Tugas Analisis Struktur V (MEKTEK) Sloop Deflection dan Rangka Batang Metode Matrix

SOAL: SISTEM PORTAL 2 DIMENSI QUIZ

Diketahui :A : 0.1 cm²E : 200000 kg/cm² = 200 t/cm²

I: 500 cm

α : 45 °L1 : (x+y) = 10 mL2 : (z) = 8 mL3 : (x) = 1 mq : (y+z) = 17 t/mP : (x+z) = 9 ton

Tentukan Kekakuan elemen Metode Matrix !

Penyelesaian :

A. ELEMEN AB(1)

AEL 0 0

−AEL 0 0

012EIL3

6EIL2 0

−12EIL3

6EIL2

06EIL2

4EIL 0

−6EIL2

EIL

[KL1 ]

=−AEL 0 0

AEL 0 0

0−12EIL3

−6EIL2 0

12EIL3

−6EIL2

06EIL2

2EIL 0

−6EIL2

4EIL

0,1.2001000 0 0

−0,1.2001000 0 0

012.200.500

100036.200.500

10002 0−12.200.500

100036.200.500

10002

i = B

j = A

α = 270

06.200.500

100024.200.500

1000 0−6.200.500

10002200.5001000

[KL1 ]

=−0,1.200

1000 0 00,1.2001000 0 0

0−12.200.500

10003−6.200.500

10002 012.200.500

10003−6.200.500

10002

06.2000.500

100022.200.500

1000 0−6.200.500

100024.2000.500

1000Cos270

Sin270 0 0 0 0

-Sin270

Cos270 0 0 0 0

0 0 1 0 0 0[ T ]= 0 0 0 Cos 270 Sin 270 0

0 0 0-

Sin270 Cos 270 0

0 0 0 0 0 1

0 1 0 0 0 0-1 0 0 0 0 00 0 1 0 0 0

[ T ]T= 0 0 0 0 1 0

0 0 0 -1 0 00 0 0 0 0 1

[Kg1 ] = [ T ] T . [KL1 ] . [ T ]

[Kg1 ] =

[Kg1 ] =

B. ELEMEN

BC (2)

AEL 0 0

−AEL 0 0

012EIL3

6EIL2 0

−12EIL3

6EIL2

06EIL2

4EIL 0

−6EIL2

EIL

i = B

j = c

α = 0

0-1 0 0 0 0 20 0 0

-20 0 0

1 0 0 0 0 0 00.001 3 0

-0.001 3

0 0 1 0 0 0 x 0 3400 0 -3

100

0 0 0 0 -1 0 -20 0 0 20 0 0

0 0 0 1 0 0 0

-0.001 -3 0

0.001 -3

0 0 0 0 0 1 0 3200 0 -3

400

0 1 0 0 0 0-1 0 0 0 0 0

x 0 0 1 0 0 00 0 0 0 1 00 0 0 -1 0 00 0 0 0 0 1

UB VB B UA VA A0 0 -1 0 0 -1 UB0 20 0 0 -20 0 VB-1 0 400 1 0 100 B0 0 1 0 0 1 UA0 -20 0 0 20 0 VA-1 0 200 1 0 400 A

[KL2 ]

=−AEL 0 0

AEL 0 0

0−12EIL3

−6EIL2 0

12EIL3

−6EIL2

06EIL2

2EIL 0

−6EIL2

4EIL

0,1.200800 0 0

−0,1.200800 0 0

012.200.500

80036.200.500

8002 0−12.200.500

80036.200.500

8002

06.2000.500

80024.200.500

800 0−6.200.500

8002200.500

800[KL

2 ]=

−0,1.200800 0 0

0,1.200800 0 0

0−12.200.500

8003−6.200.500

8002 012.200.500

8003−6.200.500

8002

06.200.500

80022.200.500

800 0−6.200.500

80024.200.500

800

Cos 0 Sin 0 0 0 0 0-Sin0 Cos 0 0 0 0 00 0 1 0 0 0

[ T ]= 0 0 0 Cos 0 Sin 0 0

0 0 0- Sin0 Cos 0 0

0 0 0 0 0 1

1 0 0 0 0 00 1 0 0 0 00 0 1 0 0 0

[ T ]T= 0 0 0 1 0 0

0 0 0 0 1 00 0 0 0 0 1

[Kg2 ] = [ T ] T . [KL2 ] . [ T ]

