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Tugas Besar Analisa struktur statis tertentu II Metode Ritter
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Tugas Besar Analisa struktur statis tertentu II Metode Ritter
penyelesaian mencari gaya gaya batang menggunakan metode ritter...
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khotibsafaat
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Tugas Besar Anstruk Metode Ritter No. 3)
P1=4t
F
5
A
G
12
7
6
1
P2=7t
DC
P3=8t
8
9
D
2
H
13
10
E
3
3m
11
4
RA
B
RB 4x3m
Mencari Reaksi ∑MB=0
RA.12m - 4t.9m - 7t.6m 7t .6m - 8t.3m=0 RA =
=
= 8.5 t Tabel Reaksi
∑MA=0
RB.12m - 8t.9m – 7t.6m - 8t.3m=0 RB =
=
= 10,5 t
Reaksi di titik A B
Reaksi (ton) 8,5 t 10,5 t
Mencari Gaya Batang Potongan I (pandang Kiri Potongan)
F
Jarak S5 - C 0
Sin 45 =
x = Sin 45 . 3m C
RA
0
x A
= 2.121303 2.1213034m 4m
∑MF=0
S1 = 8.5 t (Tarik)
∑MC = 0
S5= -12.020815 (tekan)
Potongan III (Pandang Kiri Potongan) P1=4t
∑MC=0 12
G
F
S12.3m + RA.3m = 0 S12 = -RA
7
S12 = -8.5 t (Tekan) 2 A
D
C
∑MF = 0
-S7.2,1213034m - S2.3m + RA.3m = 0 S7 =
= - 6.364011862 (tekan)
∑MG = 0
-P1.3m + RA.6m – S2.3m = 0 S2 =
= 13t (tarik)
Potongan II (Pandang Kiri Potongan) ∑MA = 0
F G
S6.3m - S7. 2,1213034m = 0 5
7
6
S6 =
S6 = 4,5t
2 A C
RA
D
Potongan IV (Pandang Kiri Potongan) P1=4t
∑MG = 0
- S3.3m – P1.3m + RA.6m = 0 F
12
G
S3 =
8
= 13t (Tarik)
-S8.3m+RA.3m-S12.3m=0
3 D
C
∑MC = 0
7
A
E
S8=
= 0 t
RA
Potongan V (Pandang Kiri Potongan) P2=7t
P1=4t
F
13
G
∑ME = 0
S13.3m + RA.9m – P1.6m – P2.3m=0 H
S13 =
S13 = -10.5t (Tekan)
9
∑MH = 0 A
C
D
3
E
-S9. 2,12m-S3.3m-P2.3m-P1.6m+RA.9m=0
S9= RA
S9 = -3,535562145t (tekan)
Potongan VI (Pandang Kiri Potongan) P1=4t
P2=7t
F
G
H
13
10 4
A
B C
D
RA
∑MH = 0
-S4.3m – P2.3m – P1.6m + RA.9m = 0 S4 =
– –
= 10,5t (tarik)
∑MG = 0
-S10.3m + RA.6m – P1.3m – S4.3m = 0 S10 =
– –
= 2.5t (Tarik)
E
Potongan VII (Pandang Kiri Potongan) P1=4t
F
P2=7t
G
P3=8t
13
H 11
10 4
A C
D
E
RA
∑ME = 0
S11. + RA.9m – P1.6m – P2.3m = 0 S11 =
= -14,8496101t (Tekan)
Tabel Gaya Batang No Batang 1 2 3 4 5 6 7 8 9 10 11 12 13
Gaya Batang (TON) 8.5 t 13 t 13t 10,5 t -12.020815 t 4,5 t - 6.364011862 t 0t -3,535562145 t 2.5 t -14,8496101 t -8.5 t -10.5 t
B
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