Contoh Soal Matriks (Update 25-01-15) No 1 s.d No. 4

7/23/2019 Contoh Soal Matriks (Update 25-01-15) No 1 s.d No. 4

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LANGKAH-LANGKAH GAMBAR BALOK

1. Gambar balok ebenarnya.

!. "atrik A# $$$$$$$$$$% &eban dihilangkan' di(lacement )#*+, dikekang

3. "atrik A#- $$$$$$$$$$% &eban dimakkan kecali yang elara dengan #*+' di(lacement dikekang

/. "atrik AR- $$$$$$$$$$% Reaki (ada trktr' di(lacement )#*+, dikekang

0. +reebody o. /

#*+ 2 #egree *f +reedom )derajat kebebaan, $$% ejah mana at joint da(at bergerakberdi(lacement

#i(lacement(er(indahan terjadi (ada joint

CONTOH SOAL 1 (Hal. 100 Buku Willian Weaver)

#*+ yg bia terjadi (ada balok ini adalah #1 dan #!' karena (erlatakannya endi

Catatan utk AL ! Be"an M #i titik B

t#k #i$a%ukkan krn

%elara% O&

Displacement/perpindahan :

Translasi (perpindahan dalam arahgaris lurus)

Rotasi (perpindahan dalam putaransudut)

A B C

2P P PM = P.L

L/2 L/2 L/2 L/2

(1).

A B C

D1

L L

(2).

D2

ADL2

L/2 L/2L/2 L/2

A B C

2P(3).

P

ADL1

P

A B C

2P(4).

P

ARL3

L/2 L/2L/2 L/2

P

ARL1

ARL2

ARL4

ADd1=[AD1AD

2]=[M0]=[PL0]

ADLd1=

[ADL

1

ADL2]=

PL4

+PL

8

PL

8

=

PL

8

PL

8

#1 2 " karena ada aki beban lar ygelara #*+ ' aki bia ber(a gaya atamomen

#! 2 4 karena aki beban lar tidak ada ygelara #*+' aki bia ber(a gaya atamomen


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(5).

ARL2

A B

2P

ARL3ARL

1

PL/4

C

PP

ARL4ARL3

B

PL/8 PL/8


3/38


'SLACMNT SS*A' O& (ROTAS' ATA* TRANSLAS' AA T'T'K TRTNT*)

L

EIS

LEI

LEI

LEIS

2

844

21

11

=

=+=

PL/4

A B

2P

PP

PL/4

C

PP

P/2P/2

B

PL/8 PL/8

EI

S11

5=1

5=1

1 2

Ard11

Ard21

Ard31

Ard41

S21

EI

L L

5=1

1 2

Ard41

S21

EI

L

Ard31

S11

EI

S11

5=1

Ard11

Ard21

Ard31

L

1

EI

S22

5=1

1 2

Ard32

Ard42

EI

S12

ARL r1=[ARL

1

ARL2

ARL 3

ARL4

]=[ P

PL

4

P+ P2

P

2

]=[ P

PL

4

3

2. P

P2

]

Ard11=6EI

L2

Ard21=

2EI

L

Ard31=

6EI

L2 +

6EI

L2 =0

Ard41=

6EI

L2


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MATR'KS KKAK*AN (S)

S12=

2EI

L

S22=

4EI

L

Ard12=0

Ard22=0

Ard32=

6EI

L2

Ard42=

6EI

L2

Sdd=[S11 S12S

21 S

22]=[

8EI

L

2EI

L

2EI

L

4EI

L]=EIL[ 8 22 4]

S1=1

EI

L (324 )

.[ 4 22 8 ]=

L

EI. 28. [

4 22 8 ]

S1=

L

28EI. 2[ 2 11 4 ]=L14EI .[ 2 11 4 ]

D=S1 .(ADADL)

D=L

14EI.[ 2 11 4 ] [PL0][PL/8PL/8 ]

D=L

14EI.[ 2 11 4 ][9/8 .PLPL/8 ]=L14EI .[ 18PL/8PL/89PL/8+4PL/8] D= L14EI .[17PL/85PL/8]= L14EI .PL8 .[175 ]= PL

2

112EI.[175]

ARDrd=

[Ard

11 Ard

12

Ard21

Ard22

Ard31

Ard32

Ard41

Ard42 ]=

[ 6EI/L

20

2EI/L 0

0 6EI/L2

6EI/L2

6EI/L2]=EI

L2

.

