Penyelesaian Soal UAS Mekanika Rekayasa IV - 7 Juli 2011 - Yoppy Soleman

Soal Belum Diganti!!!

Derajat Ketidaktentuan Kinematik (degree of kinematic indeterminacy) = 2 Translasi (Join D) dan 2 Translasi (Join C) (Komponen Aksial) Q11 = Q12 = Q21 = Q22 = 140 kN (arah positif) 0 kN 0 kN -60 kN (arah negatif) cos = sin = 0.8575 0.5145

Matriks Vektor Deformasi [A] 0.86 0 -0.86 0 0 D1 0.51 1 0.51 0 0 D2 0 0 0 1 -1 D3 0 1 0 0 0 D4 1 2 3 4 5

Matriks Vektor Deformasi Transpos [A]T 0.86 0.51 0 0 0 1 0 1 -0.86 0.51 0 0 0 0 1 0 0 0 -1 0

Matriks Kekakuan Lokal Elemen [S] 0.34 0 0 0.67 0 0 0 0 0 0.51 0 0 0 0 0 0.2 0 0 0 0 0 0.2 F1 F2 F3 F4 F5

EA

0 0 0

1

2

3

4

5

Matriks Kekakuan Struktur (Global Stiffness Matrix) [K] = [A]T [S] [A]0.86 0 -0.86 0 0 0.51 1 0.51 0 0 0 0 0 1 -1 0 1 0 0 0

x

0.34 0 0 0 0

0 0.67 0 0 0

0 0 0.51 0 0

0 0 0 0.2 0

0 0 0 0 0.2

x

0.86 0.51 0 0

0 1 0 1

-0.86 0.51 0 0

0 0 1 0

0 0 -1 0

Submatrix [K] = [A]T [S]0.29 0.18 0 0 0 0.67 0 0.67 -0.44 0.26 0 0 0 0 0.2 0 0 0 -0.2 0

[K] = [A]T [S] [A]0.63 -0.08 0 0 -0.08 0.89 0 0.67 0 0 0.4 0 0 0.67 0 0.67

EA

[K]-11.65 0.55 0 -0.55 0.55 4.59 0 -4.59 0 0 2.5 0 -0.55 -4.59 0 6.09

EA

Matriks Deformasi [D] = [K]-1 [Q]264.34 352.45 1/EA 0 -442.45 = 1.65 0.55 0 -0.55 0.55 4.59 0 -4.59 0 0 2.5 0 -0.55 -4.59 0 6.09 x 140.00 0.00 0.00 -60.00

Matriks Gaya Internal [H] = [S] [A] [D]139.94 -60 -23.32 0 0 0.34 0 0 0 0 0 0.67 0 0 0 0 0 0.51 0 0 0 0 0 0.2 0 0 0 0 0 0.2 x 0.86 0 -0.86 0 0 0.51 1 0.51 0 0 0 0 0 1 -1 0 1 0 0 0 x 264.34 352.45 0 -442.45

=

Submatrix [H] = [S] [A]0.29 0 -0.44 0 0 0.18 0.67 0.26 0 0 0 0 0 0.2 -0.2 0 0.67 0 0 0

Nilai yang dimasukkan disini adalah reaksi momen internal, sehingga berlawanan arah dengan momen internal atau -[H] atau balikkan tanda momen internal [H]

Gaya Aksial Elemen = - [H] F1 = -139.943 F2 = 60.000 F3 = 23.324 F4 = 0.000 F5 = 0.000

Plot Skema Momen-momen Ujung Elemen Frame -139.94 60 23.32 0 0 0.000 0.000 0.000 0.000 0.000kN kN kN kN kN[COMPRESSION] [TENSION] [TENSION] [NEUTRAL] [NEUTRAL]

=

+

= Perhitungan Manual - Output Aplikasi Program Komputer = 0 %

Soal Belum Diganti!!!

