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Transcript of Mekanika Rekayasa Metode Clyperon
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Mekanika Rekayasa 3
TUGAS MEKANIKA REKAYASA 3
Nama : Indra Sidik Permadi
Nim : 05111011
Kelas : KS 2A
1. Tentukan Momen Akhir dan Gambarkan diagram gaya gaya dalamnya!
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Mekanika Rekayasa 3
Asumsi tanda momen negatif
Akibat Momen di A
EI
qL
EI
LMBA
EI
LMABA
24
3
63+
=
EI
qL
EI
LMBA
EI
LMABBA
2436
3
+
=
Akibat Momen di B
EI
qL
EI
LMBCBC
243
3
+
=
EIEI
MBC
24
)30(5
3
303
+
=
EIEI
MBC
24
135000
3
30+
=
BCBA =
EIqL
EIMBC
EIqL
EILMBA
EILMA
24330
2436
33
+
=
+
EIEI
MBC
EIEI
MBA
EI
MA
24
)30(5
3
30
24
)20(5
3
20
6
2033
+
=
+
024
135000
24
40000
3
30
3
20
6
20=
+
+
EIEIEI
MBC
EI
MBA
EI
MA
EIEIEI
MBA
EI
MA240
24
175000
3
50
6
20=
+
...(1)..........4375......MB10MA2 =+
=+ 017500040080 MBMA
Karena titik A tumpuan jepit, maka AB = 0
02463
3
=+
EI
qL
EI
LMB
EI
LMA
EIEI
MB
EI
MA
24
)20(5
6
20
3
203
=
EIEIEI
MB
EI
MA24
24
40000
6
20
3
20=
...(2)..........500.......MB2MA =
= 4000080160 MBMA
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Mekanika Rekayasa 3
Substitusi dan Eliminasi Pers (1) dan (2)
+=
=+
5002
4375102
MBMA
MBMA
kFtMB
MB
555.430
38759
=
=
Maka MA
kft34.72MA =
=
+=
=
445.692
555.4305002
500555.4302
MA
MA
MA
Freebody Diagram
Diagram Momen
Mmax Lapangan A B
3
5021 =qL 50
21 =qL 75
22 =qL 75
22 =
qL
( )792.19
1
=
L
MAMB
792.19
352.1430
555.430= 352.14
RA
208.30
RB
352.89792.69
RC
648.60
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Mekanika Rekayasa 3
( )
( )
05208.30
5208.3072.34
208.3072.34
2
21
2
21
==
+=
+=
xxMx
xx
qxxMx
ft6.0416x =
( ) ( ) 22
1 0 4 2.650 4 2.62 0 8.3 07 2.3 4m a x +=M
kft56.533=
+= 264.91517.18272.34
Mmax Lapangan B C
( )
05352.89
352.89555.4302
21
==
+=
xx
Mx
qxxMx
ft17.870x =
( ) ( ) 22
1 8 7 0.1 758 7 0.1 73 5 2.8 95 5 5.4 3 0m a x +=M
kft367.823=
+= 342.798720.1596555.430
2. Tentukan Momen Akhir dan gambarkan diagram gaya gaya dalamnya!
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Mekanika Rekayasa 3
Asumsi Momen Negatif
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Mekanika Rekayasa 3
Akibat Momen di B
EIEI
MBBA
24
2000
3
10=
EI
MC
EI
MB
EIBC
6
12
3
12
24
3456=
Akibat Momen di C
EIEI
MC
EI
MBCB
24
3456
3
12
6
12+=
EI
MC
EICD
3
20
16
8000=
BCBA =
EIEIEIEI
MC
EI
MB
EI
MB
EI
MC
EI
MB
EIEIEI
MB
2424
2000
24
3456
6
12
3
12
3
10
6
12
3
12
24
3456
24
2000
3
10
+=++
=
....(1)5456......48MC176MB =+
CDCB =
EIEIEI
MC
EI
MB
EIEIEI
MC
EI
MC
EI
MB
EI
MC
EIEIEI
MC
EI
MB
4848
30912
3
32
6
12
24
3456
16
8000
3
20
3
12
6
12
3
20
16
8000
24
3456
3
12
6
12
=+
+=++
=+
)........(230912.....512MC96MB =+
Eliminasi Pers (1) dan (2)
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Mekanika Rekayasa 3
483091251296
512545648176
=+
=+
MCMB
MCMB
=+
=+
1449216245764608
27934722457690112
MCMB
MCMB
kft15.722MB =
=134425685504MB
# Substitusikan MB Ke Pers. (1)
( )5456072.2767
545648722.15176
545648176
=+
=+
=+
MCMC
MCMB
kft56.019MC =
= 072.2767545648MC
FREEBODY DIAGRAM
M max Lapangan AB
7
2
102
2
102
2
122
2
122
2
20
2
20
10
722.15
10
722.15 ( )12
722.5019.56
20
019.56
20
019.56
429.8 572.11 642.8 358.15 801.12 199.7
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Mekanika Rekayasa 3
( )
0429.8
429.82
21
==
=
qxx
Mx
qxxMx
ftx 214.42
429.8==
( ) ( )
kft
M
762.17
214.42214.4429.8max2
21
=
=
M max Lapangan B C
( )
0429.8
642.8722.152
21
==
+=
qxxMx
qxxMx
ftx 321.42
429.8==
( ) ( )
kft
M
393.34
671.18342.37722.18
321.42321.4642.8722.15max2
21
=
+=
+=
M max Lapangan C D
( )
kft
M
991.71
10801.12019.56max
=
+=
8