[ 6 0

2L 0

0 6

6 6 ]

ARr1=ARLr1+ARDrdDd1

AR=[ P

PL/4

3 P/2

P/2]+EIL2 .[

6 0

2L 0

0 6

6 6]PL2112EI .[ 175 ]

AR=[ P

PL/4

3 P/2

P/2]+P112 .[

102

34 L

30

72 ]=[

P

PL/4

3 P/2

P /2]+.[

102 P/112

34 PL/112

30 P/112

72 P/112]

AR=[ P

PL/4

3 P /2

P/2]+EIL2 .[

6 0

2L 0

0 6

6 6]PL2112EI .[ 175 ]

AR=

P

PL/4

3 P /2

+P

112.

102

34 L

30

=

P

PL/4

3 P/2

+.

102 P/112

34 PL/112

30 P/112


5/38


CK STAT'KA

A B C

2P P PM = P.L

L/2 L/2 L/2 L/2

AR1

AR2

AR3 AR4

AR=[ P

PL/4

3 P/2

P/2]+EIL2 .[

6 0

2L 0

0 6

6 6]PL2112EI .[ 175 ]

AR=[ P

PL/4

3 P/2

P/2]+P112 .[

102

34 L

30

72 ]=[

P

PL/4

3 P/2

P /2]+.[

102 P/112

34 PL/112

30 P/112

72 P/112]

AR=[112P+102 P

112

28 PL+34 PL

112

168 P30P

112

56 P72 P

112

]=[214 P

112

62PL

112

138 P

112

128P

112

]=P112 [ 21462 L138128]AR=P

1122

[ 107

31 L69

64]=P

56

[ 107

31L69

64]

AR=[ P

PL/4

3 P /2

P/2]+EIL2 .[

6 0

2L 0

0 6

6 6]PL2112EI .[ 175 ]

AR=[ P

PL/4

3 P /2

P/2]+P112 .[

102

34 L

30

72]=[

P

PL/4

3 P/2

P /2]+.[

102 P/112

34 PL/112

30 P/112

72 P/112]

AR=

[

112P+102 P

112

28 PL+34 PL

112

168 P30P

112

56 P72 P

112

]=

[

214 P

112

62PL

112

138 P

112

128P

112

]=P

112 [ 214

62 L

138

128]

AR=P

1122 [

107

31 L

69

64]=P56 [

107

31L

69

64]

V=0107P

56 +

69P

56

64P

56 2PP+P=0

112P

56

2P=0

2P2P=00=0 (OK )

MA=031PL56

PL+2P .L

2

69P

56.L+P .

3L

2 +

64P

56. 2LP . 2L=0

31PL

56

69PL

56 +

128PL

56 +

3

2PL2PL=0

28PL56

12PL=0

0=0 (OK )

MB=0

31PL56

PL+107P56

.L2P .L2+P .L

2P .L+64P

56.L=0

31PL

56 +

107PL

56 +

64PL

56 PLPL+

PL

2 PL=0

140PL

56 2. 5PL=0

MC=0

31PL

56PL+107

P

56.2L2P. 3

L

2+69

P

56.LP.

L

2=0

31PL

56+

214PL

56+

69PL

56PL3PL

PL

2=0

252PL

564.5PL=0


6/38


Selan+utn,a Ga,a ala$ (M$en Lintan/ #an Nr$al) #aat #iitun/

R'NGKASAN

Matriks ADdx1

Matriks ADLdx1

Matriks ARLrx1

Matriks Sdxd

Matriks ARDrxd

Matriks Ddx1

= S-1dxd

.[ ADdx1

- ADLdx1

]

Matriks ARrx1

= ARLrx1

+ ARDrxd

.Ddx1

56 56 . .2 .2 . 56 .

31PL

56 +

107PL

56 +

64PL

56 PLPL+

PL

2 PL=0

140PL

56 2. 5PL=0

2. 5PL2 .5PL=00=0 (OK)

56 56 . . 2 56 . . 2 =

31PL

56+

214PL

56+

69PL

56PL3PL

PL

2=0

252PL

564.5PL=0

4 .5PL4 .5PL=00=0 (OK)


7/38







0. +reebody o. /




#*+ yg bia terjadi (ada balok ini adalah #1 dan #!' karena (erlatakannya endi

Catatan utk AL !