Koefisien Kekakuan Ujung Elemen =

Momen Ujung Elemen Satu Satuan Rotasi

Momen Jepit Ujung

M AB =

1 wL 2 12

A= L1

B

M BA =

1 wL 2 12

M BC =

1 wL 2 12

B= L2

C

M CB =

1 wL 12

2

M CD =

Pab 2 L2

C

D

M DC =

Pba 2 L2

= L3 Derajat Ketidaktentuan Kinematik = Rotasi Join B, C dan D (3 DK) FEMAB = 96.000 kNm FEMBA = -96.000 kNm = 34.667 kNm FEMBC = 130.667 kNmPage 4 - Stiffness Matrix Analysis (c) Y. Soleman - 2008

FEMCB FEMCD FEMDC

= = =

-130.667 kNm 85.000 kNm -85.000 kNm

= =

-45.667 kNm -85.000 kNm

Matriks Vektor Deformasi [A] 0 1 1 0 0 0 D1 0 0 0 1 1 0 D2 0 0 0 0 0 1 D3 1 2 3 4 5 6

Matriks Vektor Deformasi Transpos [A]T 0 0 0 1 0 0 1 0 0 0 1 0 0 1 0 0 0 1

Matriks Kekakuan Lokal Elemen [S] 0.67 0.33 0 0 0 0 1 0.33 0.67 0 0 0 0 2 0 0 0.57 0.29 0 0 3 0 0 0.29 0.57 0 0 4 0 0 0 0 1.5 0.75 5 0 0 0 0 0.75 1.5 6 M1 M2 M3 M4 M5 M6

EI

Matriks Kekakuan Struktur (Global Stiffness Matrix) [K] = [A]T [S] [A]0 1 1 0 0 0 0 0 0 1 1 0 0 0 0 0 0 1 0.67 0.33 0 0 0 0 0.33 0.67 0 0 0 0 0 0 0.57 0.29 0 0 0 0 0.29 0.57 0 0 0 0 0 0 1.5 0.75 0 0 0 0 0.75 1.5 0 0 0 1 0 0 1 0 0 0 1 0 0 1 0 0 0 1

x

x

Submatrix [K] = [A]T [S]0.33 0 0 0.67 0 0 0.57 0.29 0 0.29 0.57 0 0 1.5 0.75 0 0.75 1.5

[K] = [A]T [S] [A]1.24 0.29 0 0.29 2.07 0.75 0 0.75 1.5

EI

Page 5 - Stiffness Matrix Analysis (c) Y. Soleman - 2008

Menentukan Matriks Kekakuan Invers [K]-1 yang Metoda Adjoint Metoda ADJOINT 1.24 0.29 0.29 2.07 0 0.75 Rumus : 0 0.75 1.5

[ A]dimana:

1

=

1 1+ adj A = A [ det A

]

[ A] [ A]

1

=a m triksinve rs = jo ad A intT

= d r m ete in an+

Adjo int A =[C ijA ]Cij = ( 1) i + j M ijij

transpos matriks kofaktor =

kofaktor Minor M ij =

= baris ke i kolom ke j= det er min an ( faktor penentu)

[ ] = matriksAlgoritma: 1. Hitung determinan matriks dengan rumus kofaktor Cijminor minor minor

= =

-1.8 3.03

2.07 0.75

0.75 1.5

+

0.4

0.29 0.75

0 1.5

+

0

0.29 2.07

0 0.75

2. Hitung kofaktor elemen-elemen matriks (transpos) 1.24 0.29 0 2.07 = C11 = 0.29 2.07 0.75 0.75 0 0.75 1.5

0.75 1.5

=

2.54


C12 =

1.24 0.29 0 1.24 0.29 0 1.24 0.29 0 1.24 0.29 0 1.24 0.29 0 1.24 0.29 0 1.24 0.29 0 1.24 0.29 0

0.29 2.07 0.75 0.29 2.07 0.75 0.29 2.07 0.75 0.29 2.07 0.75 0.29 2.07 0.75 0.29 2.07 0.75 0.29 2.07 0.75 0.29 2.07 0.75

0 0.75 1.5 0 0.75 1.5 0 0.75 1.5 0 0.75 1.5 0 0.75 1.5 0 0.75 1.5 0 0.75 1.5 0 0.75 1.5

=

0.29 0

0.75 1.5

=

-0.43

C13 =

=

0.29 0

2.07 0.75

=

0.21

C21 =

=

0.29 0.75

0 1.5

=

-0.43

C22 =

=

1.24 0

0 1.5

=

1.86

C23 =

=

1.24 0

0.29 0.75

=

-0.93

C31 =

=

0.29 2.07

0 0.75

=

0.21

C32 =

=

1.24 0.29

0 0.75

=

-0.93

C33 =

=

1.24 0.29

0.29 2.07

=

2.48

3. Susun adjoint matriks (adjoint = transpos matriks kofaktor) C11 C12 C13 T C21 C22 C23 [A]+ = = C31 C32 C33 4. Tentukan invers matriks 2.54 0.33 [A]-1 =

C11 C12 C13

C21 C22 C23

C31 C32 C33

=

2.54 -0.43 0.21

-0.43 1.86 -0.93

0.21 -0.93 2.48

-0.43

0.21 =

0.84

-0.14

0.07


[A]-1 =

0.33

-0.43 0.21

1.86 -0.93

-0.93 2.48

=

-0.14 0.07

0.61 -0.31

-0.31 0.82

Catatan: Cara termudah untuk mengerjakan operasi-operasi matriks (perkalian dan invers) adalah dengan menggunakan spreadsheet Microsoft Office Excel, atau program Mathcad atau program Matlab, atau menggunakan kalkulator program matriks Casio PB-1000, dan sejenisnya.

