E-06 -13
x = 0.00016861-0.0001257
maka displecement yang terjadiX2 = 0.00016861Y2 = -0.0001257
B. Menghitung gaya batang
Pi = Po + Ki Ti Xi
P1 = 0 143640 0 -143640 0 0.8 0.6 0.0 0.0 00
+0 0 0 0 -0.6 0.8 0.0 0.0 0
0 -143640 0 143640 0 0.0 0.0 0.8 0.6 2E-040 0 0 0 0 0.0 0.0 -0.6 0.8 -0.00013
P1 = 0 114912 86184 -114912 -86184 00 0 0 0 0 00 -114912 -86184 114912 86184 2E-040 0 0 0 0 -1E-04
P1 = 0 -8.54 -8.540 + 0 = 0.000 8.54 8.540 0 0.00
P1 = 0 89775 0 0 0 1 0 0 0 00 + 0 0 0 0 0 1 0 0 00 -89775 0 0 0 0 0 1 0 00 0 0 0 0 0 0 0 1 0
P1 = 0 89775 0 0 0 00 0 0 0 0 00 -89775 0 0 0 00 0 0 0 0 0
P1 = 0 0.00 0.000 + 0.00 = 0.000 0.00 0.000 0.00 0.00
P1 = 0 143640 0 -143640 0 0.8 -0.6 0.0 0.0 2E-040
+0 0 0 0 0.6 0.8 0.0 0.0 -1E-04
0 -143640 0 143640 0 0.0 0.0 0.8 -0.6 00 0 0 0 0 0.0 0.0 0.6 0.8 0
P1 = 0 114912 -86184 -114912 86184 2E-040 0 0 0 0 -1E-040 -114912 86184 114912 -86184 00 0 0 0 0 0
P1 = 0 30.21 30.210 + 0.00 = 0.000 -30.21 -30.210 0.00 0.00
C Menghitung Reaksi TumpuanSendi-sendi ada pada titik tumpuan 1 dan 3 maka :disusun dan dimasukkan nilai-nilai yang sudah diperoleh kembali ke matriks global
u1 v1 u2 v2 u3 v3Px1 181705 68947 -91930 -68947 0 0Px2 68947 51710 -68947 -51710 0 0Px3 = -181705 -68947 183859 0 -91930 68947Px4 -68947 -51710 0 103421 68947 -51710Px5 -89775 0 -91930 68947 91930 -68947Px6 0 0 68947 -51710 -68947 51710
0 0 0 0 0 0
Px1 -91930 -68947Px2 -68947 -51710Px5 = -91930 68947Px6 68947 -51710
R = Px X
-91930 -68947R = -68947 -51710 0.000169
-91930 68947 -0.0001368947 -51710
Px1 -6.83Px2 -5.13Px5 = -24.17Px6 18.12
maka reaksi tumpuan diperoleh nilai :
Px1 = Rx.1 = -6.83Px2 = Ry.1 = -5.13Px5 = Rx.3 = -24.17Px6 = Ry.3 = 18.13
7 Gambar reaksi perletakan 13
31
-5.13 -24.17
-5.13 18.13
-551577.6-413683.2551577.6413683.2
3m
3m 2m
LE 200000A 1/4(3,14)*(10)^2 = 78,54 Dimana :EA = 15708000 N/mm2
Matrik Kekauan :1 Tinjuan elemen 1 (I = 1 dan j =2)
- Transformasi koordinat lokal1
2
1 0 -1 0EA 0 0 0 0
kl = L -1 0 1 00 0 0 0
1 0 -1 0kl = EA 0 0 0 0
5830 -1 0 1 00 0 0 0
Matrik Transpor
ui 0 0 Ui 31.01vi = 0 0 x Viuj 0 0 Ujvj 0 0 -sin a Vj
√5^2+3^2 = 5830 mm
cos a sin a-sin a cos a
cos a sin acos a
0 0 200 1.732051Te = 0 0
0 0 10 0 -sin a 346.41
cos a sin a-sin a cos a