Be"an M #i titik C

t#k #i$a%ukkan krn

%elara% O&




A B C

PP

M = P.L

L 1.5L L

(1).

D

W =P/L

L/2

A B C

D1

(2).

D2

D

L 1.5L L

ADL2ADL1

A B C

(3).

D

L 1.5L L

PP W1 2

1 2

ARLP

PW

1 2ARL

AD=[AD1AD2]=[ 0M]=[ 0PL ]

ADL=

[

ADL1

ADL2

]=

[

PL8+ 3PL16

3 PL

16 +

PL

12

]=

[

2PL+3 PL16

9PL+4PL48

]=

[

PL

16

5PL48

]

#1 2 4 karena aki beban lar tidakada yang elara #*+' aki bia ber(agaya ata momen

#! 2 " karena ada aki beban lar yg elara#*+ ' aki bia ber(a gaya ata momen


8/38


*ntuk $en/itun/ ARL reak%i erletakan #aat #iitun/ %e%uai 4ree"#, "erikut !

ARL3

ARL1

ARL4

A B C

(4).

D

L 1.5L L

ARL5


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ari /a$"ar #i ata% $aka #aat #iitun/ Reak%i erletakkan.

(5).

ARL2

A B

P

ARL3

ARL1

PL/8

C

P

ARL4ARL

3

B

3PL/16 3PL/16

W

D

ARL5ARL4

C

PL/12 ARL6

W

Poii P bia di free$body A$&ata free$body &$C

C

P

ARL4ARL

3

B

3PL/16 3PL/16

W

1.5L

ARL2

A B

P

ARL3

ARL1

PL/8

L

MC=0MB+MC+ARL

31 . 5LP1 . 5LW1 . 5L1 .5L/2=0 ... Dimana MB=MC dari Tabel Momen Primer, dan =P!"

ARL332 L32 PLP

L 32 L32L

2=0

ARL3

3

2LPL(32+98)=0

ARL3

3

2LPL(12+98 )=0

MB=0

ARL1LP

L

2ARL

2+PL/8=0 .. .dimana #$"

2=PL/8 dari %abel Momen Primer

ARL1LP.

L

2PL/8+PL/8=0

ARL1=P .L

2 1

L

ARL1=P/2

V=0ARL

1+ARL

2P=0

ARL2=PARL

1

ARL2=PP/2

ARL2=P/2

V=0ARL

3+ARL

4PW1.5L=0

ARL =P+P

3L

7PL


10/38


(5). PL/8

A B

P

P/2P/2

PL/8

C

P

3P/47P/4

B

3PL/16 3PL/16

W

D

P/2P/2

C

PL/12 PL/12

W

D

ARL5ARL4

C

PL/12 ARL6

W

L

ARL3

ARL1

ARL2

ARL4

A B C

(4).

D

L 1.5L L

P

PW

1 2

ARL5

ARL6

ARL33

2LPL(32+98)=0

ARL3

3

2LPL(12+98 )=0

ARL3

3

2L

21

8 PL=0

ARL3=

21

8 PL

2

3L

ARL3=7P

4

V=0ARL

3+ARL

4PW1.5L=0

ARL4=P+

P

L

3

2L

7PL

4

ARL4=P+

3P

2

7P

4

ARL4=P(4+674

)=3P

4

MD=0ARL

4LWLL/2PL/12+ARL

6=0 . .. dimana #$"