[K]-10.84 -0.14 0.07 -0.142 0.6133 -0.307 0.07 -0.31 0.82

1/EI

Matriks Deformasi [D] = [K]-1 [Q]29.58 -6.85 1/EI -53.24 = 0.84 -0.14 0.07 -0.14 0.61 -0.31 0.07 -0.31 0.82 x 34.667 -45.667 -85.000

Matriks Gaya Internal [H] = [S] [A] [D]9.86 19.72 14.95 4.54 -50.2 -85 0.67 0.33 0 0 0 0 0.33 0.67 0 0 0 0 0 0 0.57 0.29 0 0 0 0 0.29 0.57 0 0 0 0 0 0 1.5 0.75 0 0 0 0 0.75 1.5 0 1 1 0 0 0 0 0 0 1 1 0 0 0 0 0 0 1 x 29.58 -6.85 -53.24

=

x

Submatrix [H] = [S] [A]0.33 0.67 0.57 0.29 0 0 0 0 0.29 0.57 1.5 0.75 0 0 0 0 0.75 1.5



Momen Ujung Elemen = FEMij - [H] = -[H] + FEMij MAB MBA MBC MCB MCD MDC = = = = = = 86.140 -115.720 115.720 -135.205 135.205 0.000 -9.86 -19.72 -14.95 -4.54 50.2 85 96.000 kNm -96.000 kNm 130.667 kNm -130.667 kNm 85.000 kNm -85.000 kNm

=

+

Plot Skema Momen-momen Ujung Elemen Beam


MOMEN JEPIT UJUNG (FIX-END MOMENT)

M BC =

1 wL 2 12

B

C

M CB =

1 wL 12

2

M CE =

1 wL 2 12

C

E

1 M EC = wL 12

2

M CD =

Pab 2 L2

C

D

M DC =

Pba 2 L2


F

B

1 M BF = wL 2

2

MB = 0

B

A

MA = 0

Derajat Ketidaktentuan Kinematik (degree of kinematic indeterminacy) = 2 Rotasi (Join B dan C) (Hanya pengaruh lentur) FEMEC FEMBF FEMBC FEMCB FEMCE FEMCD FEMDC FEMAB FEMAB = = = = = = = = = -105.000 kNm -40.000 kNm 142.917 kNm -142.917 kNm 105.000 kNm 16.875 kNm -50.625 kNm 0.000 kNm 0.000 kNm = = = = = = -105.000 kNm 102.917 kNm -21.042 kNm -50.625 kNm 0.000 kNm 0.000 kNm

Koefisien Kekakuan Ujung Elemen

=

Momen Ujung Elemen 1 Unit Rotasi


Matriks Vektor Deformasi [A] 1 0 0 1 1 0 0 0 0 0 D1 0 0 1 1 0 1 0 D2 2 3 4 5 6 7 8

Matriks Vektor Deformasi Transpos [A]T 0 0 1 0 1 0 0 1 0 1 0 0 0 1 0 0

Matriks Kekakuan Lokal Elemen [S] 1 0.5 0 0 0 0 0 0 1 0.5 1 0 0 0 0 0 0 2 0 0 1.71 0.86 0 0 0 0 3 0 0 0.86 1.71 0 0 0 0 4 0 0 0 0 1 0.5 0 0 5 0 0 0 0 0.5 1 0 0 6 0 0 0 0 0 0 1.33 0.67 7 0 0 0 0 0 0 0.67 1.33 8 M1 M2 M3 M4 M5 M6 M7 M8