[Kg2 ] =

X

[Kg2 ]=

1 0 0 0 0 0 25 0 0 -25 0 0

0 1 0 0 0 0 00.002 1 0

-0.002 1

0 0 1 0 0 0 x 0 1 500 0 -1 1250 0 0 1 0 0 -25 0 0 25 0 0

0 0 0 0 1 0 0

-0.002 -1 0

0.002 -1

0 0 0 0 0 1 0 1 250 0 -1 5001 0 0 0 0 00 1 0 0 0 00 0 1 0 0 00 0 0 1 0 00 0 0 0 1 00 0 0 0 0 1

UB VB B UC VC C25 0 0 -25 0 0 UB0 0 1 0 0 1 VB0 1 500 0 -1 125 B-25 0 0 25 0 0 UC0 0 -1 0 0 -1 VC0 1 250 0 -1 500 C

C. ELEMEN CD (3)

sinα=COCD

sin45=9CD

CD=9

sin45=12,728m

AEL 0 0

−AEL 0 0

012EIL3

6EIL2 0

−12EIL3

6EIL2

06EIL2

4EIL 0

−6EIL2

EIL

[KL3 ]

=−AEL 0 0

AEL 0 0

0−12EIL3

−6EIL2 0

12EIL3

−6EIL2

06EIL2

2EIL 0

−6EIL2

4EIL

i = C

j = D

α = 315

0,1.2001272,8 0 0

−0,1.2001272,8 0 0

012.200.5001272,83

6.200.5001272,82 0

−12.200.5001272,83

6.200.5001272,82

06.2000.5001272,82

4.200.5001272,8 0

−6.200.5001272,82

200.5001272,8

[KL2 ]

=−0,1.2001272,8 0 0

0,1.2001272,8 0 0

0−12.200.500

1272,83−6.200.500

1272,82 012.200.5001272,83

−6.200.5001272,82

06.200.5001272,82

2.200.5001272,8 0

−6.2000.5001272,82

4.200.5001272,8

[ T ] T=

Cos315

Sin315 0 0 0 0

-Sin315

Cos315 0 0 0 0

0 0 1 0 0 0[ T ]= 0 0 0 Cos 315 Sin 315 0

0 0 0- Sin315 Cos 315 0

0 0 0 0 0 10.707107

0.707107 0 0 0 0

-0.70710

70.70710

7 0 0 0 00 0 1 0 0 0

0 0 0 0.7071070.70710

7 0

0 0 0-

0.7071070.70710

7 00 0 0 0 0 1

[Kg3 ] = [ T ] T . [KL3 ] . [ T ]

