STATIS TAK TENTU, METODE DALIL 3 MOMEN DAN METODE CROSS

1. TUGAS BESAR
MEKANIKA REKAYASA III
FERIYAL SUMARNO / 311 11 001
2 A GEDUNG
1
A B C D E
4 m 1,5 m3,5 m1 m2 m2 m
EI EI EI
3 T 2 T q = 1,5 t/m'
Ao
t1 t2 t3
A1 A2
A3
 METODE DALIL 3 MOMEN
Penyelesaian:
 t1 = = = 3 → A1 = ½ t.l = = 6
 t2 = = = 1,556 → A2 = ½ t.l = = 3,501
 t3 = = = 3 → A3 = 2/3 t.l = = 8
 MD = = = - 1,687
 Tinjau Bentang A0 – A – B :
a1 = = = 2
MA0
MA0
8 MA + 4 MB = - 18 … (1)
 Tinjau Bentang A – B – C :
a1 = = = 2
a2 = = = 2,667

2. TUGAS BESAR
MEKANIKA REKAYASA III
FERIYAL SUMARNO / 311 11 001
2 A GEDUNG
2
4 MA + 17 MB + 4,5 MC = - 30,449 … (2)
 Tinjau Bentang B – C – D :
a2 = = = 1,833
a3 = = = 2
4,5 MB + 17 MC + 4 MD = - 32,556
4,5 MB + 17 MC + 4 X (-1,687) = - 32,556
4,5 MB + 17 MC = - 32,556 + 6,748
4,5 MB + 17 MC = - 25,808 …(3)
Eliminasi persamaan 1 dan 2 :
8 MA + 4 MB = - 18 → x 1
4 MA + 17 MB + 4,5 MC = - 30,449 → x 2
⇨ 8 MA + 4 MB = - 18
⇨ 8 MA + 34 MB + 9 MC = - 60,898 -
- 30 MB – 9 MC = 42,898 …(4)
Eliminasi persamaan 3 dan 4 :
4,5 MB + 17 MC = - 25,808 → x 6,667
- 30 MB – 9 MC = 42,898 → x 1

3. TUGAS BESAR
MEKANIKA REKAYASA III
FERIYAL SUMARNO / 311 11 001
2 A GEDUNG
3
A B C D E
4 m 1,5 m3,5 m1 m2 m2 m
EI EI EI
3 T 2 T q = 1,5 t/m'
⇨ 30 MB + 113,33 MC = - 172,062
⇨ - 30 MB – 9 MC = 42,898 +
104,333 MC = - 129,164
MC = - 1,2379
Subtitusi ke persamaan 3 :
4,5 MB + 17 MC = - 25,808
4,5 MB + 17 (- 1,2379) = - 25,808
4,5 MB = - 25,808 + 21,044
MB = - 1,058
Subtitusi ke persamaan 1 :
8 MA + 4 MB = - 18
8 MA + 4 (- 1,058) = - 18
8 MA = - 18 + 4,232
MA = - 1,721
 METODE CROSS
 Menghitung Kekakuan :
KAB = KBA = = = 1
KBC = KCB = = = 0,888
KCD = KDC = = = 1

4. TUGAS BESAR
MEKANIKA REKAYASA III
FERIYAL SUMARNO / 311 11 001
2 A GEDUNG
4
A B
F
3 T
2 m2 m
MAB MBA MBC
B
C
3,5 m1 m
2 T
MCB MCD MDC
C D
4 m
q = 1,5 t/m'
MDE
D
E
q = 1,5 t/m'
1,5 m
G
 Menghitung Koefisien Distribusi :
µAB = 0
µBA = = = 0,53
µBC = = = 0,47
µCB = = = 0,47
µBA = = = 0,53
 Menghitung Momen Primer :
 M0AB =- = - = - 1,5 t.m
 M0BA = = = 1,5 t.m
 M0BC =- = - = - 1,21 t.m
 M0CB = = = 0,346 t.m
 M0CD = - = = - 2 t.m
 M0DC = = = 2 t.m
 M0DE = - = = - 1,687 t.m

