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Tugas Besar Anstruk Metode Ritter No. 3) P1=4t P2=7t P3=8t 3m RA RB 4x3m Mencari Reaksi MB=0 RA.12m - 4t.9m - 7t.6m - 8t.3m=0 RA = 36 tm +42 tm +24 tm 12 m = 102 tm 12 m = 8.5 t MA=0 Tabel Reaksi RB.12m - 8t.9m – 7t.6m - 8t.3m=0 RB = 72 tm +42 tm +12 tm 12 m = 126 tm 12 m = 10,5 t Mencari Gaya Batang Potongan I (pandang Kiri Potongan) B A D E 2 3 4 1 5 6 7 8 9 10 11 H 13 G 12 F D Reaksi di titik Reaksi (ton) A 8,5 t B 10,5 t

Tugas Besar Analisa struktur statis tertentu II Metode Ritter

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Page 1: Tugas Besar Analisa struktur statis tertentu II Metode Ritter

Tugas Besar Anstruk Metode Ritter

No. 3) P1=4t P2=7t P3=8t

3m

RA RB

4x3m

Mencari Reaksi

∑MB=0

RA.12m - 4t.9m - 7t.6m - 8t.3m=0

RA = 36 tm+42 tm+24 tm

12m = 102tm12m = 8.5 t

∑MA=0 Tabel Reaksi

RB.12m - 8t.9m – 7t.6m - 8t.3m=0

RB = 72tm+42tm+12tm

12m = 126 tm12m = 10,5 t

Mencari Gaya Batang

Potongan I (pandang Kiri Potongan)

F Jarak S5 - C ∑MF=0

Sin 450 = x3 S1 = 8.5 t (Tarik)

x x = Sin 450 . 3m

BA D E2 3 4C1

5 6 7 8 9 10 11

H13G12F

D

Reaksi di titik Reaksi (ton)A 8,5 tB 10,5 t

Page 2: Tugas Besar Analisa struktur statis tertentu II Metode Ritter

A C = 2.1213034m

∑MC = 0

S5= -12.020815 (tekan)

Potongan III (Pandang Kiri Potongan)

P1=4t ∑MC=0

S12.3m + RA.3m = 0

S12 = -RA

S12 = -8.5 t (Tekan)

∑MF = 0

-S7.2,1213034m - S2.3m + RA.3m = 0

S7 = −13 t .3m+8,5t .3m2.1213034m = - 6.364011862 (tekan)

∑MG = 0

-P1.3m + RA.6m – S2.3m = 0

S2 = −4.3m+8.5 .6m

3m = 13t (tarik)

Potongan II (Pandang Kiri Potongan)

∑MA = 0

S6.3m - S7. 2,1213034m = 0

S6 = 6.364011862 t .2,1213034m

3m

S6 = 4,5t

GF

CA D

7

2

12

RA

2

CA

F

76

D

G5

RA

5

Page 3: Tugas Besar Analisa struktur statis tertentu II Metode Ritter

Potongan IV (Pandang Kiri Potongan)

∑MG = 0

- S3.3m – P1.3m + RA.6m = 0

S3 = −4 t .3m+8,5 t .6m

3m = 13t

(Tarik)

∑MC = 0

-S8.3m+RA.3m-S12.3m=0

S8=8,5 t .3m−8,5 t .3m

3m = 0 t

Potongan V (Pandang Kiri Potongan)

∑ME = 0

S13.3m + RA.9m – P1.6m – P2.3m=0

S13 = −8.5 t .9m+4 t .6m+7 t .3m

3m

S13 = -10.5t (Tekan)

∑MH = 0

-S9. 2,12m-S3.3m-P2.3m-P1.6m+RA.9m=0

S9=−13 t .3m−7 t .3m−4 t .6m+8,5 t .9m

2,1213034m

S9 = -3,535562145t (tekan)

RA

A

F G

D

7

83

12

P1=4t

EC

C

P1=4t

D

RA

A 3

GF H

E

P2=7t

9

13

Page 4: Tugas Besar Analisa struktur statis tertentu II Metode Ritter

13

10

4

P2=7tP1=4t

RA

A

C D E

Potongan VI (Pandang Kiri Potongan)

∑MH = 0

-S4.3m – P2.3m – P1.6m + RA.9m = 0

S4 = – 7 t .3m– 4 t .6m+8.5 t .9m

3m = 10,5t (tarik)

∑MG = 0

-S10.3m + RA.6m – P1.3m – S4.3m = 0

S10 = 8.5 t .6m– 4 t .3m –10,5.3m

3m = 2.5t (Tarik)

GF H

B

Page 5: Tugas Besar Analisa struktur statis tertentu II Metode Ritter

G H13

P2=7tP1=4t

RA

A

C D E

B

P3=8t

Potongan VII (Pandang Kiri Potongan)

∑ME = 0

S11.2,1213034m + RA.9m – P1.6m – P2.3m = 0

S11 = −8,5 t .9m+4 t .6m+7 t .3m

2,1213034m = -14,8496101t (Tekan)

Tabel Gaya Batang

No Batang Gaya Batang (TON)1 8.5 t2 13 t3 13t4 10,5 t5 -12.020815 t6 4,5 t7 - 6.364011862 t8 0 t9 -3,535562145 t

10 2.5 t11 -14,8496101 t12 -8.5 t

11

F

10

4

Page 6: Tugas Besar Analisa struktur statis tertentu II Metode Ritter

13 -10.5 t