EI

Matriks Kekakuan Struktur (Global Stiffness Matrix) [K] = [A]T [S] [A]0 1 1 0 0 0 0 1 1 0.5 0 0 0.5 1 0 0 0 0 1.71 0.86 0 0 0.86 1.71 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 1 0 1

x

x


x0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0.5 0 0 0.5 1 0 0 0 0 1.33 0.67 0 0 0.67 1.33

x

Submatrix [K] = [A]T [S]0.5 0 1 0 1.71 0.86 0.86 1.71 0 1 0 0.5 0 1.33 0 0.67

[K] = [A]T [S] [A]2.71 0.86 0.86 4.05

EI

Matriks Kekakuan Invers [K]-1 yang berukuran ordo 2 dapat ditentukan lagi dengan mencari determinannnya saja. Metoda DETERMINAN 2.71 0.86 0.86 4.05Rumus :

[ A]

1

=

1 adjo det A

Aint =

1+ A

[ ]

= (2.5143*2.5143 - 0.8571*0.8571) = 10.25 [A]-1 = 0.1 2.71 -0.86 -0.86 4.05 = 0.265 -0.084 -0.084 0.395

[K]-10.395 -0.084 -0.084 0.265

1/EI

Matriks Deformasi [D] = [K]-1 [Q]42.39 1/EI -14.18 = 0.39 -0.08 -0.08 0.26 x 102.917 -21.042Page 10 - Stiffness Matrix Analysis (c) Y. Soleman - 2008

Matriks Gaya Internal [H] = [S] [A] [D]21.2 42.39 60.52 12.04 -14.18 -7.09 -18.9 -9.45 1 0.5 0 0 0 0 0 0 0.5 1 0 0 0 0 0 0 0 0 1.71 0.86 0 0 0 0 0 0 0.86 1.71 0 0 0 0 0 0 0 0 1 0.5 0 0 0 0 0 0 0.5 1 0 0 0 0 0 0 0 0 1.33 0.67 0 0 0 0 0 0 0.67 1.33 0 1 1 0 0 0 0 0 0 0 0 1 1 0 1 0 x 42.39 -14.18

=

x

Submatrix [H] = [S] [A]0.5 1 1.71 0.86 0 0 0 0 0 0 0.86 1.71 1 0.5 1.33 0.67


Momen Ujung Elemen = FEMij - [H] = -[H] + FEMij MAB = -21.197 -21.2 0.000kNm


MBA MBC MCB MCD MDC MCE MEC MBF

= = = = = = = =

-42.393 82.393 -154.952 31.051 -43.537 123.901-95.549 -40.000

=

-42.39 -60.52 -12.04 14.18 7.09 18.9 9.450

+

0.000 142.917 -142.917 16.875 -50.625 105.000

kNm kNm kNm kNm kNm kNm

-105.000 kNm -40.000 kNm

Plot Skema Momen-momen Ujung Elemen Frame


METODA CROSSJOIN ELEMEN Kekakuan Relatif, EI DISTRIBUTION FACT FIXED-END MOMENT Balancing CARRY-OVER M. FEM Adjusting Bal Siklus 1 Com Bal Siklus 2 Com Bal Siklus 3 Com Bal Siklus 4 Com Bal Siklus 5 Com Bal Siklus 6 Com Bal Siklus 7 Com Bal Siklus 8 Com Bal Siklus 9 Com 34.516 -2.294 -0.085 -0.043 -0.254 -0.127 -0.005 -0.002 -0.014 -0.007 0.000 0.000 -0.001 0.000 0.000 0.000 0.000 0.000 -34.516 -4.588 46.413 -7.310 0.222 -0.136 0.660 -0.405 0.012 -0.008 0.037 -0.022 0.001 0.000 0.002 -0.001 0.000 0.000 0.000 0.000 0.000 -46.413 0.443 -3.655 1.319 -0.068 0.025 -0.203 0.073 -0.004 0.001 -0.011 0.004 0.000 0.000 -0.001 0.000 0.000 0.000 0.000 22.59 45.185 0.785 2.335 0.044 0.129 0.002 0.007 0.000 0.000 0.000 0.392 -0.392 1.168 -1.168 0.022 -0.022 0.065 -0.065 0.001 -0.001 0.004 -0.004 0.000 0.000 0.000 0.000 0.000

AAB0.85

BBA0.85

CBC1.36

DCD2.40

CB1.36

DC2.40

034.52

0.386-34.52

0.61446.41

0.361-46.41

0.63922.59

1.000-45.19 45.19

FINAL MOMENTSSelisih Momen Ujung -

32.04

-39.46

39.46

-48.49

48.49-

0.000

0.0000

0.0000



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Penyelesaian Soal UAS Mekanika Rekayasa IV - 7 Juli 2011 - Yoppy Soleman