5. TUGAS BESAR
MEKANIKA REKAYASA III
FERIYAL SUMARNO / 311 11 001
2 A GEDUNG
5
 Tabel Cross
TITIK A B C D
BATANG AB BA BC CB CD DC DE
SIKLUS μ 0 -0.53 -0.47 -0.47 -0.53 -1 0
1
M -1.5 1.5 -1.21 0.346 -2 2 -1.687
BAL 0 -0.154 -0.136 0.7774 0.8766 -0.313 0
2
CO -0.077 0.000 0.389 -0.068 -0.157 0.438 0
BAL 0.000 -0.206 -0.183 0.106 0.119 -0.438 0.000
3
CO -0.103 0.000 0.053 -0.091 -0.219 0.060 0
BAL 0.000 -0.028 -0.025 0.146 0.165 -0.060 0.000
4
CO -0.014 0.000 0.073 -0.012 -0.030 0.082 0
BAL 0.000 -0.039 -0.034 0.020 0.022 -0.082 0.000
5
CO -0.019 0.000 0.010 -0.017 -0.041 0.011 0
BAL 0.000 -0.005 -0.005 0.027 0.031 -0.011 0.000
6
CO -0.003 0.000 0.014 -0.002 -0.006 0.015 0
BAL 0.000 -0.007 -0.006 0.004 0.004 -0.015 0.000
7
CO -0.004 0.000 0.002 -0.003 -0.008 0.002 0
BAL 0.000 -0.001 -0.001 0.005 0.006 -0.002 0.000
8
CO 0.000 0.000 0.003 0.000 -0.001 0.003 0
BAL 0.000 -0.001 -0.001 0.001 0.001 -0.003 0.000
9
CO -0.001 0.000 0.000 -0.001 -0.001 0.000 0
BAL 0.000 0.000 0.000 0.001 0.001 0.000 0.000
10
CO 0.000 0.000 0.000 0.000 0.000 0.001 0
BAL 0.000 0.000 0.000 0.000 0.000 -0.001 0.000
11
CO 0.000 0.000 0.000 0.000 0.000 0.000 0
BAL 0.000 0.000 0.000 0.000 0.000 0.000 0.000
12
CO 0.000 0.000 0.000 0.000 0.000 0.000 0
BAL 0.000 0.000 0.000 0.000 0.000 0.000 0.000
MOMEN AKHIR -1.721 1.058 -1.058 1.237 -1.237 1.687 -1.687

6. TUGAS BESAR
MEKANIKA REKAYASA III
FERIYAL SUMARNO / 311 11 001
2 A GEDUNG
6
A B
F
3 T
2 m2 m
MAB MBA
VA VB
MBC
B
C
3,5 m1 m
2 T
MCB
G
VB VC
MCD MDC
C D
4 m
q = 1,5 t/m'
VC VD
 REAKSI PERLETAKAN
 Free Body AB :
∑MB ⇨ VA . 4 – MAB – P . 2 + MBA = 0
VA . 4 – 1,721 – 3 . 2 + 1,058 = 0
VA =
VA = 1,666 Ton (↑)
∑MA ⇨ -VB . 4 – MAB + P . 2 + MBA = 0
-VB . 4 – 1,721 + 3 . 2 + 1,058 = 0
VB =
VB = 1,334 Ton (↑)
 Free Body BC :
∑MC ⇨ VB . 4,5 – MBC – P . 3,5 + MCB = 0
VB . 4,5 – 1,058 – 2 . 3,5 + 1,237= 0
VB =
VB = 1,516 Ton (↑)
∑MB ⇨ -VC . 4,5 – MBC + P . 1 + MCB = 0
-VC . 4,5 – 1,058 + 2 . 1 + 1,237= 0
VC =
VC = 0,484 Ton (↑)
 Free Body CD :
∑MD ⇨ VC . 4 – MCD –1/2qL2 + MDC = 0
VC . 4 – 1,237 –1/2.1,5.42 + 1,687= 0
VC =
VC = 2,887 Ton (↑)