cos a sin acos a
0.857077397 0 0 0T1 = 0.000 0.857 0 0
0 0 0.857077397 00 0 0 0.8570773965
1.166755773 0.000 0 00 1.16675577 0 0
= 0 0 1.166755773 00 0 0 1.1667557726
Kekauan Batang elemen 1 transformasi global (kg)
kg = kl
1.16675577 0.000 0 0 1 0 -1 0 0.85708 0 0 015708000 0 1.166755773 0 0 0 0 0 0 0 0.8571 0 0
= 5830 0 0 1.1667557726 0 x -1 0 1 0 x 0 0 0.85708 00 0 0 1.1667557726 0 0 0 0 0 0 0 0.85708
m4x4 m4x4 m4x4
= 2694.339623 1.16675577 0 -1.166755773 0 0.857077397 0 0 00 0 0 0 x 0 0.85708 0 0
-1.16675577 0 1.1667557726 0 0 0 0.85708 00 0 0 0 0 0 0 0.857077
1 0 -1.00 0.00 2694.34 0 -2694.34 0= 2694.339623 0 0 0.00 0.00 0 0 0 0
-1 0 1 0.00 -2694.34 0 2694.34 00 0 0 0.00 0 0 0 0
T1 T
T1 T T1
Matrik Kekauan :2 Tinjuan elemen 2 (I = 2 dan j =3) A 5000*3000= 15000
- Transformasi koordinat lokal EA 3000000000L 5385.16
uivi
vi uj
1 0 -1 0EA 0 0 0 0
kl = L -1 0 1 00 0 0 0
1 0 -1 0kl = 3E+09 0 0 0 0
5385.16 -1 0 1 00 0 0 0
Matrik Transpor
ui 0 0 Uivi = 0 0 x Viuj 0 0 Ujvj 0 0 -sin a Vj
21.77
0 0 21.87Te = 0 0
0 00 0 -sin a
0.928 0.37088163 0 0T2 = -0.371 0.92803142 0 0
0 0 0.928031423 0.37088163060 0 -0.37088163 0.9280314227
0.929 -0.371 0 00.371 0.92915058 0 0
= 0 0 0.92915058 -0.3710 0 0.371328894 0.9291505795
Kekauan Batang elemen 2 transformasi global (kg)
cos a sin a-sin a cos a
cos a sin acos a
cos a sin a-sin a cos a
cos a sin acos a
T2 T
kg = kl
0.92915058 -0.371 0 0 1 0 -1 0 0.92803 0.3709 0 03000000000 0.371 0.92915058 0 0 0 0 0 0 -0.371 0.928 0 0
= 5385.16 0 0 0.9291505795 -0.371 x -1 0 1 0 x 0 0 0.92803 0.3710 0 0.3713288943 0.9291505795 0 0 0 0 0 0 -0.371 0.92803
m4x4 m4x4 m4x4
= 557086.5118 0.92915058 0 -0.92915058 0 0.928031423 0.37088 0 00.37132889 0 -0.371328894 0 x -0.37088163 0.92803 0 0
-0.92915058 0 0.9291505795 0 0 0 0.92803 0.370882-0.37132889 0 0.3713288943 0 0 0 -0.37088 0.928031
T2 T T2
0.86228093 0.344604882 -0.862280934 -0.34 480365.08 ### ### ###= 557086.5118 0.34460488 0.137719066 -0.344604882 -0.14 191974.73 ### ### ###
-0.86228093 -0.34460488 0.8622809342 0.34 -480365.08 ### ### ###-0.34460488 -0.13771907 0.344604882 0.14 -191974.73 ### ### 76721.43
Overall Stiffness matrixGaya Batang
F = Te Kg U
0.857077397 0 0 00 0.8570774 0 0