[Kg3 ]=

16 0 0 -16 0 00.7071

0.7071 0 0 0 0

00.001 0 0

-0.001 0

-0.7071

0.7071 0 0 0 0

0 0 314 0 0 79 x 0 0 1 0 0 0

-16 0 0 16 0 0 0 0 00.707

10.7071 0

0

-0.001 0 0

0.001 0 0 0 0

-0.707

10.7071 0

0 0 157 0 0 314 0 0 0 0 0 1

[Kg3 ]=

D. GLOBAL

0.707-

0.707 0 0 0 00.707 0.707 0 0 0 00 0 1 0 0 0

0 0 0 0.707-

0.707 00 0 0 0.707 0.707 00 0 0 0 0 1

x

UC VC C UD VD D8 8 0 -8 -8 0 UC8 8 0 -8 -8 0 VC0 0 314 0 0 79 c-8 -8 0 8 8 0 UD-8 -8 0 8 8 0 VD0 0 157 0 0 314 D

UA VA ѲA UB VB ѲB UC VC ѲC UD VD ѲD0 0 1 0 0 1 0 0 0 0 0 0 UA0 20 0 0 -20 0 0 0 0 0 0 0 VA1 0 400 -1 0 200 0 0 0 0 0 0 ѲA0 0 -1 25 0 -1 -25 0 0 0 0 0 UB0 -20 0 0 20 1 0 0 1 0 0 0 VB1 0 100 -1 1 900 0 -1 125 0 0 0 ѲB0 0 0 -25 0 0 33 8 0 -8 -8 0 UC0 0 0 0 0 -1 8 8 -1 -8 -8 0 VC0 0 0 0 1 250 0 -1 814 0 0 79 ѲC0 0 0 0 0 0 -8 -8 0 8 8 0 UD0 0 0 0 0 0 -8 -8 0 8 8 0 VD0 0 0 0 0 0 0 0 157 0 0 314 ѲD

KONDISI BATAS

[F ] = [ K ]. [U ]

{FA

GA

MA

FB

GBMBFCGC

MC

FD

GD

MD

} {UA

VA

θA

UB

VBθBUCVC

θC

UD

VD

θD

}RE-ARRANGEMENT

UB VB ѲB UC VC ѲC UA VA ѲA UD VD ѲD0.9 25 0 -1 -25 0 0 0 0 -1 0 0 0 UB-68 0 20 1 0 0 1 0 -20 0 0 0 0 VB-

89.8567

-1 1 900 0 -1 125 1 0 100 0 0 0 ѲB

0 -25 0 0 33 8 0 0 0 0 -8 -8 0 UC-68 0 0 -1 8 8 -1 0 0 0 -8 -8 0 VC

90.6667 0 1 250 0 -1 814 0 0 0 0 0 79 ѲC

RHA+8.1 0 0 1 0 0 0 0 0 1 0 0 0 0

RVA 0 -20 0 0 0 0 0 20 0 0 0 0 0MA-7.29 -1 0 200 0 0 0 1 0 400 0 0 0 0RHD 0 0 0 -8 -8 0 0 0 0 8 8 0 0

RVD 0 0 0 -8 -8 0 0 0 0 8 8 0 0MD 0 0 0 0 0 157 0 0 0 0 0 314 0

REDUKSI

0.9 25 0 -1 -25 0 0 UB-68 0 20 1 0 0 1 VB-

89.8567 -1 1 900 0 -1 125 ѲB0 -25 0 0 33 8 0 UC-68 0 0 -1 8 8 -1 VC

90.6667 0 1 250 0 -1 814 ѲC

INVERS

UB 226 0 0 226 -226 0 0.9VB 0 0 0 0 0 0 -68

ѲB 0 0 0 0 0 0-

89.857UC 226 0 0 226 -226 0 0VC -226 0 0 -226 226 0 -68ѲC 0 0 0 0 0 0 90.667

RHA+8.1 0 0 1 0 0 0RVA 0 -20 0 0 0 0MA-7.29 -1 0 200 0 0 0

RHD 0 0 0 -8 -8 0RVD 0 0 0 -8 -8 0MD 0 0 0 0 0 157

=

RAH = -21,69 + 8,1 = -13,59

RAV = 93,85

MA = -10340,12- 7,29 = -10347,41

RHD = 20,79

RVD = 42.15

MD = -9122,90

CROSS CHECK

ƩV = RAV + RDV – (q.l)

= 93,85+ 42.15– (17 . 8)

= 0

ƩH = RAH +RDH + P

= -13,59+20,79+9

= 0

-21.6993.85

-10340.1

220.7942.15

-9122.90

SOAL: SISTEM RANGKA BATANG QUIZ

Nama : Viorenza Everlyn

Npm : 123 110 198

P1 = 10 TONP2 = 5 TONP3 = 2 TONL = 10 mα = 60Tentukan Kekakuan elemen Metode Matrix !