6=PL/12

ARL4L

P

LL

L

2PL

12+PL

12=0

ARL4L

PL

2=0

ARL4=PL

2

1

L=P

2

V=0ARL

4+ARL

5WL=0

ARL5=WLARL

4

ARL5=P

LLP/2

ARL4=PP/2=P/2


11/38



ari /a$"ar #i ata% itun/ Reak%i erletakan #ari $a%in/-$a%in/ 4ree"#,

ari eritun/an reak%i erletakan $aka #aat #iitun/ $atrik% S #an Ar# %""5

EI

S11

5=1

5=

1

1 2

Ard11

Ard21

Ard31

Ard41

S21

EI

L 1.5L L

EI

5=1

1

Ard11

Ard21

Ard31

L

S11

LC D

+R99$&*# S

11

5=1

2

Ard41

S21

EI

1.5L

Ard31

Ard51

=0Ard41

=0

S21

=0

Ard61

=0

S11=

4EI

L+

4EI

3L/2=

4EI

L+

8EI

3L=

12EI+8EI3L

=20EI

3L

S21=

2EI=

4EI

ARL=

[

ARL1

ARL2

ARL3

ARL4

ARL5

ARL6

]=[

P/2

PL /8

P/2+7P /4

3P /4+P/2

P/2

PL /12]=[

P/2

PL /8

9P /4

5P /4

P/2

PL /12]


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EI

S22

5=1Ard

32

S12

L

5=1

1 2

Ard52

Ard62

EI

1.5LArd

42L

Ard12

=0

Ard22

=0

L

A B

EI

S22

5=1Ard

32

1 2

1.5LArd

42

S12

Ard42

5=1

Ard52

Ard62

EI

L

S22+R99$&*#

S11=

4EI

L+

4EI

3L/2=

4EI

L+

8EI

3L=

12EI+8EI3L

=20EI

3L

S21=

2EI

3L/2=

4EI

3

Ard11=

6EI

L2

Ard21=2EIL

Ard31=

6EI

L2 +

6EI

( 3L/2)2=

6EI

L2 +

24EI

9L2 =

54EI+24EI

9L2

=30EI

9L2 =

10EI

3L2

Ard41=

6EI

(3L2)2=

24EI

9L2


13/38




MATR'KS KKAK*AN (S)

D=S1 .(ADADL)

S12=

2EI

3L/2=

4EI

L

S22=4EI

3L/2+4EIL

=8EI3L

+4EIL

=8EI+12EI3L

=20EI3L

Ard32=

6EI

(3L/2 )2=

24EI

9L2

Ard42=

6EI

(3L /2 )2+

6EI

L2 =

24EI

9L2 +

6EI

L2 =

24EI+54EI

9L2

=30EI

9L2

Ard52=

6EI

L2

Ard62=

2EI

L

Sdd=[S11 S12S

21 S

22]=[20EI/3L 4EI/3L4EI/3L 20EI/3L ]

S=4EI

3L[5 11 5 ]S1=3L

4EI(1251[

5 11 5 ])=

3L4EI

.124[

5 11 5 ]=

L32EI[

5 11 5 ]

ARDrd=[

Ard11

Ard12

Ard21

Ard22

Ard31

Ard32

Ard41

Ard42

Ard51

Ard52

Ard61

Ard62

]=[ 6EI/L2 0

2EI/L 010EI/3L2 24EI/9L2

24EI/9L2 10EI/3L2

0 6EI/L2

0 2EI/L2]= EI9L2 [

54 0

18L 0

30 2424 30

0 540 18L

]

D=L

32EI[ 5 11 5 ][[ 0PL][ PL/165PL/48]]D=

L

32EI[ 5 11 5 ][PL/1643PL/48 ]=L32EI[5PL/16+43PL/48PL/16215PL/ 48 ]D=L

32EI[28PL/48

212P/48]=L

32EI[7PL/12

53P /12]=PL

2

384EI[7

53]


14/38

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CK STAT'KA

AR1

AR2

AR3

AR4

A B C

PP

M = P.L

L 1.5L L

(1).

D

W =P/L

L/2

AR5

AR6

93PL/576

A B C

PP

M = P.L

(1).

D

W =P/L

L/2

207PL/576

ARr1=ARLr1+ARDrdDd1

AR=[ P/2PL/89P/45P/4P/2

PL/12]+EI9L2 [ 54 0

18L 0

30 2424 30

0 540 18L ]PL

2

384EI[ 753]

AR=

[

P/2PL/89P/45P/4

P/2PL/12

]+P

3456

[

378

126L

14821758

2862

954L

]=

[

P/2PL/89P/45P/4

P/2PL/12

]+

[

378P /3456126PL/3456

1482P/34561758P /3456

2862P/3456954PL/3456

]=

[

351P/57693PL/5761049P /576427P /576

765P /576207PL/576

]


15/38


351P/576

1049P/576 427P/576

L 1.5L L

765P/576

V=0351P

576+

1049P

576+

427P

576+

765P

5762P

P

L(3L2 +2L2)=04 .5P2P5 /2P=00=0

MA=093PL

576 +PL

+

207PL

576 +PL

/2+PL

1049P

576 .L

+

P

L .