T1 = 0 0 0.857077397 00 0 0 0.8570773965
0.928031423 0.37088163 0 0-0.370881631 0.92803142 0 0
T2 = 0 0 0.928031423 0.37088163060 0 -0.37088163 0.9280314227
0.857077397 0 0 00 1.78510882 0.370881631 0 0
Te = 0 -0.37088163 1.785108819 0 0.37088163060 0 0 1.7851088192 0.92803142270 0 0 0 0
2694.34 0.00 -2694.34 0.00 u1 u1Kg1 = AE 0.00 0.00 0.00 0.00 v1 v1
0 -2694.34 0.00 2694.34 0.00 u2 u30.00 0.00 0.00 0.00 v2 v3
480365.08 191974.73 -480365.08 -191974.73 u1Kg2 = AE 191974.73 76721.43 -191974.73 -76721.43 v1
0 -480365.08 -191974.73 480365.08 191974.73 u2-191974.73 -76721.43 191974.73 76721.43 v2
u1 v1 u2 v2483059.42 191974.73 -483059.42 -191974.73191974.73 76721.43 -191974.73 -76721.43
Kg = AE -483059.42 -191974.73 483059.42 191974.730 -191974.73 -76721.43 191974.73 76721.43
0 0 0 00 0 0 0
fokuskan gaya-gaya yang bekerja pada struktur tersebut
u1 v1 u2 v2483059.42 191974.73 -483059.42 -191974.73191974.73 76721.43 -191974.73 -76721.43-483059.42 -191974.73 483059.42 191974.73
Kg = -191974.73 -76721.43 191974.73 76721.430 0 0 00 0 0 0
Menghitung displacement yang terjadi
4 5 6 7P = Kg U 6 8 9 10
V1 U21 76721.43 -191974.73 V12 = -191974.73 483059.42 U2
U = Kg U
V1 76721.43 -191974.73 1U2 -191974.73 483059.42 2
V1 0.002 0.001 1U2 = 0.001 0.000 2
V1 0.00419425U2 = 0.001671
di susun dan dimasukkan niilai-nilai yang sudah diperoleh kembali ke matriks global
Px1 483059.42 191974.73 -483059.42 -191974.73 0Px2 191974.73 76721.43 -191974.73 -76721.43 0Px3 = -483059.42 -191974.73 483059.42 191974.73 0.0042Px4 -191974.73 -76721.43 191974.73 76721.43 0.0017
Px1 -2346.861828
Px2 -933.3915684Px3 = 2346.861828Px4 933.3915684
maka reaksi tumpuan diperoleh nilai :
Px1 = Rx.1 = -2346.86183Px2 = Ry.1 = -933.391568
20 5 0 0 0 0 0.25 0 05 20 0 0 0 0 0.25 1 00 0 30 10 0 0 -0.50 1 00 0 10 30 0 0 -0.50 0 10 0 0 0 12 6 0.25 0 10 0 0 0 6 12 0.25 0 0
6.25 5 0 1 2 36.25 20 0 4 5 6-20 30 10-20 10 304.5 0 124.5 0 6
30036.87
400
1 3 5 3 2 12 4 6 2 7 63 6 8 7 6 5
44 53 4456 68 56
77 96 79
STRUKTUR RANGKA BATANG
Penyelesaian :
E = 2100 t/cm2A = 35 cm2
Pencarian panjang bentang setiap elemen dan besar sudut diukur menggunakan AutoCad secara langsung :
Matrik Kekauan :1 Tinjuan elemen 1 (I = 1 dan j =2)