Penyelesain:

ELEMEN AB (1) = ELEMEN CD (5)

[KL1] = AE10 [ 1 0 −1 00 0 0 0

−1 0 1 00 0 0 0]

[T1] =

[ cos60 sin60 0 0−sin60 cos60 0 0

0 0 cos60 sin600 0 −sin60 cos60]

ELEMEN AB (1)

[KG1] = [T1]T . [KL1] . [T1]

0.5-0.866 0 0 1 0 -1 0 0.5

0.866 0 0

0.866 0.5 0 0 AE 0 0 0 0 X -0.866 0.5 0 00 0 0.5 -0.866 10 -1 0 1 0 0 0 0.5 0.866

0 00.866 0.5 0 0 0 0 0 0 -0.866 0.5

UA VA UB VB0.25 0.433 -0.25 -0.433 UA0.433 0.75 -0.433 -0.75 VA-0.25 -0.433 0.25 0.433 UB-0.433 -0.75 0.433 0.75 VB

0.5 0.866025 0 0-0.866025 0.5 0 0

0 0 0.5 0.866025

0 0-

0.86603 0.5

ELEMEN CD (5)

[KG5] = [T5]T . [KL5] . [T5]

0.5-0.866 0 0 1 0 -1 0 0.5

0.866 0 0

0.866 0.5 0 0 AE 0 0 0 0 X -0.866 0.5 0 00 0 0.5 -0.866 10 -1 0 1 0 0 0 0.5 0.866

0 00.866 0.5 0 0 0 0 0 0 -0.866 0.5

UC VC UD VD0.25 0.433 -0.25 -0.433 UC0.433 0.75 -0.433 -0.75 VC-0.25 -0.433 0.25 0.433 UD-0.433 -0.75 0.433 0.75 VD

ELEMEN AC(2) = ELEMEN BD(4) = ELEMEN CE (6)

i = A ,

B , C

j = C ,

D , E

[KL2] = AE10 [ 1 0 −1 00 0 0 0

−1 0 1 00 0 0 0]

[T2] = [ cos0 sin0 0 0−sin0 cos0 0 0

0 0 cos0 sin00 0 −sin0 cos0]=¿

Jadi [KL2] =

Jadi [KL4]=

Jadi [KL6] =

ELEMEN BC (3) = ELEMEN DE (7)

i = C , E

j = B , D

α = 300

1 0 0 00 1 0 00 0 1 00 0 0 1

UA VA UC VC1 0 -1 0 UA0 0 0 0 VA-1 0 1 0 UC0 0 0 0 VC

UB VB UD VD1 0 -1 0 UB0 0 0 0 VB-1 0 1 0 UD0 0 0 0 VD

UC VC UE VE1 0 -1 0 UC0 0 0 0 VC-1 0 1 0 UE0 0 0 0 VE

[KL2] = AE10 [ 1 0 −1 00 0 0 0

−1 0 1 00 0 0 0]

[T2] =

ELEMEN BC (3)

[KG3] = [T3]T . [KL3] . [T3]

0.50.866 0 0 1 0 -1 0 0.5

-0.866 0 0

-0.8660.5 0 0

AE 0 0 0 0 X

0.866 0.5 0 0

0 0 0.50.866

10 -1 0 1 0 0 0 0.5 -0.866

0 0-

0.866 0.5 0 0 0 0 0 00.866 0.5

UC VC UB VB0.25 -0.433 -0.25 0.433 UC-0.433 0.75 0.433 -0.75 VC-0.25 0.433 0.25 -0.433 UB0.433 -0.75 -0.433 0.75 VB

ELEMEN DE (7)

0.5 -0.866 0 00.866 0.5 0 00 0 0.5 -0.8660 0 0.866 0.5

[KG7] = [T7]T . [KL7] . [T7]

