contoh soal matriks (update 25-01-15) no 1 s.d no. 4
TRANSCRIPT
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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
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'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
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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
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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)
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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 2 (Hal. 103 Buku Willian Weaver)
#*+ 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&
Displacement/perpindahan :
Translasi (perpindahan dalam arahgaris lurus)
Rotasi (perpindahan dalam putaransudut)
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
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*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
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(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 #$"
6=PL/12
ARL4L
P
LL
L
2PL
12+PL
12=0
ARL4L
PL
2=0
ARL4=PL
2
1
L=P
2
V=0ARL
4+ARL
5WL=0
ARL5=WLARL
4
ARL5=P
LLP/2
ARL4=PP/2=P/2
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'SLACMNT SS*A' O& (ROTAS' ATA* TRANSLAS' AA T'T'K TRTNT*)
ari /a$"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$&*#
S11=
4EI
L+
4EI
3L/2=
4EI
L+
8EI
3L=
12EI+8EI3L
=20EI
3L
S21=
2EI
3L/2=
4EI
3
Ard11=
6EI
L2
Ard21=2EIL
Ard31=
6EI
L2 +
6EI
( 3L/2)2=
6EI
L2 +
24EI
9L2 =
54EI+24EI
9L2
=30EI
9L2 =
10EI
3L2
Ard41=
6EI
(3L2)2=
24EI
9L2
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ari /a$"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
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]
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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
]
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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)
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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
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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 6 (Hal. 107 Buku Willian Weaver)
Gided ((ort )tm(an (enntn, hanya menahan momen dan tidak menahan reaki hori;ontal )hori;ontal terhada( (erletakan,.
Displacement/perpindahan :
Translasi (perpindahan dalam arahgaris lurus)
Rotasi (perpindahan dalam putaransudut)
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
]
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*ntuk $en/itun/ ARL reak%i erletakan #aat #iitun/ %e%uai 4ree"#, "erikut !
ARL1
ARL3
L/2 L/2L/2 L/2
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AR' GAMBAR ' ATAS MAKA AAT 'ROLH !
'SLACMNT SS*A' O& (ROTAS' ATA* TRANSLAS' AA T'T'K TRTNT*)
(5).ARL
2
A B
2P
ARL3
ARL1
PL/4
EI
S11
5=1
5=1
1 2
Ard11
Ard21
Ard31
S21
Ard41
EI
L L
A B
P
ARL
ARL3
PL/8
P/2
A B
2P
PP
PL/4
A B
P/2P/2
PL/8 PPL/8PL/4
A#A P=>. S=#=>R*>AS? 22%A
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ari /a$"ar #i ata% itun/ Reak%i erletakan #ari $a%in/-$a%in/ 4ree"#,
ari eritun/an reak%i erletakan $aka #aat #iitun/ $atrik% S #an Ar# %""5
ari /a$"ar #i ata% itun/ Reak%i erletakan #ari $a%in/-$a%in/ 4ree"#,
ari eritun/an reak%i erletakan $aka #aat #iitun/ $atrik% S #an Ar# %""5
SLAN8*TN9A 'S*S*N MATR'KS KKAK*AN SBB5
MATR'KS KKAK*AN (S)
EI
5=1
1
Ard11
Ard31
L
2
Ard31
L
S21
+R99$&*#
LL
EI
S22Ard
32
S12
1
2
Ard42
Ard12
=0
Ard22
=0
A#A -9#=>A>RAS-AS? 22%A "9A@A GAA
9R>?
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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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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
]
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)
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Catatan kecil Sondra Raharja Powered by : Android Page !3 of 38
Matriks ARrx1
= ARLrx1
+ ARDrxd
.Ddx1
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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 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&
Displacement/perpindahan :
Translasi (perpindahan dalam arahgaris lurus)
Rotasi (perpindahan dalam putaransudut)
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
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L L
2L/3
D
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Catatan kecil Sondra Raharja Powered by : Android Page !6 of 38
&ree"#, untuk itun/ AL
&ree"#, untuk itun/ ARL
AL = > AL1? = > -L ?
ARL2
ARL3
Gambar AR- tidak dibat
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ARL5
ARL4=0
MD=0ARL
6=0
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'SLACMNT SS*A' O& (ROTAS' ATA* TRANSLAS' AA T'T'K TRTNT*)
&RBO9 SS*A' O& (ROTAS' ATA* TRANSLAS' AA T'T'K TRTNT*)
ari eritun/an reak%i erletakan $aka #aat #iitun/ $atrik% S #an Ar# %""5
EI
5=
1
1
Ard21
Ard31
6EI/L2
S11
L
+R99$&*#
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)
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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
]
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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
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Selan+utn,a Ga,a ala$ (M$en Lintan/ #an Nr$al) #aat #iitun/
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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>?
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"_=+
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C9< S>A>?
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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
]
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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
]