tugas analisa

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 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


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


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



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



cos a sin acos a


cos a sin acos a

T2 T

T2 T T2


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


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 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


- 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 acos a

0 0 200 1.732051Te = 0 0

0 0 10 0 -sin a 346.41


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


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


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



cos a sin acos 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


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 :



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


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




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



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




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


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





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


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






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


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


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

tugas analisa

Documents