0.50.866 0 0 1 0 -1 0 0.5

-0.866 0 0

-0.8660.5 0 0

AE 0 0 0 0 X

0.866 0.5 0 0

0 0 0.50.866

10 -1 0 1 0 0 0 0.5 -0.866

0 0-

0.866 0.5 0 0 0 0 0 00.866 0.5

UE VE UD VD0.25 -0.433 -0.25 0.433 UE-0.433 0.75 0.433 -0.75 VE-0.25 0.433 0.25 -0.433 UD0.433 -0.75 -0.433 0.75 VD

GLOBAL

UA VA UB VB UC VC UD VD UE VE1.25 0.43 -0.25 -0.43 -1 0 0 0 0 0 UA0.43 0.75 -0.43 -0.75 0 0 0 0 0 0 VA-0.25 -0.43 1.50 0.00 -0.25 0.43 -1 0 0 0 UB-0.43 -0.75 0.00 1.50 0.43 -0.75 0 0 0 0 VB-1 0 -0.25 0.43 2.50 0.00 -0.25 -0.43 -1 0 UC0 0 0.43 -0.75 0.00 1.50 -0.43 -0.75 0 0 VC0 0 -1 0 -0.25 -0.43 1.50 0.00 -0.25 0.43 UD0 0 0 0 -0.43 -0.75 0.00 1.50 0.43 -0.75 VD0 0 0 0 -1 0 -0.25 0.43 0.25 -0.43 UE0 0 0 0 0 0 0.43 -0.75 -0.43 0.75 VE

KONDISI BATAS

[F ] = [ K ]. [U ]

{FAGAFBGBFCGCFDGDFEGE

} {RHARVA

P=10ton00

P=−2ton0

P=5ton0RVE

} {UAVAUB

VB

UC

VC

UD

VD

UEFE

}RE-ARRANGEMENT

UB VB UC VC UD VD UE UA VA VE10 1.50 0 -0.25 0.43 -1 0 0 -0.25 -0.43 0 UB0 0 1.50 0.43 -0.75 0 0 0 -0.43 -0.75 0 VB0 -0.25 0.43 2.50 0 -0.25 -0.43 -1 -1 0 0 UC-2 0.43 -0.75 0 1.50 -0.43 -0.75 0 0 0 0 VC0 AE -1.00 0 -0.25 -0.43 1.50 0 -0.3 0 0 0.43 UD5 10 0 0 -0.43 -0.75 0 1.50 0.4 0 0 -0.75 VD0 0 0 -1 0 -0.25 0.43 0.25 0 0 -0.43 UE

RHA -0.25 -0.43 -1 0 0 0 0 1.25 0.43 0 UARVA -0.43 -0.75 0 0 0 0 0 0.43 0.75 0 VARVE 0 0 0 0 0 -1 0 0 0 0.75 VE

REDUKSI MATRIX

10 1.50 0 -0.25 0.43 -1 0 0 UB0 0 1.50 0.43 -0.75 0 0 0 VB

0 AE-

0.25 0.43 2.50 0 -0.25 -0.43 -1 UC-2 10 0.43 -0.75 0 1.50 -0.43 -0.75 0 VC0 -1 0 -0.25 -0.43 1.50 0 -0.25 UD5 0 0 -0.43 -0.75 0 1.50 0.43 VD0 0 0 -1 0 -0.25 0.43 0.25 UE