- Transformasi koordinat lokal
1 0 -1 0EA 0 0 0 0
kl = L -1 0 1 00 0 0 0
1 0 -1 0kl = EA 0 0 0 0
354 -1 0 1 00 0 0 0
Matrik Transpor
ui 0 0 Uivi = 0 0 x Viuj 0 0 Ujvj 0 0 Vj
0 0Te = 0 0
0 00 0
-1 6.12E-17 0 0T1 = 6.12E-17 1 0 0
0 0 -1 6.12E-170 0 6.12E-17 1
-1 6.12E-17 0 06.12E-17 1 0 0
= 0 0 -1 6.12E-170 0 6.12E-17 1
Kekauan Batang elemen 1 transformasi global (kg)
kg = kl
-1 6.12E-17 0 0EA 6.12E-17 1 0 0
= 354 0 0 -1 6.12E-17 x
-sin a cos acos a sin a
-sin a cos acos a sin a
-sin a cos acos a sin a
-sin a cos acos a sin a
T1 T
T1 T T1
0 0 6.12E-17 1
m4x4
= EA -1 0 1 0354 6.12E-17 0 -6.1E-17 0 x
1 0 -1 0-6.1E-17 0 6.12E-17 0
1 -6.1E-17 -1 0.00= EA -6.1E-17 3.75E-33 6.12E-17 0.00
354 -1 6.12E-17 1 0.006.12E-17 -3.7E-33 -6.1E-17 0.00
Matrik Kekauan :2 Tinjuan elemen 2 (I = 2 dan j =3)
- Transformasi koordinat lokal
1 0 -1 0EA 0 0 0 0
kl = L -1 0 1 00 0 0 0
1 0 -1 0kl = EA 0 0 0 0
354 -1 0 1 00 0 0 0
Matrik Transpor
ui 0 0 Uivi = 0 0 x Viuj 0 0 Ujvj 0 0 Vj
-sin a cos acos a sin a
-sin a cos acos a sin a
0 0Te = 0 0
0 00 0
-0.70711 0.707107 0 0T2 = 0.707107 0.707107 0 0
0 0 -0.70711 0.7071070 0 0.707107 0.707107
-0.70711 0.707107 0 00.707107 0.707107 0 0
= 0 0 -0.70711 0.7071070 0 0.707107 0.707107
Kekauan Batang elemen 2 transformasi global (kg)
kg = kl
-0.70711 0.707107 0 0EA 0.707107 0.707107 0 0
= 354 0 0 -0.70711 0.707107 x0 0 0.707107 0.707107
m4x4
= EA -0.70711 0 0.707107 0354 0.707107 0 -0.70711 0 x
0.707107 0 -0.70711 0-0.70711 0 0.707107 0
0.5 -0.5 -0.5 0.50= EA -0.5 0.5 0.5 -0.50
354 -0.5 0.5 0.5 -0.500.5 -0.5 -0.5 0.50
Matrik Kekauan :3 Tinjuan elemen 3 (I = 1 dan j =3)
- Transformasi koordinat lokal
-sin a cos acos a sin a
-sin a cos acos a sin a
T2 T
T2 T T2
1 0 -1 0EA 0 0 0 0
kl = L -1 0 1 00 0 0 0
1 0 -1 0kl = EA 0 0 0 0
500 -1 0 1 00 0 0 0
Matrik Transpor
ui 0 0 Uivi = 0 0 x Viuj 0 0 Ujvj 0 0 Vj
0 0Te = 0 0
0 00 0
0 1 0 0T3 = 1 0 0 0
0 0 0 10 0 1 0
0 1 0 01 0 0 0
= 0 0 0 10 0 1 0
Kekauan Batang elemen 3 transformasi global (kg)
-sin a cos acos a sin a
-sin a cos acos a sin a
-sin a cos acos a sin a
-sin a cos acos a sin a
T3 T
kg = kl
0 1 0 0EA 1 0 0 0
= 500 0 0 0 1 x0 0 1 0
m4x4
= EA 0 0 0 0500 1 0 -1 0 x
0 0 0 0-1 0 1 0