7. TUGAS BESAR
MEKANIKA REKAYASA III
FERIYAL SUMARNO / 311 11 001
2 A GEDUNG
7
MDE
D
E
q = 1,5 t/m'
1,5 m
VD
∑MC ⇨ -VD . 4 – MCD +1/2qL2 + MDC = 0
-VD . 4 – 1,237 +1/2.1,5.42 + 1,687= 0
-VD =
-VD = 3,113 Ton (↑)
 Free Body DE :
∑V = 0 ⇨ VD – q . L = 0
VD = 1,5 X 1,5
VD = 2,25 Ton (↑)
Jadi VTotal :
VA = 1,666 Ton
VB = 1,334 + 1,516 = 2,85 Ton
VC = 0,484 + 2,887 = 3,371 Ton
VD = 3,113 + 2,25 = 5,363 Ton
KONTROL :
∑V = 0 ⇨ ( VA + VB + VC + VD ) – ( P1 + P2 + q . L ) = 0
(1,666 + 2,85 + 3,371 + 5,363) – (3 + 2 + 8,25) = 0
13,25 – 13,25 = 0 … OK!!!

8. TUGAS BESAR
MEKANIKA REKAYASA III
FERIYAL SUMARNO / 311 11 001
2 A GEDUNG
8
A B
F
3 T
2 m2 m
MAB MBA
VA VB
MBC
B
C
3,5 m1 m
2 T
MCB
G
VB VC
4. GAYA DALAM
 Free Body AB :
Momen :
MA = MAB = - 1,721 t.m
MF = MAB + VA . 2
= - 1,721 + 1,666 X 2 = 1,611 t.m
MB = MAB + VA . 4 – P1 . 2
= - 1,721 + 1,666 X 4 – 3 X 2
= - 1,057 t.m
Lintang :
DA – F = VA = 1,666 Ton
DF – B = VA – P1
= 1,666 – 3 = - 1,334 Ton
DB = VA – P1 + VC
= 1,666 – 3 + 1,334 = 0 Ton
 Free Body BC :
Momen :
MB = MBC = - 1,058 t.m
MG = MBC + VB . 1
= - 1,058 + 1,516 X 1 = 0,458 t.m
MC = MBC + VB . 4,5 – P2 . 3,5
= - 1,058 + 1,516 X 4,5 – 2 X 3,5
= - 1,236 t.m
Lintang :
DB – G = VB = 1,516 Ton
DG – C = VB – P2
= 1,516 – 2 = - 0,484 Ton
DC = VB – P2 + VC
= 1,516 – 2 + 0,484 = 0 Ton

9. TUGAS BESAR
MEKANIKA REKAYASA III
FERIYAL SUMARNO / 311 11 001
2 A GEDUNG
9
MCD MDC
C D
4 m
q = 1,5 t/m'
VC VD
MDE
D
E
q = 1,5 t/m'
1,5 m
VD
 Free Body CD :
Momen :
( ≤ X ≤ )
MX = VC . X – ½ q . X2 - MCD
= 2,887 . X – ½ . 1,5 . X2 – 1,237
Lintang :
DX = VC - q . X
= 2,887 – 1,5 . X
X = = 1,925 m
X 0 1,925 4
MX - 1,237 1,541 -1,689
DX 2,887 0 -3,113
 Free Body DE :
Momen :
( ≤ X ≤ 1,5 )
MX = VD . X – ½ q . X2 - MDE
= 2,25 . X – ½ 1,5 . X2 – 1,687
Lintang :
DX = VD - q . X
= 2,25 – 1,5 . X
X = = 1,5 m
X 0 1,5
MX - 1,687 0
DX 2,25 0

10. TUGAS BESAR
MEKANIKA REKAYASA III
FERIYAL SUMARNO / 311 11 001
2 A GEDUNG
10