INVERS

UB 1.875 -0.505 0.750 -0.577 1.375 -0.361 1 10VB -0.505 1.292 -0.433 1 -0.217 0.542 -0.577 0

UC10 0.750 -0.433 1 -0.289 0.750 -0.144 1 0

VCAE -0.577 1 -0.289 1.833 0 1 -0.577 -2

UD 1.375 -0.217 0.750 0 1.875 -0.072 1 0VD -0.361 0.542 -0.144 1 -0.072 1.292 -0.577 5UE 1 -0.577 1 -0.577 1 -0.577 2 0

UA -0.25-

0.43 -1 0 0 00

VA AE -0.43-

0.75 0 0 0 0 0

VE10

0 0 0 0 0.43-

0.75-

0.43

-10

=-

2.080199.08018

9

CROSS CHECK

ΣV= 0 ΣH= 0

RVA + RVE –P2 – P3 = 0 RHA + P1 = 0

-2.08019 + 9.080189 – 5 – 2= 0 -10 + 10 = 0

0 = 0 OK! 0 = 0 OK!

GAYA BATANG

ELEMEN 1

i = A

j = B

{FAGAFBGB} = [T1] [KG1] {UAVAUBVB }

{FAGAFBGB} = [ 0,5 0,87 0 0−0,87 0,5 0 0

0 0 0,5 0,870 0 −0,87 0,5 ] AE5

[ 0,25 0,435 −0,25 −0,4350,435 0,757 −0,435 −0,757−0,25 −0,435 0,25 0,435−0,435 −0,757 0,435 0,757 ] 5

AE { 00

21,709−9,760

}¿

ELEMEN 2

i = A

j = C

{FAGAFCGC} = [T2] [KG2] {UAVAUCVC }{FAGAFCGC} = [1 0 0 0

0 1 0 00 0 1 00 0 0 1 ] AE5 [ 1 0 −1 0

0 0 0 0−1 0 1 00 0 0 0] 5AE { 0

08,799

−14,440}=

ELEMEN 3

i = C

j = B

-2.400

2.400

{FCGCFBGB} = [T3] [KG3] {UCVCUBVB}{FBGBFCGC} = [ 0,5 −0,87 0 0

0,87 0,5 0 00 0 0,5 −0,870 0 0,87 0,5 ] AE5

[ 0,25 −0,435 −0,25 0,435−0,435 0,757 0,435 −0,757−0,25 0,435 0,25 −0,4350,435 −0,757 −0,435 0,757 ] 5AE { 8,799

−14,44021.709−9.760

}=

ELEMEN 4

i = B

j = D

2.400

-2.400

{FBGBFDGD} = [T4] [KG4] {UBVBUDVD}{FBGBFDGD} = [1 0 0 0

0 1 0 00 0 1 00 0 0 1 ] AE5 [ 1 0 −1 0

0 0 0 0−1 0 1 00 0 0 0] 5AE { 21.709−9.760

14.111−12.067

} = ELEMEN 5

i = C

j = D

{FCGCFDGD} = [T5] [KG5] {UCVCUDVD}{FCGCFDGD} = [ 0,5 0,87 0 0

−0,87 0,5 0 00 0 0,5 0,870 0 −0,87 0,5 ] AE

5

[ 0,25 0,435 −0,25 −0,4350,435 0,757 −0,435 −0,757−0,25 −0,435 0,25 0,435−0,435 −0,757 0,435 0,757 ] 5

AE

=-4.710

4.710

ELEMEN 6

i = C

j = E

{FCGCFE¿

} = [T6] [KG6] {UCVCUEVE}{FCGCFE

¿} = [1 0 0 0

0 1 0 00 0 1 00 0 0 1 ] AE5 [ 1 0 −1 0

0 0 0 0−1 0 1 00 0 0 0] 5AE =

ELEMEN 7

i = D

j = E

{FDGDFE¿

} = [T7] [KG7] {UDVDUEVE}

-5.240

5.240

{FDGDFE¿

} = [ 0,5 −0,87 0 00,87 0,5 0 00 0 0,5 −0,870 0 0,87 0,5 ] AE5

[ 0,25 −0,435 −0,25 0,435−0,435 0,757 0,435 −0,757−0,25 0,435 0,25 −0,4350,435 −0,757 −0,435 0,757 ] 5AE

= 18.350

-18.350