0 0 0 0= EA 0 1 0 -1
500 0 0 0 00 -1 0 1
Overall Stiffness matrixGaya Batang
F = Te Kg U
-1 6.12E-17 0 06.12E-17 1 0 0
T1 = 0 0 -1 6.12E-17 T30 0 6.12E-17 1
-0.70711 0.707107 0 00.707107 0.707107 0 0
T2 = 0 0 -0.70711 0.7071070 0 0.707107 0.707107
-1 6.12E-17 0 0 0 06.12E-17 0.292893 0.707107 0 0 0
Te = 0 0.707107 -0.29289 1 0 00 0 1 0.292893 0.707107 00 0 0 0.707107 0.707107 00 0 0 0 0 0
T3 T T3
1 -6.1E-17 -1 6.12E-17Kg1 = AE -6.1E-17 3.75E-33 6.12E-17 -3.7E-33
354 -1 6.12E-17 1 -6.1E-176.12E-17 -3.7E-33 -6.1E-17 3.75E-33
0.5 -0.5 -0.5 0.5Kg2 = AE -0.5 0.5 0.5 -0.5
354 -0.5 0.5 0.5 -0.50.5 -0.5 -0.5 0.5
u1 v1 u2 v2 u31 -6.1E-17 -1 6.12E-17 0
-6.1E-17 3.75E-33 0.5 -0.5 0Kg = AE -1 6.12E-17 0.50 0.5 -0.5
354 0.5 -0.5 -0.5 0.5 0.50 0 -0.5 0.5 0.50 0 0.5 -0.5 -0.5
=
0 0 00 0 00 0 00 0 00 0 00 0 0
Hitung Gaya Batang yang terjadi pada bentukstruktur di atas
Pencarian panjang bentang setiap elemen dan besar sudut diukur menggunakan AutoCad secara langsung :
1 0 -1 0 -1 6.12E-17 0 00 0 0 0 6.12E-17 1 0 0-1 0 1 0 x 0 0 -1 6.12E-17
0 0 0 0 0 0 6.12E-17 1
m4x4 m4x4
-1 6.12E-17 0 06.12E-17 1 0 0
0 0 -1 6.12E-170 0 6.12E-17 1
4 -2.4E-16 -4 2.45E-16-2.4E-16 1.5E-32 2.45E-16 -1.5E-32
-4 2.45E-16 4 -2.4E-162.45E-16 -1.5E-32 -2.4E-16 1.5E-32
1 0 -1 0 -0.70711 0.707107 0 00 0 0 0 0.707107 0.707107 0 0-1 0 1 0 x 0 0 -0.70711 0.7071070 0 0 0 0 0 0.707107 0.707107
m4x4 m4x4
-0.70711 0.707107 0 00.707107 0.707107 0 0
0 0 -0.70711 0.7071070 0 0.707107 0.707107
2 -2 -2 2-2 2 2 -2-2 2 2 -22 -2 -2 2
1 0 -1 0 0 1 0 00 0 0 0 1 0 0 0-1 0 1 0 x 0 0 0 10 0 0 0 0 0 1 0
m4x4 m4x4
0 1 0 01 0 0 00 0 0 10 0 1 0
0 0 0 0= EA 0 1.412429 0 -1.41243
#DIV/0! 0 0 0 00 -1.41243 0 1.412429
0 1 0 01 0 0 0
= 0 0 0 10 0 1 0
u1 0 0 0 0v1 Kg1 = AE 0 1.412429 0 -1.41243u2 #DIV/0! 0 0 0 0v2 0 -1.41243 0 1.412429
u2v2u3v3
v30
-1.412430.5-0.5-0.50.5
-0.00380.0000
0 0 00 0 00 0 00 0 00 0 00 0 0
u1v1u3v3
Kekauan Batang elemen 3 transformasi global (kg)
kg = kl
1 0 0 0 140 0 1 0 0 0
= 8 0 0 1 0 x -10 0 0 1 0
m4x4 m4x4
= 5 1 0 -1 0 10 0 0 0 x 0
-1 0 1 0 00 0 0 0 0
1 0 -1 0= 5 0 0 0 0 =
-1 0 1 00 0 0 0
T3 T T3
0 -1 0 1 0 0 00 0 0 0 1 0 00 1 0 x 0 0 1 00 0 0 0 0 0 1
m4x4
0 0 01 0 00 1 00 0 1
5 0 -5 0EA 0 0 0 00 -5 0 5 0
0 0 0 0