5

2L

(5

4L

+

4

4 L

)427P

576 .

5

2L

765P

576 .

7

2L

=0

4680PL

576 +

65PL

8 =0

8.125PL+8.125PL=00=0 (OK )

MB=0351P

576

.L93PL

576

PL

2

+P

L

.5

2

L

(5

4

L

)

427P

576

.1.5L+PL765P

576

.2.5L+207PL

576

=0

2088PL

576 +

29

8 L=0

3.625PL+3.625PL=00=0 (OK)

MC=0351P

576.2,5L

93PL

576 P . 2PLP .1,5L+

1049P

576.1,5L

P

L.5

2L (32 L54 L)+PL765P576 .L+207PL576 =0

1800PL

576

25PL

8 =0

3,125PL3,125PL=00=0 (OK)

MD=0351P

576.3,5L

93PL

576 P .3LP.2,5L+

1049P

576.2,5L

P

L.5

2L .(54 L)+PL+427P576 .L+207PL576 =0

4392PL576

61PL8

=0

7,625PL7,625PL=00=0 (OK)


16/38



R'NGKASAN

Matriks ADdx1

Matriks ADLdx1

Matriks ARLrx1

Matriks Sdxd

Matriks ARDrxd

Matriks Ddx1

= S-1dxd

.[ ADdx1

- ADLdx1

]

Matriks ARrx1

= ARLrx1

+ ARDrxd

.Ddx1


17/38







0. +reebody o. /




Gided ((ort )tm(an (enntn, hanya menahan momen dan tidak menahan reaki hori;ontal )hori;ontal terhada( (erletakan,.




ARL2

ARLA

B

P1= 2P P

2= P

(3). C

A B

P1= 2P P

2= P

(3). C

L/2 L/2L/2 L/2

ADL1 ADL2

1 2

A B

P1= 2P P

2= P

(1).

C

L/2 L/2L/2 L/2gided ((ort

A B C

D1

(2).

D2

1 2

L/2 L/2L/2 L/2

karena gayahori;ontal th((erletakan tidakbia ditahan)beba,

ADd1=[AD1AD

2]=[00 ]

ADLd1=

[

ADL1

ADL2

]


18/38


*ntuk $en/itun/ ARL reak%i erletakan #aat #iitun/ %e%uai 4ree"#, "erikut !

ARL1

ARL3

L/2 L/2L/2 L/2


19/38


AR' GAMBAR ' ATAS MAKA AAT 'ROLH !


(5).ARL

2

A B

2P

ARL3

ARL1

PL/4

EI

S11

5=1

5=1

1 2

Ard11

Ard21

Ard31

S21

Ard41

EI

L L

A B

P

ARL

ARL3

PL/8

P/2

A B

2P

PP

PL/4

A B

P/2P/2

PL/8 PPL/8PL/4

A#A P=>. S=#=>R*>AS? 22%A


20/38

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SLAN8*TN9A 'S*S*N MATR'KS KKAK*AN SBB5

MATR'KS KKAK*AN (S)

EI

5=1

1

Ard11

Ard31

L

2

Ard31

L

S21

+R99$&*#

LL

EI

S22Ard

32

S12

1

2

Ard42

Ard12

=0

Ard22

=0

A#A -9#=>A>RAS-AS? 22%A "9A@A GAA

9R>?


21/38


CK STAT'KA

D=S1 .(ADADL)

ARr1=ARLr1+ARDrdDd1

Sdd=[S11 S12S

21 S

22]=[

8EI

L

6EI

L2

6EI

L2

12EI

L3]=2EIL3 [ 4L2 3L3L 6]

S1=

L3

2EI{124L29L2[6 3L3L 4L2 ]}=L3

2EI(115L2[6 3L3L 4L2 ])S1

=

L

30EI

[6 3L

3L 4L2

]

ARDrd=[

Ard11

Ard12

Ard21

Ard21

Ard31

Ard32

Ard41

Ard42

]=[6 EI

L2

0

2 EI

L 0

0 12 EI

L

2 EI

L

6 EI

L2

]= 2 EIL3 [3 L 0L2 00 6L2 3L ]

D=L

30EI[6 3L3L 4L2 ]([00 ][PL/8P/2])=L30EI[6 3L3L 4L2 ][ PL/8P/2]D=PL

240EI[6 3L

3L 4L2 ][

L4]=

PL240EI[

6L12L3L

216L2]=PL

240EI[6L13L2]

=PL2

240EI[6

13L]

ARr1=

P

8

[

8

2L

12

L

]+

2EI

L3

[

3L 0

L2

0

0 6L

2 3L

]PL

2

240EI

[

613L

]=P

8

[

8

2L

12

L

]+P

120L

[

18L6L2

78L

33L2

]AR

r1=[ P

18P

120

PL

4

6PL

120

3P

2+

78P

120

PL

8+

33PL

120

]=[120P18P120

30PL6PL120

180P+78P120

15PL+33PL120

]=[102P/12024PL/120258P/12018PL/120 ]=P20 [174L433L ]


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R'NGKASAN

Matriks ADdx1

Matriks ADLdx1

Matriks ARLrx1

Matriks Sdxd

Matriks ARDrxd

Matriks Ddx1

= S-1dxd

.[ ADdx1

- ADLdx1

]

AR1

2

AR3

AR4

A B

1 2=

C

L/2 L/2L/2 L/2

17P/20

4PL/20

43P/20

3PL/20

A B

P1= 2P P

2= P

C

L/2 L/2L/2 L/2

V=017P

20+

43P

203P=0

60

20P3P=0

3P3P=00=0 (OK)

MA=0

4PL

20+2P .

L

2

43P

20.L+P.

3

2L

3PL

20=0

50

20PL+

5

2PL=0

2,5PL+2,5PL=00=0 (OK)

MB=0

4PL

20+

17P

20.L2PL.

L

2+P .

L

2

3PL

20=0

10

20PL

PL

2=0

0,5PL0,5PL=00=0 (OK)

MC=0

4PL

20+

17PL

20.2L2P .

3

2LP .

L

2+

43P

20.L

3PL

20=0

70

20PL

7

2PL=0

3,5PL3,5PL=0

0=0 (OK)


23/38


Matriks ARrx1

= ARLrx1

+ ARDrxd

.Ddx1


24/38

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0. +reebody o. /



CONTOH SOAL 6.6-2:. (Hal. 1;; Buku Willian Weaver)

Hitun/ reak%i erletakkan #i A #an (e4r$a%i a 1? = > 0 ? ---@ Karena ti#ak a#a ak%i (/a,a luar) ,an/ %elara% #en/an O&




L=3H/2

L/2

P

L L

H

A B C

D

P

H=2L/3

L L

2L/3

A B C

D

(1). MARI!S D"#

A B C

(2). MARI!S ADL

D1

PP

L/2


25/38


L L

2L/3

D


26/38


&ree"#, untuk itun/ AL

&ree"#, untuk itun/ ARL

AL = > AL1? = > -L ?

ARL2

ARL3

Gambar AR- tidak dibat


27/38


ARL5

ARL4=0

MD=0ARL

6=0


28/38



&RBO9 SS*A' O& (ROTAS' ATA* TRANSLAS' AA T'T'K TRTNT*)


EI

5=

1

1

Ard21

Ard31

6EI/L2

S11

L

+R99$&*#

S11

5=1

EI

L

EI

S11

5=1

5=1

1

Ard21

Ard31

Ard51

EI

L L

5=1

Ard11

Ard61

Ard41

A C

D

B

EI 2L/3

Ard11

6EI/L2

Ard51

Ard61

Ard41

D

EI

5=1

2L/3

S11

3'(2L6)2

3'(2L6)2

2 4

H=0

Ard11=

6EI

(23 L)2=

27EI

2L2

V=0

Ard21=

6EI

L2

Ard31=

2EI

L (dari %abel)

V=0

Ard51=

6EI

L2

6EI

L2 =0

H=0

Ard41=

6EI

(23 L)2=

27EI

2L2

Ard61=

2EI

2

3L

=3EI

L (dari %abel)


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Ard21=

6EI

L2

Ard31=

2EI

L (dari %abel)

3

Ard61=

2EI

2

3L

=3EI

L (dari %abel)


30/38


Matrik% Kekakuan >S?

Reaki dititik C belm dihitng

S = > S11

? = ;'L ;'L ;'(2L6) = 'L 3'L = 1;'L

S-1= L1;'

D=S1

.(ADADL)

ARr1=ARLr1+ARDrdDd1

ARD=

[

Ard11

Ard21

Ard31

Ard41

Ard51

Ard61

]=

[

27 EI

2 L2

6EI

L2

2 EI

L

27EI

2 L2

0

3 EI

L

]D=

L

14EI.( [0 ][PL8])= L14EI PL8 = PL

2

112EI

AR=[ 0

P/2

PL/8

0

3 P/2

0

]+[27EI

2L2

6EI

L2

2EI

L

27 EI

2L2

0

3EI

L

] PL2112EI=[ 0P/2PL/803 P /20 ]+[27 P

224

6P

112

PL

56

27 P

224

0

3 PL

112

]

AR=[ 0+

27P

224

P

2+6P

112

PL

8+PL

56

0+27P224

3P

2

+0

0+3PL

112

]=[ 27P

224

(56+6 )P112

(7+1 )PL

56

27P224

3P

2

3PL

112

]=[ 27P

224

31P

56

PL

7

27P224

3P

2

3PL

112

]


31/38



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Catatan kecil Sondra Raharja Powered by : Android Page 3! of 38

Reaki dititik C belm dihitng

CK STAT'KA

L/2

P

L L

2L/3

A B C

D

P

31P/56

27P/224

PL/7

3P/2

27P/224

3PL/112

MC=(31P56 2L)(P

3L

2 )(PL)+(3P

2L)+(27P224

2L

3 )PL

7

3PL

112

MC=(31PL28 )PL+(

9PL

112)PL

7

3PL

112

MC=(

31PL28

28PL28

4PL28 )+(

9PL112

3PL112)

MC=

PL

28+

6PL

112

MC=4PL+6PL

112=

2PL

112=PL

56

Mc=0

(31P56 2L)(P3L2 )(PL )+(3P2 L)+(27P224 2L3 )PL7 3PL112 MC=0(31PL28 )PL+(9PL112)PL7 3PL112 PL56 =0(31PL28 28PL28 4PL28 )+(9PL112 3PL112)PL56 =0PL

28+

6PL

112PL

56=0

4PL+6PL2PL

112 =0


33/38




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Reaki di titik C belm dihitng' namn reaki di titik lain dh d(t dihitng' hal ini dikarenakan cara kekakan langng )matrik, d(t dilakkan hal

Selanjtnya reaki dititik C da(at dihitng terakhir dg cara tatika berikt ini :

@itng c dan @c :

C9< S>A>?


35/38


"_=+


36/38


C9< S>A>?


37/38

A B C

2P P PM = P.L

L/2 L/2 L/2 L/2

(1).

A B C

D1

L L

(2).

D2

ADL2

L/2 L/2L/2 L/2

A B C

2P(3).

P

ADL1

P

A B C

2P(4).

P

ARL3

L/2 L/2L/2 L/2

P

ARL1

ARL2

ARL4

(5).ARL2

A B

2P

ARL3

ARL1

PL/4

C

P

P

ARL4ARL3

B

PL/8 PL/8

PL/4

A B

2P

PP

PL/4

C

PP

P/2P/2

B

PL/8 PL/8

ADd1=[AD1AD

2]=[M0]=[PL0]

ADLd1=[ADL1ADL

2]=[

PL4

+PL

8

PL

8]=[

PL

8

PL

8]

ARLr1=[ARL

1

ARL2

ARL3

ARL4

]=[ P

PL

4

P+ P2

P

2

]=[ P

PL

4

3

2. P

P2

]


38/38

ARLr1=[ARL

1

ARL2

ARL3

ARL4

]=[ P

PL

4

P+ P2

P

2

]=[ P

PL

4

3

2. P

P2

]]02P/1124PL/1120P/1122P/112

]

Documents

Contoh Soal Matriks (Update 25-01-15) No 1 s.d No. 4